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18.06 course at MIT

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Folders and files, repository files navigation, mit 18.06, spring 2023 linear algebra.

Welcome to MIT 18.06: Linear Algebra! The Spring 2023 course information, materials, and links are recorded below. Course materials for previous semesters are archived in the other branches of this repository. You can dive in right away by reading this introduction to the course by Professor Strang.

NOTICE OF VIDEO RECORDING: The Spring 2023 lectures for 18.06 will be recorded for OCW . See the notice of recording for more information.

Catalog Description: Basic subject on matrix theory and linear algebra, emphasizing topics useful in other disciplines, including systems of equations, vector spaces, determinants, eigenvalues, singular value decomposition, and positive definite matrices. Applications to least-squares approximations, stability of differential equations, networks, Fourier transforms, and Markov processes. Uses linear algebra software. Compared with 18.700, more emphasis on matrix algorithms and many applications.

Lectures : Monday, Wednesday, and Friday at 11am in 26-100.

Instructors : Prof. Gilbert Strang and Dr. Andrew Horning .

Textbook : Introduction to Linear Algebra: 6th Edition . Professor Strang will explain more about this new 6th edition in class (it is not yet on Amazon). It now ends with two chapters on deep learning. Professor Strang plans to make the textbook available for students to purchase at a discount. Here again is a link to the preface and contents .

Recitations : Tuesday at the following times and locations.

• R01: 9am with M. Chen (66-144)
• R02: 10am with M. Chen (66-144)
• R03: 10am with V. Krylov (2-136)
• R04: 11am with D. Kliuev (4-159)
• R05: 11am with V. Krylov (2-136)
• R06: 12pm with M. Harris (4-159)
• R07: 12pm with D. Kliuev (2-105)
• R08: 12pm with I. Ganguly (2-136)
• R09: 1pm with M. Harris (4-149)
• R10: 1pm with I. Ganguly (2-136)
• R11: 1pm with Z. Chen (2-135)
• R12: 2pm with Z. Chen (2-132)
• R13: 2pm with K. Vashaw (2-136)
• R14: 3pm with K. Vashaw (2-136)

Office Hours: Please make arrangements to meet Professor Strang before or after class if you wish. The rest of the instructional team will hold weekly office hours. You may attend any office hours that fit your schedule.

• A. Horning: Friday 12:30 - 1:30 in 66-144
• M. Chen: Friday 4-6 in 66-154
• V. Krylov: Thursday 5 - 7 in 2-361
• D. Kliuev: Thursday 11 - 1 in 2-341
• M. Harris: Wednesday 4 - 6 in 2-333
• I. Ganguly: Monday 12:30-1:30 in 2-146 and Thursday 1-2 in 2-139
• Z. Chen: Wednesday 12 - 2 in 2-340
• K. Vashaw: Monday 1:30 - 2:30 in 2-131 and Friday 1:30 - 2:30 in 66-144
• S. Bentley: Friday 3 - 4 in 2-147
• A. Lu: Friday 7 - 9 in 2-139

Exams : We will have 3 exams during the semester: February 22, March 20, April 19. Final Exam on May 22 (official schedule here ). For conflicts and/or accomodations, please contact Sapphire Tang in Academic Services as soon as possible (1-2 weeks in advance).

Homework: Due weekly on Sunday at midnight (except exam weeks and spring break). Upload a .pdf of your clearly written or typed solutions on Gradescope . Late homework will not be accepted and extensions will not be granted within 48 hours of the due date except in cases of genuine emergency (a letter from S^3 is required).

Collaboration: Collaboration on homework is highly encouraged! However, please maintain academic integrity by writing up your solutions individually and by naming all collaborators and information sources consulted for the assignment (you don't need to cite the textbook).

Grading: 20% Homeworks (equally weighted, lowest dropped) + 45% Quizzes (3 midterms, each worth 15%) + 35% Final Exam

Resources : In addition to this repository, we will use the following online resources.

• Canvas - course announcements will be posted on Canvas.
• Gradescope - submit Psets and check grades through Gradescope.
• Piazza - ask questions in the course discussion forum.

MIT also has excellent study resources: math learning center , TSR^2 study/resource room , pset partners .

Problem sets

Pset 1 is due on Sunday February 12 at 11:59pm.

Pset 2 is due on Sunday February 19 at 11:59pm.

Pset 3 is due on Sunday March 5 at 11:59pm.

Pset 4 is due on Sunday March 12 at 11:59pm. Extended to Wednesday March 15 at 11:59pm.

Pset 5 is due on Sunday April 9 at 11:59pm. Problem 9 is OPTIONAL for EXTRA CREDIT.

Pset 6 is due on Sunday April 16 at 11:59pm.

Pset 7 is due on Sunday April 30 at 11:59pm.

Pset 8 is due on Sunday May 7 at 11:59pm.

Exam 1 will be held on Wednesday, February 22 between 11am-12pm. Last names beginning with A-L will be in 50-340, while last names beginning with M-Z will be in 26-100.

Exam 2 will be held on Monday, March 20 between 11am-12pm. Last names beginning with A-L will be in 50-340, while last names beginning with M-Z will be in 26-100.

Exam 3 will be held on Wednesday, April 19 between 11am-12pm. Last names beginning with A-L will be in 50-340, while last names beginning with M-Z will be in 26-100.

Lecture Material and Summaries

Lecture 1 (february 6).

Two ways to do matrix-vector multiplication are the "row way" and the "column way."

• "Row way" = dot products between rows of matrix and the vector.
• "Column way" = linear combination of columns in the matrix. The coefficients in the linear combination are the entries of the vector.

The column way leads to the column space of a matrix, the space of all possible linear combinations of the columns. Does the column space of a 3 x 3 matrix form a line, a plane, or fill the whole space? It depends on how many linearly independent columns there are. The geometric picture of a matrix's column space is the first key idea of linear algebra.

The dot product tells us how to do matrix-vector multiplication the "row way." The remarkable cosine formula for the dot product also shows how to use it to measure the length of vectors and the angles between them. In particular, perpendicular vectors have dot product = 0. Two useful inequalities follow quickly from the cosine formula: the Cauchy-Schwarz and triangle inequalities. Beyond matrix multiplication, the dot product reveals the geometry of vectors and their linear combinations.

Further Reading: Textbook, chapter 1.1 and 1.2.

Lecture 2 (February 8)

The number of linearly indepdent columns in a matrix is the column rank . The number of linearly independent rows is the row rank . The remarkable fact is that these two numbers are the same! In the rank one case, this means that all the columns are multiples of a single vector. The rows are also multiples of a single vector! We can write the rank one matrix A as the product of a column vector and row vector: the row vector tells us what multiples we use to get the columns of A.

Beyond rank one, we can select linearly independent columns of A by moving from left to right. We ask, is the next column a linear combination of the previous columns? The linearly independent columns of A become the columns of a matrix C. The columns of a matrix R tell us how to combine the columns of C to get ALL columns of A: this is the factorization A = C R. R always contains the identity matrix (unless A is zero) and its rows are linearly independent. With A = C R, we are very close to understanding why the column and row rank of A have to be the same!

Further Reading: Textbook, chapter 1.3 and 1.4.

Lecture 3 (February 10)

There are four useful ways to organize matrix-matrix multiplication:

• dot product between row i of A and column j of B gives entry (i,j) of AB,
• matrix A times column j of B gives column j of AB,
• row i of A times matrix B gives row i of AB,
• AB is a sum of rank 1 matrices: columns of A times rows of B.

The middle two are particularly useful for understanding the column or row space of the product AB.

With these conceptual ways to organize matrix-matrix multiplication, we see that A = CR means that the columns of A are combinations of columns of C and the rows of A are combinations of the rows of R. The factors C and R reveal the column space and the row space of the matrix A.

Further Reading: Textbook, review chapter 1. You may find this review sheet helpful.

Lecture 4 (February 13)

At the heart of linear algebra is the equation Ax = b. We have n equations in n unknowns and A is an n by n matrix.

• If A has rank n, there is a unique solution.
• If A has rank less than n, then there is either (a) no solution or (b) infinitely many solutions!

In a simple 2 by 2 example, the second scenario happens when the equations define parallel lines that (a) never intersect or (b) are the same line.

To solve Ax = b, we combine equations to eliminate unknowns. This is elimination: elementary row operations on A and b convert A to upper triangular form. Once A is in triangular form, we can solve one variable at a time with backsubstitution .

Further Reading: Textbook, chapter 2.1.

Lecture 5 (February 15)

Elimination converts A to an upper triangular matrix U. We can write the elimination steps as multiplication on the left of A by elimination matrices . Gathering the product of the elimination matrices on the left of A gives us EA = U. Inverting E, we arive at the famous factorization A = LU: A is the product of a lower (L) and an upper triangular matrix (U).

Elimination and backsubstitution implicitly invert A to solve Ax = b. In practice, we rarely compute the inverse of A directly (e.g., calculate its entries). However, it's a useful theoretical tool: studying the properties of inverses will allow us to connect the ideas of elimination to the ideas of column and row space from week one.

Further Reading: Textbook, chapter 2.2 and 2.3.

Lecture 6 (February 17)

The inverse of a matrix gives us a direct way to think about Ax = b. An invertible matrix has a unique inverse, a unique solution to Ax = b for any b, and its columns are linearly independent. A triangular matrix is invertible if and only if its diagonal entries are nonzero. The inverse of a product of matrices is the product of the inverses in reverse order!

We rarely compute inverses explicitly on the computer. Instead, we solve Ax = b with elimination and backsubstitution. The inverse of the product of elimination matrices has a special structure. The diagonal entries are all one and the subdiagonal entries are the multipliers from elimination! This is the lower triangular matrix L in A = LU. If we encounter unwanted zeros on the diagonal during elimination, we can (when A is invertible) remedy the situation by interchanging rows to move the offending zero and replace it with a nonzero pivot. This leads us to elimination with row-interchanges: PA = LU.

Further Reading: Textbook, chapter 2.2, 2.3, and 2.4.

Lecture 7 (February 21: Tuesday with Monday Schedule)

The algebra of row interchanges is captured in a very special family, or group , of matrices called permutatation matrices. The product, inverse, and transpose of a permutation matrix are all also permutation matrices! When permuation matrices multiply a matrix (or vector) from the left, they exchange rows. When they multiply from the right, the exchange columns.

The tranpose of a matrix switches the rows and the columns. The tranpose of a column vector is a row vector and vice versa. The transpose of a product is the product of the tranposes, in reverse order. The dot product is a special case of matrix multiplication: the tranpose of x multiplies y. Going back to the law of associativity, the dot product of A times x with y is the dot product of x with A transpose times y.

A very special class of matrices has A transpose equal to A. The rows and columns of A are the same! Professor Strang's favorite matrix is the second central difference matrix: symmetric, tridiagonal, and invertible. It is often used to approximate the second derivative of a function, sampled at equispaced points, with the method of finite-differences.

Further Reading: Textbook, chapter 2.4 and 2.5.

Lecture 8 (February 22)

EXAM 1: NO LECTURE.

Lecture 9 (February 24)

Finite difference matrices approximate derivatives of a function from its samples on a finite grid: these are the difference quotients of calculus in the language of linear algebra. The second central different matrix is special: it is symmetric, tridiagonal, and invertible (with the right boundary conditions). It's LU factorization reflects this symmetry: it is a product of forward and backward difference matrices.

We can use difference quotients and finite difference matrices to solve differential equations on the computer! Discretizing the heat equation with finite differences and a backward difference quotient in time (implicit time-stepping), we have to solve a linear system of equations to step forward in time: each new time step gives a new right-hand side that depends on the solution from the previous time step. The coefficient matrix stays the same, so it is best to factor once (A=LU) and only solve triangular systems after that!

Further Reading: Textbook, Chapter 2.5.

Lecture 10 (February 27)

A VECTOR SPACE has 2 operations: sum x + y of "vectors" and multiplication cx by "scalars." Then 8 rules like c(x + y) = cx + cy must hold. A "SUBSPACE" is a vector space INSIDE another vector space, as in these examples of subspaces of R^n = n-dimensional space:

• all combinations of k given vectors
• all solutions to Ax = 0 for a given matrix A

Similarly, we have subspaces of C^n with complex numbers and matrix spaces like all m by n real matrices and function spaces like all continuous functions f(x) for 0 <= x <= 1. All diagonal matrices and all symmetric matrices would be subspaces of the vector space of n by n matrices.

Further Reading: Textbook, Chapter 3.1.

Lecture 11 (March 1)

Finding a complete set of solutions to Ax = zero vector for a given m by n matrix A. This is the "nullspace of A " - a subspace of n-dimensional space. We need elimination to simplify Ax = 0 to a "reduced echelon form Rx = 0." Suppose A is m by n of rank r (r independent rows and columns). Use row operations on A to produce R0 (m by n) and then R (m by r, the same R as in A = CR of Chapter 1).

Delete the zero row and now A = CR.

Elimination is complete and we easily solve Rx = 0 (same nullspace as A). There is a special solution for each column of R apart from the r=2 columns of I. Set x3 = 1 and x4 = 0 to find the special solution x = ( -3, -4, 1 , 0) to Ax = 0. Set x3 = 0 and x4 = 1 to find the special solution x = ( -5, -6, 0, 1) to Ax = 0. The combinations of those special solutions fill the nullspace of A. We will soon show that this nullspace is perpendicular to the row space because Ax = 0.

Further Reading: Textbook, Chapter 3.2.

Lecture 12 (March 3)

We're now ready to describe the complete solution to A * x=b. If A is square and invertible, there is a unique solution. If A is not invertible and b is not in the column space of A, there is no solution. If A is not invertible, but b is in the column space of A, we have infinitely many solutions ! How do we describe them? The nullspace of A plays the key role.

If xp solves A * xp = b and xn solves A * xn = 0 (xn is in the nullspace of A), then A * (xp + xn) = b so x = xp + xn also solves A * x = b! Every solution of A * x = b can be written in this form: a particular solution xp, which solves A * xp = b, plus any vector xn in the nullspace of A. When A is invertible, the nullspace is trivial and the solution is unique. Otherwise, we can write down an infinite set of solutions because the nullspace contains infinitely many vectors!

** Further Reading:** Textbook, Chapter 3.3.

Lecture 13 (March 6)

The big picture of linear algebra is the four fundamental subspaces of an m x n matrix A with rank r: the column spaces of A and its transpose (row space), and the nullspaces of A and its transpose. The row space (dimension r) and the nullspace (dimension n-r) of A are orthogonal complements in R^n. Every vector in R^n can be written as the sum of two orthogonal vectors: one in the row space and one in the null space. There is a beautiful symmetry here because the same picture holds for the column space of A and the nullspace of A^T!

A basis for a vector space is a spanning set of linearly independent vectors: any vector in the space can be written as a unique linear combination of basis vectors. The dimension of a vector space is the number of vectors in (any and every) basis for the space. The factorization A = C*R gives us a basis for the column space of A (columns of C), row space of A (rows of R), and the nullspace of A (constructed by solving Rx = 0: easy since R is already the result of elimination). How can we find a basis for the nullspace of A^T?

Further Reading: Textbook, Chapter 3.4, 3.5, and 4.1.

Lecture 14 (March 8)

An incidence matrix A describes the flow of information on a directed graph, e.g., the flow of electricity in an electical circuit. The four fundamental subspaces of A provide clear insight into the structure and behavior of the circuit. Kirchoff's laws of current, voltage, and Euler's formula relating the number of nodes, edges, small loops in the graph: these can all be seen in the column spaces and nullspaces of A and A^T.

So far, we have learned how to solve and analyze Ax = b when A is square and invertible or when b is in the column space of A. But what should we do when b is not in the column space of A? This situation is typical of regression problems in classical statistics and data-science, when each data-point leads to an equation and one hopes to find model parameters that fit the data well. The first thing to try is to make Ax as close to b as possible: this means choosing x so that Ax-b is orthogonal to Ax. To find x and Ax-b, we need to study orthogonal projections onto the column space of A.

Further Reading: Textbook, Chapter 3.5, 4.1, and 4.2.

Lecture 15 (March 10)

To make Ax-b as small as possible, we choose x so that Ax - b is orthogonal to the column space of A. Then, Pb = Ax is the orthogonal projection of b onto the column space of A. The error in Ax = b is b - Ax = b - Pb = (I-P)b. The projections P and (I-P) are called orthogonal projectors: they are square, symmetric, and their ranks are equal to the subspaces they project onto. The projector P projects onto the column space of A and I-P projects onto the orthogonal complement, the nullspace of A^T.

To find P and I - P, we have to find x! The condition Ax - b orthogonal to columns of A leads to the normal equations : (A^TA)x=A^Tb. This is a square linear system, and it is invertible when the columns of A are linearly independent.

Further Reading: Textbook, Chapter 4.1 and 4.2.

Lecture 16 (March 13)

If the columns of A are indepedent (full column rank), then the normal equations (A^TA)x = A^Tb have a solution x that makes the vector A x -b as short as possible! The vector p = A x is the projection of b onto the column space of A: p = A * (A^T * A)^(-1) A^T * b. The matrix multiplying b from the left is the orthogonal projection matrix P. Applying the projection twice is the same as applying the projection once: P^2 = P.

We can use the normal equations to solve regression problems in statistics, like finding a line that best fits the data. Each data point gives us an equation, a row of Ax = b. The slope and intercept of the line are the unknowns, entries of x. To choose the slope and intercept of the best fitting line, we minimize ||Ax-b|| - we solve the normal equations! The error in our fit is ||A x -b||=||b - A x || = ||(I-P)b||: this is how much of b is orthogonal to the column space of A.

Further Reading: Textbook, Chapter 4.2 and 4.3.

Lecture 17 (March 15)

Orthogonal matrices are a beautiful family of square matrices with orthonormal columns: the columns are orthogonal to each other and have unit length. Examples of orthgonal matrices come from permutations, rotations, reflections, and Hadamard matrices (entires are plus and minus one). Orthogonal matrices preserve the length of a vector when they mutliply it. When Q is an orthogonal matrix, ||Qx||=||x||! Orthogonal matrices and orthogonal bases are the key to make orthogonal projections and least-squares work reliably on the computer.

Further Reading: Textbook, Chapter 4.4. Check out the least-squares notebook for more applications of least-squares.

Lecture 18 (March 17)

To find an orthonormal basis (orthogonal and normalized to unit length) for the column space of A, we can perform the Gram-Schmidt orthogonalization procedure. The first k basis vectors from Gram-Schmidt form an orthonormal basis for the first k linearly independent columns of A. If A has linearly independent columns, we get the factorization A = QR. The columns of Q are the orthonormal basis for the column space of A and the columns of R tell us how to recover the columns of A from Q!

Further Reading: Textbook, Chapter 4.4.

Lecture 18 (March 22)

Gramd-Schmidt orthogonalization provdes the factorization A = QR. The columns of Q form an orthonormal basis for the column space of A. The columns of R link the columns of Q to the columns of A. The entries of R are the dot products we compute during Gram-Schmidt. Once we have computed an orthonormal basis for the column space of A, we can use the factors Q and R to solve the least-squares problem. The new (simpler) equation for the least-squares solution is the upper triangular system Rx = Q^Tb. High-quality numerical linear algebra software usually works with Rx = Q^Tb instead of the normal equations in order to reduce the accumulation of rounding errors made during computer arithmetic.

Lecture 19 (April 3)

It's determinant week! The determinant of a square matrix provides a useful test for invertibility: it is zero exactly when the matrix is not invertible. The determinant of the identity is one, exchanging rows (or columns) changes the sign of the determinant, and the determinant is linear in each separate row and column . This last statement means that scaling a single row (or column) scales the whole determinant. The familiar and remarkable formualas for the determinant follow from these three properties!

Although historically important and theoretically insightful, we rarely use the determinant (or explicit formulas) for computing inverses: it is almost always best to solve linear systems by elimination. When the determinant must be computed, it is usually done via elimination and A=LU. Then, det(A) = det(L) * det(U), and the determinants of the triangular matrices L and U are the products of their diagonal elements.

Further Reading: Textbook, Chapter 5.1.

Lecture 20 (April 5)

The three definining properties of the determinant lead to the famous product rule, det(AB) = det(A) * det(B). They also lead to the cofactor expansions, which reduce the determinant calculation to a combination of "one size smaller" determinants: the cofactors . The cofactors of a matrix are the key to explicit formulas for its inverse! The inverse of a matrix A is the matrix of cofactors C transposed and divided by the determinant: A^(-1) = C^T / det(A).

The cofactor formula for the inverse is often useful theoretically, but it does not lead to efficient or stable methods for numerical inversion or the solution of linear systems. However, the cofactor expansion also reveals that the determinant of a triangular matrix is the product of its diagonal elements. This leads to an elegant and practical method for computing determinants using A = L * U because det(A) = det(L) * det(U) = (product of pivots). When computing determinants numerically is explicilty required (rarely) in an application, we return to elimination and A = LU.

Further Reading: Textbook, Chapter 5.1 and 5.2.

Lecture 21 (April 7)

The determinant has an elegant connection to geometry that makes it indispensible in certain areas of geometry and calculus. If we use the columns of a square n-by-n matrix A to fill out the sides of a parallelpiped in n dimensions, the determinant of A is equal to the area of that parallelpiped.

The eigenvectors of a matrix identify extremely special directions. When a matrix multiplies its eigenvector, the direction doesn't change! The result is a scaled ("strectched" or "shrunk") version of the same eigenvector. The scaling factor is the eigenvalue: if the eigenvalue is less than 1 in magnitude, the eigenvector shrinks. If the eigenvalue is greater than 1 in magnitude, the eigenvector stretches. A zero eigenvalue means the matrix is singular, while an eigenvalue with magnitude one means the matrix doesn't stretch or shrink the eigenvector at all.

Further Reading: Textbook, Chapter 5.3 and 6.1.

Lecture 22 (April 10)

To compute eigenvalues and eigenvectors, we first use the det(A - lambda * I)=0 to find the eigenvalues. Once we have the eigenvalues lambda that make A - lambda * I singular (zero determinant), we solve for vectors in the nullspace of A - lambda * I. These are the eigenvectors, the solutions of Ax = lambda x.

The eigenvalues of a triangular matrix are just the values on the diagonal. The eigenvectors of a rank one matrix uv^T are u (eigenvalue = v^Tu) and all vectors orthogonal to v (eigenvalues = 0). The eigenvalues of an orthogonal matrix always have |lambda|=1 because orthogonal matrices don't change the lengths of vectors! The eigenvalues of the familiar 2 x 2 "rotation-by-theta" matrix might be a surprise: they are the complex numbers exp(i theta). The eigenvalues of a 2 x 2 matrix can be expressed simply in terms of the trace and determinant of the matrix using the quadratic formula. In any dimension, the trace of a matrix = the sum of the eigenvalues and the determinant of a matrix = the product of the eigenvalues.

Further Reading: Textbook, Chapter 6.1 and 6.2.

Lecture 23 (April 12)

When the eigenvectors of A form a basis (n linearly independent vectors for n x n A), A acts like a diagonal matrix in that basis! We can switch to the eigenvector basis to break difficult problems into simpler pieces. For example, coupled linear differential equations in multiple variables become a collection of uncoupled linear differential equations that can be solved with the tools of 1D calculus. That is the power of eigenvalues and eigenvectors at work! In matrix notation, diagonalization is expressed as A = X D X^(-1), where the columns of X are the eigenvectors of A and the diagonal matrix D has the eigenvalues of A on its diagonal. In the eigenvector basis A becomes X A X^(-1) = D - a diagonal matrix! We say that A is diagonalizable. Every matrix with distinct eigenvalues (no repeats) is diagonalizable.

Further Reading: Textbook, Chapter 6.2 and 6.3.

Lecture 24 (April 14)

Matrices that are symmetric, S^T = S, are a very special type of diagonalizable matrix. They have a full basis of orthonormal eigenvectors! And their eigenvalues are always real (remember that many matrices can have complex eigenvalues). Life is good when we are dealing with symmetric matrices.

Further Reading: Textbook, Chapter 6.3.

Lecture 25 (April 21)

Although this course focuses primarily on matrices and vectors whose entries are real numbers, complex matrices and vectors play an incredibly important role in applied mathematics. Most of the important facts about real linear algebra have a mirror image in complex linear algebra, as long as we replace the transpose operation for real vectors and matrices by the conjugate transpose operation. Then, real symmetric matrices become complex Hermitian matrices, real orthogonal matrices become complex unitary matrices, and so on.

Further Reading: Textbook, Chapter 6.4.

Lecture 26 (April 24)

Eigenvalues and eigenvectors play a key role in the solution and analysis of linear systems of differential equations. The eigenvectors define important invariant directions in the phase space. Any initial condition aligned with an eigenvector stays in that direction for all time! The eigenvalues reveal the dynamics along these invariant directions: positive eigenvalues lead to solutions that grow expontially in time, while negative eigenvalues lead to solutions that decay exponentially in time. The general solutions are linear combinations of the solutions along the invariant directions.

Further Reading: Textbook, Chapter 6.5.

Lecture 27 (April 26)

The singular value decomposition (SVD) factors every matrix (square or rectangular) into the product of three simpler matrices: A = USV^T. U and V have orthonormal columns called left and right singular vectors, respectively, while S is the diagonal matrix of singular values. Geometrically, the SVD that every matrix can be diagonalized by rotating (or reflecting) its input and output spaces, i.e., with orthogonal transformations. At the heart of the SVD, one finds the eigenvalues and eigenvectors of the square symmetric positive definite matrices AA^T and A^TA.

Further Reading: Textbook, Chapter 7.1.

Lecture 28 (April 28)

The SVD is constructed from eigenvalues and eigenvectors of the square symmetric positive definite matrices A^TA and AA^T. Symmetry means these matrices have complete orthogonormal sets of eigenvectors, which can be collected into orthogonal matrices U and V. Postive definite means that the eigenvalues are real and postive - in fact, both matrices have the same nonzero eigenvalues! The key to the SVD is the link between these two orthonormal sets of eigenvectors: they lead to the decomposition AV = US and then A = USV^T. The columns of U and V are orthogonal bases for the column and row spaces of A, respectively, and the singular values S describes how A stretches and shrinks vectors along these coupled input-output directions.

Further Reading: Textbook, Chapter 7.1 and 7.2.

Lecture 29 (May 1)

The singular values of A describe the coupling strength between special input directions (columns of V - right singular vectors) and special output directions (columns of U - left singular vectors). The singular values reveal the important directions in the row and column spaces of A. We can build a low-rank opproximation to A by dropping the small singular values and associated singular vectors from the SVD. Storing the largest few singular values and vectors can provide very good approximations to matrices with rapidly decaying singular values at a fraction of the cost of storing A.

As a first example, consider the task of compressing an array of pixels in a digital image without losing too much image quality. The SVD of the array can provide remarkable compression power! As a second example, consider the problem of identifying directions of large variance in student or consumer data. Which combinations of variables account for the largest differences among students? The SVD leads to principle componant analysis (PCA), which identifies these directions and uses them to decouple covariant parameters.

Further Reading: Textbook, Chapter 7.2 and 7.3. Experiment wit the SVD for image compression !

Lecture 30 (May 3)

On the computer, eigenvalues (and singular values) are not computed from det(A - lambda I). Instead, they are computed with an iterative method called the QR algorithm. The idea of the QR algorithm is simple: compute the QR factorization A=QR, reverse the factors and compute AA = RQ, compute the new QR factorization of AA, reverse the factors again, ...., and repeat until convergence! This extroardinary strategy reveals the eigenvalues of A along the diagonal of the upper triangular factors. With the right "bells and whistles," the QR iterations can be computed efficiently and each eigenvalues is revealed rapidly within a few iterations.

Many problems in applied math boil down to finding the minimum value of a function that depends on many variables. Often, we don't know the function explicitly, but we can evaluate it and (perhaps approximate) its partial derivatives. How should we go about finding the minimum? One strategy is to simply walk downhill . The gradient (vector of partial derivatives) tells us which way to walk - the direction of steepest descent. Gradient descent (also called steepest descent) is the basic building block for many first-order optimization algorithms that are used to solve engineering design problems, train neural networks, and much more.

Further Reading: Textbook, Chapter 9.1.

Lecture 31 (May 5)

Gradient descent is the prototypical "first-order" optimization algorithm: the algorithm tries to find local minima by evaluating the objective function and its derivative to "walk downhill." Pure gradient descent can get stuck in a criss-crossing pattern in narrow valley's, where the function is ill-conditioned, and converge slowly. Adding an inertial term that accounts for the previous search direction can help mitigate slow convergence due to ill-conditioning.

In deep neural networks, complex and high-dimensional functions are approximated by a layered network of "neurons:" Each neuron has an "activation function" that governs its response to inputs. The inputs and ouputs of neurons in each layer are connected by linear (or affine) transformations. The weights of a neural network govern the strength of connections between individual neurons. The weights are adjusted while "training" the network, i.e., minimizing a loss function that describes how closely the neural network's predictions match the training data.

Further Reading: Textbook, Chapter 9.2

Lecture 32 (May 8)

Deep neural networks are constructed by composing relatively simple parts (linear transformations and simple nonlinear activation functions) and composing them in complex architectures. Althuough classical statistics warns against overparametrizing models and "overfitting" noisy data, neural nets have led to the discovery that nonlinear models can actually thrive in the overparametrized regime. Understanding the approximation power of deep neural networks and their ability to generalize from training data to test data is a fascinating area of active research.

Further Reading: Explore the basics with this playground developed by Daniel Smilkov and Shan Carter.

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18.06 Linear Algebra, Spring 2020

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Video Lectures - Lecture 1

These video lectures of Professor Gilbert Strang teaching 18.06 were recorded in Fall 1999 and do not correspond precisely to the current edition of the textbook. However, this book is still the best reference for more information on the topics covered in each lecture.

Instructor/speaker: Prof. Gilbert Strang

Transcript - Lecture 1

Hi. This is the first lecture in MIT's course 18.06, linear algebra, and I'm Gilbert Strang. The text for the course is this book, Introduction to Linear Algebra.

And the course web page, which has got a lot of exercises from the past, MatLab codes, the syllabus for the course, is web.mit.edu/18.06.

And this is the first lecture, lecture one.

So, and later we'll give the web address for viewing these, videotapes. Okay, so what's in the first lecture? This is my plan.

The fundamental problem of linear algebra, which is to solve a system of linear equations.

So let's start with a case when we have some number of equations, say n equations and n unknowns.

So an equal number of equations and unknowns.

That's the normal, nice case.

And what I want to do is -- with examples, of course -- to describe, first, what I call the Row picture. That's the picture of one equation at a time. It's the picture you've seen before in two by two equations where lines meet.

So in a minute, you'll see lines meeting.

The second picture, I'll put a star beside that, because that's such an important one.

And maybe new to you is the picture -- a column at a time.

And those are the rows and columns of a matrix.

So the third -- the algebra way to look at the problem is the matrix form and using a matrix that I'll call A.

Okay, so can I do an example? The whole semester will be examples and then see what's going on with the example.

So, take an example. Two equations, two unknowns. So let me take 2x -y =0, let's say. And -x +2y=3.

Okay. let me -- I can even say right away -- what's the matrix, that is, what's the coefficient matrix? The matrix that involves these numbers -- a matrix is just a rectangular array of numbers. Here it's two rows and two columns, so 2 and -- minus 1 in the first row minus 1 and 2 in the second row, that's the matrix.

And the right-hand -- the, unknown -- well, we've got two unknowns. So we've got a vector, with two components, x and x, and we've got two right-hand sides that go into a vector 0 3.

I couldn't resist writing the matrix form, right -- even before the pictures. So I always will think of this as the matrix A, the matrix of coefficients, then there's a vector of unknowns.

Here we've only got two unknowns.

Later we'll have any number of unknowns.

And that vector of unknowns, well I'll often -- I'll make that x -- extra bold. A and the right-hand side is also a vector that I'll always call b.

So linear equations are A x equal b and the idea now is to solve this particular example and then step back to see the bigger picture. Okay, what's the picture for this example, the Row picture? Okay, so here comes the Row picture.

So that means I take one row at a time and I'm drawing here the xy plane and I'm going to plot all the points that satisfy that first equation. So I'm looking at all the points that satisfy 2x-y =0. It's often good to start with which point on the horizontal line -- on this horizontal line, y is zero.

The x axis has y as zero and that -- in this case, actually, then x is zero. So the point, the origin -- the point with coordinates (0,0) is on the line. It solves that equation.

Okay, tell me in -- well, I guess I have to tell you another point that solves this same equation.

Let me suppose x is one, so I'll take x to be one.

Then y should be two, right? So there's the point one two that also solves this equation.

And I could put in more points. But, but let me put in all the points at once, because they all lie on a straight line. This is a linear equation and that word linear got the letters for line in it.

That's the equation -- this is the line that ...

of solutions to 2x-y=0 my first row, first equation.

So typically, maybe, x equal a half, y equal one will work. And sure enough it does.

Okay, that's the first one. Now the second one is not going to go through the origin. It's always important.

Do we go through the origin or not? In this case, yes, because there's a zero over there. In this case we don't go through the origin, because if x and y are zero, we don't get three. So, let me again say suppose y is zero, what x do we actually get? If y is zero, then I get x is minus three.

So if y is zero, I go along minus three.

So there's one point on this second line.

Now let me say, well, suppose x is minus one -- just to take another x. If x is minus one, then this is a one and I think y should be a one, because if x is minus one, then I think y should be a one and we'll get that point. Is that right? If x is minus one, that's a one.

If y is a one, that's a two and the one and the two make three and that point's on the equation.

Okay. Now, I should just draw the line, right, connecting those two points at -- that will give me the whole line. And if I've done this reasonably well, I think it's going to happen to go through -- well, not happen -- it was arranged to go through that point. So I think that the second line is this one, and this is the all-important point that lies on both lines. Shall we just check that that point which is the point x equal one and y was two, right? That's the point there and that, I believe, solves both equations.

Let's just check this. If x is one, I have a minus one plus four equals three, okay. Apologies for drawing this picture that you've seen before. But this -- seeing the row picture -- first of all, for n equal 2, two equations and two unknowns, it's the right place to start. Okay.

So we've got the solution. The point that lies on both lines. Now can I come to the column picture? Pay attention, this is the key point. So the column picture.

I'm now going to look at the columns of the matrix.

I'm going to look at this part and this part.

I'm going to say that the x part is really x times -- you see, I'm putting the two -- I'm kind of getting the two equations at once -- that part and then I have a y and in the first equation it's multiplying a minus one and in the second equation a two, and on the right-hand side, zero and three. You see, the columns of the matrix, the columns of A are here and the right-hand side b is there. And now what is the equation asking for? It's asking us to find -- somehow to combine that vector and this one in the right amounts to get that one. It's asking us to find the right linear combination -- this is called a linear combination.

And it's the most fundamental operation in the whole course.

It's a linear combination of the columns.

That's what we're seeing on the left side.

Again, I don't want to write down a big definition.

You can see what it is. There's column one, there's column two. I multiply by some numbers and I add. That's a combination -- a linear combination and I want to make those numbers the right numbers to produce zero three. Okay.

Now I want to draw a picture that, represents what this -- this is algebra. What's the geometry, what's the picture that goes with it? Okay. So again, these vectors have two components, so I better draw a picture like that. So can I put down these columns? I'll draw these columns as they are, and then I'll do a combination of them.

So the first column is over two and down one, right? So there's the first column.

The first column. Column one.

It's the vector two minus one. The second column is -- minus one is the first component and up two.

It's here. There's column two.

So this, again, you see what its components are. Its components are minus one, two. Good.

That's this guy. Now I have to take a combination. What combination shall I take? Why not the right combination, what the hell? Okay. So the combination I'm going to take is the right one to produce zero three and then we'll see it happen in the picture. So the right combination is to take x as one of those and two of these.

It's because we already know that that's the right x and y, so why not take the correct combination here and see it happen? Okay, so how do I picture this linear combination? So I start with this vector that's already here -- so that's one of column one, that's one times column one, right there.

And now I want to add on -- so I'm going to hook the next vector onto the front of the arrow will start the next vector and it will go this way. So let's see, can I do it right? If I added on one of these vectors, it would go left one and up two, so we'd go left one and up two, so it would probably get us to there.

Maybe I'll do dotted line for that.

Okay? That's one of column two tucked onto the end, but I wanted to tuck on two of column two. So that -- the second one -- we'll go up left one and up two also.

It'll probably end there. And there's another one.

So what I've put in here is two of column two.

Added on. And where did I end up? What are the coordinates of this result? What do I get when I take one of this plus two of that? I do get that, of course.

There it is, x is zero, y is three, that's b. That's the answer we wanted.

And how do I do it? You see I do it just like the first component. I have a two and a minus two that produces a zero, and in the second component I have a minus one and a four, they combine to give the three.

But look at this picture. So here's our key picture.

I combine this column and this column to get this guy.

That was the b. That's the zero three.

Okay. So that idea of linear combination is crucial, and also -- do we want to think about this question? Sure, why not.

What are all the combinations? If I took -- can I go back to xs and ys? This is a question for really -- it's going to come up over and over, but why don't we see it once now? If I took all the xs and all the ys, all the combinations, what would be all the results? And, actually, the result would be that I could get any right-hand side at all.

The combinations of this and this would fill the whole plane.

You can tuck that away. We'll, explore it further.

But this idea of what linear combination gives b and what do all the linear combinations give, what are all the possible, achievable right-hand sides be -- that's going to be basic. Okay.

Can I move to three equations and three unknowns? Because it's easy to picture the two by two case.

Let me do a three by three example.

Okay, I'll sort of start it the same way, say maybe 2x-y and maybe I'll take no zs as a zero and maybe a -x+2y and maybe a -z as a -- oh, let me make that a minus one and, just for variety let me take, -3z, -3ys, I should keep the ys in that line, and 4zs is, say, 4. Okay.

That's three equations. I'm in three dimensions, x, y, z. And, I don't have a solution yet. So I want to understand the equations and then solve them. Okay.

So how do I you understand them? The row picture one way. The column picture is another very important way. Just let's remember the matrix form, here, because that's easy. The matrix form -- what's our matrix A? Our matrix A is this right-hand side, the two and the minus one and the zero from the first row, the minus one and the two and the minus one from the second row, the zero, the minus three and the four from the third row. So it's a three by three matrix. Three equations, three unknowns. And what's our right-hand side? Of course, it's the vector, zero minus one, four. Okay.

So that's the way, well, that's the short-hand to write out the three equations. But it's the picture that I'm looking for today. Okay, so the row picture.

All right, so I'm in three dimensions, x, y and z. And I want to take those equations one at a time and ask -- and make a picture of all the points that satisfy -- let's take equation number two.

If I make a picture of all the points that satisfy -- all the x, y, z points that solve this equation -- well, first of all, the origin is not one of them.

x, y, z -- it being 0, 0, 0 would not solve that equation. So what are some points that do solve the equation? Let's see, maybe if x is one, y and z could be zero. That would work, right? So there's one point.

I'm looking at this second equation, here, just, to start with. Let's see.

Also, I guess, if z could be one, x and y could be zero, so that would just go straight up that axis. And, probably I'd want a third point here. Let me take x to be zero, z to be zero, then y would be minus a half, right? So there's a third point, somewhere -- oh my -- okay. Let's see.

I want to put in all the points that satisfy that equation.

Do you know what that bunch of points will be? It's a plane. If we have a linear equation, then, fortunately, the graph of the thing, the plot of all the points that solve it are a plane.

These three points determine a plane, but your lecturer is not Rembrandt and the art is going to be the weak point here.

So I'm just going to draw a plane, right? There's a plane somewhere. That's my plane.

That plane is all the points that solves this guy.

Then, what about this one? Two x minus y plus zero z.

So z actually can be anything. Again, it's going to be another plane. Each row in a three by three problem gives us a plane in three dimensions.

So this one is going to be some other plane -- maybe I'll try to draw it like this. And those two planes meet in a line. So if I have two equations, just the first two equations in three dimensions, those give me a line. The line where those two planes meet. And now, the third guy is a third plane. And it goes somewhere.

Okay, those three things meet in a point.

Now I don't know where that point is, frankly.

But -- linear algebra will find it.

The main point is that the three planes, because they're not parallel, they're not special.

They do meet in one point and that's the solution.

But, maybe you can see that this row picture is getting a little hard to see. The row picture was a cinch when we looked at two lines meeting.

When we look at three planes meeting, it's not so clear and in four dimensions probably a little less clear.

So, can I quit on the row picture? Or quit on the row picture before I've successfully found the point where the three planes meet? All I really want to see is that the row picture consists of three planes and, if everything works right, three planes meet in one point and that's a solution.

Now, you can tell I prefer the column picture.

Okay, so let me take the column picture.

That's x times -- so there were two xs in the first equation minus one x is, and no xs in the third.

It's just the first column of that.

And how many ys are there? There's minus one in the first equations, two in the second and maybe minus three in the third.

Just the second column of my matrix.

And z times no zs minus one zs and four zs.

And it's those three columns, right, that I have to combine to produce the right-hand side, which is zero minus one four.

Okay. So what have we got on this left-hand side? A linear combination.

It's a linear combination now of three vectors, and they happen to be -- each one is a three dimensional vector, so we want to know what combination of those three vectors produces that one.

Shall I try to draw the column picture, then? So, since these vectors have three components -- so it's some multiple -- let me draw in the first column as before -- x is two and y is minus one. Maybe there is the first column. y -- the second column has maybe a minus one and a two and the y is a minus three, somewhere, there possibly, column two.

And the third column has -- no zero minus one four, so how shall I draw that? So this was the first component. The second component was a minus one. Maybe up here.

That's column three, that's the column zero minus one and four. This guy.

So, again, what's my problem? What this equation is asking me to do is to combine these three vectors with a right combination to produce this one. Well, you can see what the right combination is, because in this special problem, specially chosen by the lecturer, that right-hand side that I'm trying to get is actually one of these columns. So I know how to get that one.

So what's the solution? What combination will work? I just want one of these and none of these.

So x should be zero, y should be zero and z should be one. That's the combination.

One of those is obviously the right one.

Column three is actually the same as b in this particular problem. I made it work that way just so we would get an answer, (0,0,1), so somehow that's the point where those three planes met and I couldn't see it before. Of course, I won't always be able to see it from the column picture, either.

It's the next lecture, actually, which is about elimination, which is the systematic way that everybody -- every bit of software, too -- production, large-scale software would solve the equations.

So the lecture that's coming up.

If I was to add that to the syllabus, will be about how to find x, y, z in all cases. Can I just think again, though, about the big picture? By the big picture I mean let's keep this same matrix on the left but imagine that we have a different right-hand side. Oh, let me take a different right-hand side. So I'll change that right-hand side to something that actually is also pretty special.

Let me change it to -- if I add those first two columns, that would give me a one and a one and a minus three.

There's a very special right-hand side.

I just cooked it up by adding this one to this one.

Now, what's the solution with this new right-hand side? The solution with this new right-hand side is clear.

took one of these and none of those.

So actually, it just changed around to this when I took this new right-hand side.

Okay. So in the row picture, I have three different planes, three new planes meeting now at this point. In the column picture, I have the same three columns, but now I'm combining them to produce this guy, and it turned out that column one plus column two which would be somewhere -- there is the right column -- one of this and one of this would give me the new b. Okay.

So we squeezed in an extra example.

But now think about all bs, all right-hand sides.

Can I solve these equations for every right-hand side? Can I ask that question? So that's the algebra question.

Can I solve A x=b for every b? Let me write that down.

Can I solve A x =b for every right-hand side b? I mean, is there a solution? And then, if there is, elimination will give me a way to find it.

I really wanted to ask, is there a solution for every right-hand side? So now, can I put that in different words -- in this linear combination words? So in linear combination words, do the linear combinations of the columns fill three dimensional space? Every b means all the bs in three dimensional space.

Do you see that I'm just asking the same question in different words? Solving A x -- A x -- that's very important. A times x -- when I multiply a matrix by a vector, I get a combination of the columns. I'll write that down in a moment. But in my column picture, that's really what I'm doing. I'm taking linear combinations of these three columns and I'm trying to find b.

And, actually, the answer for this matrix will be yes. For this matrix A -- for these columns, the answer is yes. This matrix -- that I chose for an example is a good matrix. A non-singular matrix.

An invertible matrix. Those will be the matrices that we like best. There could be other -- and we will see other matrices where the answer becomes, no -- oh, actually, you can see when it would become no. What could go wrong? How could it go wrong that out of these -- out of three columns and all their combinations -- when would I not be able to produce some b off here? When could it go wrong? Do you see that the combinations -- let me say when it goes wrong. If these three columns all lie in the same plane, then their combinations will lie in that same plane. So then we're in trouble.

If the three columns of my matrix -- if those three vectors happen to lie in the same plane -- for example, if column three is just the sum of column one and column two, I would be in trouble. That would be a matrix A where the answer would be no, because the combinations -- if column three is in the same plane as column one and two, I don't get anything new from that.

All the combinations are in the plane and only right-hand sides b that I could get would be the ones in that plane.

So I could solve it for some right-hand sides, when b is in the plane, but most right-hand sides would be out of the plane and unreachable.

So that would be a singular case.

The matrix would be not invertible.

There would not be a solution for every b.

The answer would become no for that.

Okay. I don't know -- shall we take just a little shot at thinking about nine dimensions? Imagine that we have vectors with nine components.

Well, it's going to be hard to visualize those.

I don't pretend to do it. But somehow, pretend you do. Pretend we have -- if this was nine equations and nine unknowns, then we would have nine columns, and each one would be a vector in nine-dimensional space and we would be looking at their linear combinations. So we would be having the linear combinations of nine vectors in nine-dimensional space, and we would be trying to find the combination that hit the correct right-hand side b. And we might also ask the question can we always do it? Can we get every right-hand side b? And certainly it will depend on those nine columns. Sometimes the answer will be yes -- if I picked a random matrix, it would be yes, actually. If I used MatLab and just used the random command, picked out a nine by nine matrix, I guarantee it would be good.

It would be non-singular, it would be invertible, all beautiful. But if I choose those columns so that they're not independent, so that the ninth column is the same as the eighth column, then it contributes nothing new and there would be right-hand sides b that I couldn't get.

Can you sort of think about nine vectors in nine-dimensional space an take their combinations? That's really the central thought -- that you get kind of used to in linear algebra. Even though you can't really visualize it, you sort of think you can after a while. Those nine columns and all their combinations may very well fill out the whole nine-dimensional space. But if the ninth column happened to be the same as the eighth column and gave nothing new, then probably what it would fill out would be -- I hesitate even to say this -- it would be a sort of a plane -- an eight dimensional plane inside nine-dimensional space.

And it's those eight dimensional planes inside nine-dimensional space that we have to work with eventually.

For now, let's stay with a nice case where the matrices work, we can get every right-hand side b and here we see how to do it with columns. Okay.

There was one step which I realized I was saying in words that I now want to write in letters.

Because I'm coming back to the matrix form of the equation, so let me write it here. The matrix form of my equation, of my system is some matrix A times some vector x equals some right-hand side b. Okay.

So this is a multiplication. A times x.

Matrix times vector, and I just want to say how do you multiply a matrix by a vector? Okay, so I'm just going to create a matrix -- let me take two five one three -- and let me take a vector x to be, say, 1and 2. How do I multiply a matrix by a vector? But just think a little bit about matrix notation and how to do that in multiplication.

So let me say how I multiply a matrix by a vector.

Actually, there are two ways to do it.

Let me tell you my favorite way.

It's columns again. It's a column at a time.

For me, this matrix multiplication says I take one of that column and two of that column and add.

So this is the way I would think of it is one of the first column and two of the second column and let's just see what we get. So in the first component I'm getting a two and a ten. I'm getting a twelve there.

In the second component I'm getting a one and a six, I'm getting a seven. So that matrix times that vector is twelve seven. Now, you could do that another way. You could do it a row at a time. And you would get this twelve -- and actually I pretty much did it here -- this way.

Two -- I could take that row times my vector.

This is the idea of a dot product.

This vector times this vector, two times one plus five times two is the twelve. This vector times this vector -- one times one plus three times two is the seven.

So I can do it by rows, and in each row times my x is what I'll later call a dot product.

But I also like to see it by columns.

I see this as a linear combination of a column.

So here's my point. A times x is a combination of the columns of A. That's how I hope you will think of A times x when we need it.

Right now we've got -- with small ones, we can always do it in different ways, but later, think of it that way. Okay.

So that's the picture for a two by two system.

And if the right-hand side B happened to be twelve seven, then of course the correct solution would be one two.

Okay. So let me come back next time to a systematic way, using elimination, to find the solution, if there is one, to a system of any size and find out -- because if elimination fails, find out when there isn't a solution. Okay, thanks.

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This course parallels the combination of theory and applications in Professor Strang’s textbook Introduction to Linear Algebra . The course picks out four key applications in the book: Graphs and Networks; Systems of Differential Equations; Least Squares and Projections; and Fourier Series and the Fast Fourier Transform.

This is a basic subject on matrix theory and linear algebra. Emphasis is given to topics that will be useful in other disciplines, including systems of equations, vector spaces, determinants, eigenvalues, similarity, and positive definite matrices.

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