University:
University of Michigan
Course:
MATH 217 | Linear Algebra
Academic year:

2024
Views:

321

Pages:

19
Author:

solcellep7g5

Eigenvectors of a symmetric matrix We have seen two kinds of bases for Rn that are particularly convenient: • Orthonormal bases, where our intuition from Euclidean geometry is relevant • Eigenbases, which are good for computing powers of matrices. What happens if we put them together? Let A be a square matrix. Suppose that A has an orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn , with eigenvalues λ1 , λ2 , . . . , λn . What can we say about A? Let Q be the matrix with columns ~v1 , ~v2 , . . . , ~vn . Let D be the diagonal matrix with diagonal λ1 , λ2 , . . . , λn . Then we have A = QDQ−1 . Let A be a square matrix. Suppose that A has an orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn , with eigenvalues λ1 , λ2 , . . . , λn . What can we say about A? Let Q be the matrix with columns ~v1 , ~v2 , . . . , ~vn . Let D be the diagonal matrix with diagonal λ1 , λ2 , . . . , λn . Then we have A = QDQ−1 . But, since ~v1 , ~v2 , . . . , ~vn is orthonormal, we know that Q is orthogonal, and thus Q−1 = QT . So we have A = QDQT . Let A be a square matrix. Suppose that A has an orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn , with eigenvalues λ1 , λ2 , . . . , λn . What can we say about A? Let Q be the matrix with columns ~v1 , ~v2 , . . . , ~vn . Let D be the diagonal matrix with diagonal λ1 , λ2 , . . . , λn . Then we have A = QDQ−1 . But, since ~v1 , ~v2 , . . . , ~vn is orthonormal, we know that Q is orthogonal, and thus Q−1 = QT . So we have A = QDQT . We deduce that AT = (QDQT )T = (QT )T DT QT = QDQT = A. In other words, A is symmetric! We have thus shown: Let A be a square matrix. If A has an orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn then A is symmetric, meaning that A = AT . We have thus shown: Let A be a square matrix. If A has an orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn then A is symmetric, meaning that A = AT . Symmetric matrices come up in many places.    T1 (t + 1) 0.9 0.1    T2 (t + 1) = 0.1 0.8    T3 (t + 1) 0 0.1   0 T1 (t)     0.1 T2 (t)  0.9 T3 (t) We have thus shown: Let A be a square matrix. If A has an orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn then A is symmetric, meaning that A = AT . The big result about symmetric matrices is that the reverse is true: The spectral theorem: If A is a symmetric n × n matrix, then A has an orthonormal eigenbasis. We have thus shown: Let A be a square matrix. If A has an orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn then A is symmetric, meaning that A = AT . The big result about symmetric matrices is that the reverse is true: The spectral theorem: If A is a symmetric n × n matrix, then A has an orthonormal eigenbasis. The easy part of this is that, if A has an eigenbasis, then it has an orthonormal eigenbasis. Let’s see why, if A is a symmetric matrix with an eigenbasis, then A has an orthonormal eigenbasis. Let ~v and w ~ be any two vectors. Since A is symmetric, ~v T Aw ~ = ~v T AT w ~ = (A~v )T w. ~ In other words, ~v · (Aw) ~ = (A~v ) · w. ~ Let’s see why, if A is a symmetric matrix with an eigenbasis, then A has an orthonormal eigenbasis. Let ~v and w ~ be any two vectors. Since A is symmetric, ~v T Aw ~ = ~v T AT w ~ = (A~v )T w. ~ In other words, ~v · (Aw) ~ = (A~v ) · w. ~ Now, let ~v and w ~ be two eigenvectors of A, with distinct eigenvalues α and β. ~v · (β w) ~ =w ~ · (α~v ). β~v · w ~ = α~v · w. ~ (β − α)~v · w ~ =0 ~v · w ~ = 0. Let’s see why, if A is a symmetric matrix with an eigenbasis, then A has an orthonormal eigenbasis. Let ~v and w ~ be any two vectors. Since A is symmetric, ~v T Aw ~ = ~v T AT w ~ = (A~v )T w. ~ In other words, ~v · (Aw) ~ = (A~v ) · w. ~ Now, let ~v and w ~ be two eigenvectors of A, with distinct eigenvalues α and β. ~v · (β w) ~ =w ~ · (α~v ). β~v · w ~ = α~v · w. ~ (β − α)~v · w ~ =0 ~v · w ~ = 0. So ~v and w ~ are orthogonal! We have shown that any eigenbasis of A will be orthogonal, and we can rescale such a basis to be orthonormal. The spectral theorem: If A is a symmetric n × n matrix, then A has an orthonormal eigenbasis. So far, we have seen that • If A has an orthonormal eigenbasis, then A is symmetric. • If A is symmetric and has an eigenbasis, it has an orthonormal eigenbasis. That symmetric matrices have eigenbases at all is much harder. We’ll prove that later, after we’ve also talked about singular value decomposition. For now, two other ways to think about the spectral theorem. The spectral theorem gives us a geometric way to think about the condition that a matrix is symmetric: A is symmetric if and only if there is an orthonormal bases on which A acts by dilation. The spectral theorem gives us a geometric way to think about the condition that a matrix is symmetric: A is symmetric if and only if there is an orthonormal bases on which A acts by dilation. Here is a picture, and what happens to it under [ 54 45 ]. The spectral theorem gives us a geometric way to think about the condition that a matrix is symmetric: A is symmetric if and only if there is an orthonormal bases on which A acts by dilation. Here is a picture, and what happens to it under [ 54 45 ]. Here is one more important way to think about the eigenvectors of a symmetric matrix. Let A be a symmetric matrix, with orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn and eigenvalues λ1 , λ2 , . . . , λn . Sort them so that λ1 > λ2 > · · · > λn . If w ~ is a vector of length 1, how long can Aw ~ be? How short can it be? Here is one more important way to think about the eigenvectors of a symmetric matrix. Let A be a symmetric matrix, with orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn and eigenvalues λ1 , λ2 , . . . , λn . Sort them so that λ1 > λ2 > · · · > λn . If w ~ is a vector of length 1, how long can Aw ~ be? How short can it be? Write w ~ = c1~v1 + c2~v2 + · · · + cn~vn . The condition that |w| ~ = 1 is that c21 + c22 + · · · + c2n = 1. We have (Aw) ~ · (Aw) ~ = (c1 λ1~v1 + · · · + cn λn~vn ) · (c1 λ1~v1 + · · · + cn λn~vn ) = c21 λ21 + ··· + c2n λ2n . Here is one more important way to think about the eigenvectors of a symmetric matrix. Let A be a symmetric matrix, with orthonormal eigenbasis ~v1 , ~v2 , . . . , ~vn and eigenvalues λ1 , λ2 , . . . , λn . Sort them so that λ1 > λ2 > · · · > λn . If w ~ is a vector of length 1, how long can Aw ~ be? How short can it be? Write w ~ = c1~v1 + c2~v2 + · · · + cn~vn . The condition that |w| ~ = 1 is that c21 + c22 + · · · + c2n = 1. We have (Aw) ~ · (Aw) ~ = (c1 λ1~v1 + · · · + cn λn~vn ) · (c1 λ1~v1 + · · · + cn λn~vn ) = c21 λ21 + ··· + c2n λ2n The largest possible value is to take c1 = 1 and the other to be 0; the smallest is to take cn = 1 and the others to be 0. So the largest value of |Aw| ~ is λ1 , by taking w ~ = ~v1 , and the smallest value of |Aw| ~ is λn , by taking w ~ = ~vn . .

Eigenvectors of a symmetric matrix

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