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

2022
Views:

326

Pages:

12
Author:

Clarence Dougherty

Eigenbases and diagonalization Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . We saw before that, if w ~ is in the span of ~v1 , ~v2 , . . . , ~vk , then we have a simple formula for An w. ~ Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . We saw before that, if w ~ is in the span of ~v1 , ~v2 , . . . , ~vk , then we have a simple formula for An w. ~ An (c1~v1 + c2~v2 + · · · + ck~vk ) = c1 λn1 ~v1 + c2 λn2 ~v2 + · · · + ck λnk~vk . Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . We saw before that, if w ~ is in the span of ~v1 , ~v2 , . . . , ~vk , then we have a simple formula for An w. ~ An (c1~v1 + c2~v2 + · · · + ck~vk ) = c1 λn1 ~v1 + c2 λn2 ~v2 + · · · + ck λnk~vk . If ~v1 , ~v2 , . . . , ~vn is a basis of eigenvectors, then we can explain this using our previous ideas about coordinates. A basis of eigenvectors is called an eigenbasis. We’ll discuss in the next lecture when eigenbases exist. For now, let’s assume that we have one. Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vn be an eigenbasis of A, with A~vi = λi~vi . We write B for the basis (~v1 , ~v2 , . . . , ~vn ). Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vn be an eigenbasis of A, with A~vi = λi~vi . We write B for the basis (~v1 , ~v2 , . . . , ~vn ). So 0 0 [~vj ]B  ..    =  1.  . .. 0 with the 1 in the j-th row. If w ~ = c1~v1 + c2~v2 + · · · + cn~vn then  c1  c2 [w] ~ B =  ..  . . cn Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vn be an eigenbasis of A, with A~vi = λi~vi . We write B for the basis (~v1 , ~v2 , . . . , ~vn ). The equation A~vj = λj ~vj  tells us that the j-th column of [A]B 0 0   ..    is  λ.j    .. . 0 So  λ1 [A]B  =  λ2 λ3 ..  . . λn Earlier, we had the key equation A (c1~v1 + c2~v2 + · · · + cn~vn ) = c1 λ1~v1 + c2 λ2~v2 + · · · + cn λn~vn In B-coordinates, this looks like  λ1   c1  λ2    c1 λ1  c2 λ3 .. . λn c2 λ2   c3   c3 λ3   .  =  . . .. .. cn cn λn And we just have  [At ]B   =  λt1  λt2   .  λt3 .. . λtn We can rewrite these ideas without using the notion of coordinates. Let   | | |    S = ~v1 ~v2 · · · ~vn  . | | | We can rewrite these ideas without using the notion of coordinates. Let " | | # | S= ~ v1 ~ v2 ··· ~ vn | | | | | | . We can turn the n equations A~vj = λj ~vj into a single matrix equation: " | | # " | | # " | # | | | | AS = A ~ v1 ~ v2 ··· ~ vn | | | | | | " = So | ~ v1 | | = A~ v1 A~ v2 ··· A~ vn | | | | | | | | ~ v2 ··· ~ vn | | | | #  λ1  λ2 ..  .  λ2 ..  S −1 . . λn λ1 ~ v1 λ2 ~ v2 ··· λn ~ vn | | | | | |  λ1 =S λn  λ1 A=S =  λ2 ..  . λn  λ1 I’ll abbreviate the diagonal matrix  A = SDS −1 .  λ2 ..  to D. So . λn This gives us another way to think about taking powers of A: At = (SDS −1 )t = (SDS −1 )(SDS −1 ) · · · (SDS −1 ) = SDS −1 SDS −1 · · · SDS −1 = SDD · · · DS −1 = SDt S −1 .  λ1 I’ll abbreviate the diagonal matrix  A = SDS −1 .  λ2 ..  to D. So . λn And to think about the equation A~vj = λj ~vj : We have 0 0 0 0 .. .. .. .. . . .  .  −1       λj  = λj ~ ~vj = S  1  so A~vj = (SDS ) S  1  = SD  1  = S  vj .   .. .. .. .. . . . . 0 0 0 0 In general, writing A = SDS −1 gives a good way to think about any equation with eigenvectors.

Eigenbases and diagonalization

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock