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

2023
Views:

330

Pages:

17
Author:

bballsumo31a87g

Eigenbases Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be nonzero 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 nonzero 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 nonzero 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 . So it is useful to express vectors as linear combinations of eigenvectors. The ideal situation would be is A has a basis of eigenvectors. We call this an eigenbasis. Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be nonzero eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . Let’s talk about whether or not ~v1 , ~v2 , . . . , ~vk could be a basis. There are two issues: 1. Are ~v1 , ~v2 , . . . , ~vk linearly independent? 2. Do ~v1 , ~v2 , . . . , ~vk span Rn ? Are ~v1 , ~v2 , . . . , ~vk linearly independent? The first question has a very clear answer! Theorem: Let A be an n × n square matrix. Suppose that λ1 , λ2 , . . . , λk are scalars, no two of them equal, and that ~vi is a nonzero λi -eigenvector of A. Then the vectors ~v1 , ~v2 , . . . , ~vk are linearly independent. Are ~v1 , ~v2 , . . . , ~vk linearly independent? The first question has a very clear answer! Theorem: Let A be an n × n square matrix. Suppose that λ1 , λ2 , . . . , λk are scalars, no two of them equal, and that ~vi is a nonzero λi -eigenvector of A. Then the vectors ~v1 , ~v2 , . . . , ~vk are linearly independent. Let’s first see why it works for k = 2. Are ~v1 , ~v2 , . . . , ~vk linearly independent? The first question has a very clear answer! Theorem: Let A be an n × n square matrix. Suppose that λ1 , λ2 , . . . , λk are scalars, no two of them equal, and that ~vi is a nonzero λi -eigenvector of A. Then the vectors ~v1 , ~v2 , . . . , ~vk are linearly independent. Let’s first see why it works for k = 2. To say that ~v1 and ~v2 are linearly independent is to say that they are both nonzero, and they are not proportional. We assumed that ~v1 and ~v2 are nonzero. Suppose that they were proportional, so that ~v2 = c~v1 . Then A~v2 = A(c~v1 ) = cA~v1 = cλ1~v1 and also A~v2 = λ2~v2 = cλ2~v1 . Are ~v1 , ~v2 , . . . , ~vk linearly independent? The first question has a very clear answer! Theorem: Let A be an n × n square matrix. Suppose that λ1 , λ2 , . . . , λk are scalars, no two of them equal, and that ~vi is a nonzero λi -eigenvector of A. Then the vectors ~v1 , ~v2 , . . . , ~vk are linearly independent. Let’s first see why it works for k = 2. To say that ~v1 and ~v2 are linearly independent is to say that they are both nonzero, and they are not proportional. We assumed that ~v1 and ~v2 are nonzero. Suppose that they were proportional, so that ~v2 = c~v1 . Then A~v2 = A(c~v1 ) = cA~v1 = cλ1~v1 and also A~v2 = λ2~v2 = cλ2~v1 . So cλ1~v1 = cλ2~v1 and c(λ1 − λ2 )~v1 = ~0. But λ1 − λ2 is not zero, since the lambdas are different, the scalar c is nonzero since ~v2 6= ~0, and the vector ~v1 is also nonzero. This gives a contradiction. Theorem: Let A be an n × n square matrix. Suppose that λ1 , λ2 , . . . , λk are scalars, no two of them equal, and that ~vi is a nonzero λi -eigenvector of A. Then the vectors ~v1 , ~v2 , . . . , ~vk are linearly independent. We will prove the general case by induction on k. So, suppose we have already shown that ~v1 , ~v2 , . . . , ~vk−1 are linearly independent. We want to show that adding in the vector ~vk doesn’t disturb the linear independence. Suppose, for the sake of contradiction, that ~vk is redundant. So ~vk = c1~v1 + c2~v2 + · · · + ck−1~vk−1 . Suppose we have already shown that ~v1 , ~v2 , . . . , ~vk−1 are linearly independent. Suppose, for the sake of contradiction, that ~vk is redundant. So ~vk = c1~v1 + c2~v2 + · · · + ck−1~vk−1 . We have: A~vk = A (c1~v1 + c2~v2 + · · · + ck−1~vk−1 ) = c1 A~v1 + c2 A~v2 + · · · + ck−1 A~vk−1 = c1 λ1~v1 + c2 λ2~v2 + · · · + ck−1 λk−1~vk−1 . But also: A~vk = λk~vk = λk (c1~v1 + c2~v2 + · · · + ck−1~vk−1 ) = c1 λk~v1 + c2 λk~v2 + · · · + ck−1 λk~vk−1 . So c1 λ1~v1 + · · · + ck−1 λk−1~vk−1 = c1 λk~v1 + · · · + ck−1 λk~vk−1 and c1 (λ1 − λk )~v1 + · · · + ck−1 (λk−1 − λk )~vk−1 = ~0. Suppose we have already shown that ~v1 , ~v2 , . . . , ~vk−1 are linearly independent. Suppose, for the sake of contradiction, that ~vk is redundant. So ~vk = c1~v1 + c2~v2 + · · · + ck−1~vk−1 · · · after some computations · · · c1 (λ1 − λk )~v1 + · · · + ck−1 (λk−1 − λk )~vk−1 = ~0. But we assumed that ~v1 , ~v2 , . . . , ~vk−1 were linearly independent, so this means that c1 (λ1 − λk ) = c2 (λ2 − λk ) = · · · = ck−1 (λk−1 − λk ) = 0. And the λ’s are all different, so c1 = c2 = · · · = ck−1 = 0. Suppose we have already shown that ~v1 , ~v2 , . . . , ~vk−1 are linearly independent. Suppose, for the sake of contradiction, that ~vk is redundant. So ~vk = c1~v1 + c2~v2 + · · · + ck−1~vk−1 · · · after some computations · · · c1 (λ1 − λk )~v1 + · · · + ck−1 (λk−1 − λk )~vk−1 = ~0. But we assumed that ~v1 , ~v2 , . . . , ~vk−1 were linearly independent, so this means that c1 (λ1 − λk ) = c2 (λ2 − λk ) = · · · = ck−1 (λk−1 − λk ) = 0. And the λ’s are all different, so c1 = c2 = · · · = ck−1 = 0. But then ~vk = 0 + 0 + · · · + 0 = 0, contradicting that the ~vj are nonzero eigenvectors. QED Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be nonzero eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . Let’s talk about whether or not ~v1 , ~v2 , . . . , ~vk could be a basis. There are two issues: 1. Are ~v1 , ~v2 , . . . , ~vk linearly independent? 2. Do ~v1 , ~v2 , . . . , ~vk span Rn ? Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be nonzero eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . Let’s talk about whether or not ~v1 , ~v2 , . . . , ~vk could be a basis. There are two issues: 1. Are ~v1 , ~v2 , . . . , ~vk linearly independent? ◦ Yes, assuming all the λi are different from each other. 2. Do ~v1 , ~v2 , . . . , ~vk span Rn ? Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be nonzero eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . Let’s talk about whether or not ~v1 , ~v2 , . . . , ~vk could be a basis. There are two issues: 1. Are ~v1 , ~v2 , . . . , ~vk linearly independent? ◦ Yes, assuming all the λi are different from each other. 2. Do ~v1 , ~v2 , . . . , ~vk span Rn ? ◦ This is trickier. If the characteristic polynomial of A has n distinct roots, then the answer must be yes, since n linearly independent vectors must be a basis. Let A be an n × n square matrix. Let ~v1 , ~v2 , . . . , ~vk be nonzero eigenvectors of A, with eigenvalues λ1 , λ2 , . . . , λk . Let’s talk about whether or not ~v1 , ~v2 , . . . , ~vk could be a basis. There are two issues: 1. Are ~v1 , ~v2 , . . . , ~vk linearly independent? ◦ Yes, assuming all the λi are different from each other. 2. Do ~v1 , ~v2 , . . . , ~vk span Rn ? ◦ This is trickier. If the characteristic polynomial of A has n distinct roots, then the answer must be yes, since n linearly independent vectors must be a basis. ◦ To discuss the general case, we introduce the notions of geometric and algebraic multiplicity . These are the topic of the next lecture.

Lecture 9b - Eigenbases

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