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

2024
Views:

35

Pages:

15
Author:

tifletzphv

Algebraic and Geometric Multiplicity Let A be an n × n square matrix. We learned how to compute the eigenvalues and eigenvectors of A: 1. Compute the characteristic polynomial, det(A − tId), and find its roots. These are the eigenvalues. 2. For each eigenvalue λ, compute Ker(A − λId). This is the λ-eigenspace, the vectors in the λ-eigenspace are the λ-eigenvectors. Let A be an n × n square matrix. We learned how to compute the eigenvalues and eigenvectors of A: 1. Compute the characteristic polynomial, det(A − tId), and find its roots. These are the eigenvalues. 2. For each eigenvalue λ, compute Ker(A − λId). This is the λ-eigenspace, the vectors in the λ-eigenspace are the λ-eigenvectors. We learned that it is particularly nice when A has an eigenbasis, because then we can diagonalize A. An eigenbasis is a basis of eigenvectors. Let’s see what can happen when we carry out this algorithm. The simplest case: The characteristic polynomial has n distinct roots  1  1  4  A= −2  4−t  det(A − tId) = det −2  1 1−t = (4 − t)(1 − t) − (−2) = (t2 − 5t + 4) + 2 = t2 − 5t + 6 = (t − 2)(t − 3). 2-eigenspace: Ker (A − 2Id) = Ker 1 kernel is Span −2 . 4−2 1 −2 1−2 3-eigenspace: Ker (A − 3Id) = Ker 1 kernel is Span −1 . 4−3 1 −2 1−3 = Ker 2 1 −2 −1 The = Ker 1 1 −2 −2 The The simplest case: The characteristic polynomial has n distinct roots  4  A= −2  1  1 λ1 = 2, ~v1 = 1 −2 λ2 = 3, ~v2 = 1 −1 . The vectors ~v1 and ~v2 form a basis. There are n of them, and λ1 and λ2 are distinct. The characteristic polynomial may not have roots   1  0 0  A= −2   −t 1  = t2 − (−2) = t2 + 2.  det(A − tId) = det −2 −t No eigenvalues! Or, at least, not real ones. We’ll talk about complex numbers in the next lecture. The characteristic polynomial can have repeated roots Here are two examples:    3−t 3 0    det(A. − tId) = det A= 0 0 3  3  B= 0  1  3  3−t  det(B. − tId) = det 0 0   = (3 − t)2 . 3−t 1 3−t   = (3 − t)2 .   0 0   = R2 . Ker(A − 3Id) = Ker 0 0     0 1 1    . Ker(B − 3Id) = Ker = Span 0 0 0  3  A= 0  0  3  3  B= 0  1 . 3 A and B have the same characteristic polynomial: (3 − t)2 . But A has an eigenbasis – any basis for R2 is an eigenbasis – and B doesn’t. We introduce two vocabulary words to describe what we have seen. Let A be a square matrix and λ a scalar. • The geometric multiplicity of λ is the dimension of the λ-eigenspace. In other words, dim Ker (A − λId). • The algebraic multiplicity of λ is the number of times (λ − t) occurs as a factor of det(A − tId). We introduce two vocabulary words to describe what we have seen. Let A be a square matrix and λ a scalar. • The geometric multiplicity of λ is the dimension of the λ-eigenspace. In other words, dim Ker (A − λId). • The algebraic multiplicity of λ is the number of times (λ − t) occurs as a factor of det(A − tId). For example, take B = [ 30 13 ]. Then Ker(B − 3Id) = Ker [ 00 10 ] is one dimensional, so the geometric multiplicity is 1. 3−t 1 But det(B − tId) = det 0 3−t = (3 − t)2 , so the algebraic multiplicity is 2. By contrast, for the matrix A = [ 30 03 ], both the algebraic and geometric multiplicity are 2. Let A be an n × n square matrix with eigenvalues λ1 , λ2 , . . . , λk . Let ai be the algebraic multiplicity of λi . Let Vi be the λi -eigenspace and let gi = dim Vi , the geometric multiplicity of λi . P Theorem 1: We have gi = n if and only if A has an eigenbasis. So Theorem 1 gives us an answer to when we have an eigenbasis. But it requires computing eigenspaces. What can we see from just the characteristic polynomial? Let A be an n × n square matrix with eigenvalues λ1 , λ2 , . . . , λk . Let ai be the algebraic multiplicity of λi . Let Vi be the λi -eigenspace and let gi = dim Vi , the geometric multiplicity of λi . P Theorem 1: We have gi = n if and only if A has an eigenbasis. So Theorem 1 gives us an answer to when we have an eigenbasis. But it requires computing eigenspaces. What can we see from just the characteristic polynomial? Theorem 2: We have gi ≤ ai . P P So gi ≤ ai . But we can’t compute gi from ai : The matrices [ 30 03 ] and [ 30 13 ] both have algebraic multiplicity 2, but one has geometric multiplicity 2 and the other has geometric multiplicity 1. Let A be an n × n square matrix with eigenvalues λ1 , λ2 , . . . , λk . Let ai be the algebraic multiplicity of λi . Let Vi be the λi -eigenspace and let gi = dim Vi , the geometric multiplicity of λi . P Theorem 1: We have gi = n if and only if A has an eigenbasis. So Theorem 1 gives us an answer to when we have an eigenbasis. But it requires computing eigenspaces. What can we see from just the characteristic polynomial? Theorem 2: We have gi ≤ ai . P P So gi ≤ ai . But we can’t compute gi from ai : The matrices [ 30 03 ] and [ 30 13 ] both have algebraic multiplicity 2, but one has geometric multiplicity 2 and the other has geometric multiplicity 1. P P Theorem 3: We have ai ≤ n, and ai = n if and only if det(A − tId) factors completely into linear terms. Let A be an n × n square matrix with eigenvalues λ1 , λ2 , . . . , λk . Let ai be the algebraic multiplicity of λi . Let Vi be the λi -eigenspace and let gi = dim Vi , the geometric multiplicity of λi . P Theorem 1: We have gi = n if and only if A has an eigenbasis. So Theorem 1 gives us an answer to when we have an eigenbasis. But it requires computing eigenspaces. What can we see from just the characteristic polynomial? Theorem 2: We have gi ≤ ai . P P So gi ≤ ai . But we can’t compute gi from ai : The matrices [ 30 03 ] and [ 30 13 ] both have algebraic multiplicity 2, but one has geometric multiplicity 2 and the other has geometric multiplicity 1. P P Theorem 3: We have ai ≤ n, and ai = n if and only if det(A − tId) factors completely into linear terms. So we have an eigenbasis if and only if gi = ai and det(A − tId) factors completely into linear terms. Let A be an n × n square matrix with eigenvalues λ1 , λ2 , . . . , λk . Let ai be the algebraic multiplicity of λi . Let Vi be the λi -eigenspace and let gi = dim Vi , the geometric multiplicity of λi . P Theorem 1: We have gi = n if and only if A has an eigenbasis. Theorem 2: We have gi ≤ ai . P P Theorem 3: We have ai ≤ n, and ai = n if and only if det(A − tId) factors completely into linear terms. . . . to be continued.

Algebraic and Geometric Multiplicity

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