Math 21b Eigenvectors and Eigenbases
Eigenvectors and Eigenbases. Let A be an n × n matrix. Recall that a nonzero vector ~v is an eigenvector for A with eigenvalue λ if A~v = λ~v.
An eigenbasis is a basis of Rn consisting of eigenvectors of A.
Eigenvectors and Linear Independence. Eigenvectors with different eigenvalues are automatically linearly independent.
If an n × n matrix A has n distinct eigenvalues then it has an eigenbasis. Eigenspaces. If A is an n × n matrix and λ is a scalar, the λ-eigenspace of A (usually denoted Eλ) is the set of all vectors ~v in Rn so that A~v = λ~v.
The non-zero vectors in Eλ are exactly the eigenvectors of A with eigenvalue λ.
1 1 Example. Let A = , which represents a shear. Find all eigenvalues and eigenvectors of A. Is there 0 1 an eigenbasis for A?
Solution. To find the eigenvalue we compute the characteristic polynomial fA(λ) = det(A − λI)
1 − λ 1 f (λ) = det = (1 − λ)2. A 0 1 − λ
The eigenvalues of A are then the roots of fA(λ), so we get one eigenvalue, namely λ = 1.
To find the eigenvectors we need to find the eigenspace Eλ for each eigenvalue λ. Here we only have one eigenspace E1 to compute. To compute E1 we need to solve the system A~x = ~x, which is equivalent to (A − I)~x = ~0. That is, we just need to compute ker(A − I). 0 1 The coefficient matrix A − I = is already in reduced row echelon form, so we get one free 0 0 variable x . 1 x1 x1 1 = = x1 x2 0 0 1 so E = ker(A − I) = span . 1 0 1 So, the eigenvectors for A with eigenvalue 1 are all of the form c where c 6= 0 (eigenvectors 0 are the nonzero vectors in the eigenspaces).
We don’t have an eigenbasis in this case, since the only eigenspace is 1-dimensional. Computing Eigenspaces. Let A be an n × n matrix and λ a scalar. Then the λ-eigenspace is
Eλ = ker(A − λI).
In particular, the λ-eigenspace is a linear subspace of Rn.
Example. Let A be projection onto a plane V through the origin in R3. 1. What are the eigenspaces for A?
Solution. Any vector in the plane V is an eigenvector with eigenvalue 1, so E1 = V . Any ⊥ vector perpendicular to V is an eigenvector with eigenvalue 0, so E0 = V .
2. Is it possible to find an eigenbasis for A?
Solution. Yes. We can find such a basis by taking a basis ~v1,~v2 for V = E1 and a basis ~v3 ⊥ 3 for V = E0 and combining them into a basis for R .
3. Generalize your conclusion from part (b).
Solution. Suppose A is an n × n matrix with eigenvalues λ1, . . . , λk. We can try to find an eigenbasis for Rn consisting of eigenvectors of a linear transformation A by taking a basis for
each eigenspace Eλi and putting them all together. Each eigenspace Eλi will give us dim Eλi basis vectors. For this to be a basis for Rn we need to get a total of n vectors, so we need
dim Eλ1 + dim Eλ2 + ··· + dim Eλk = n. Geometric Multiplicity. Let A be an n × n matrix and λ an eigenvalue of A. The geometric multiplicity of λ is the dimension of the eigenspace Eλ.
Geometric Multiplicity and Eigenbases. An n × n matrix A has an eigenbasis if and only if the sum of the geometric multiplicities of A is n.
Algebraic Multiplicity. Let A be an n × n matrix and λ an eigenvalue of A. The algebraic multiplicity of λ is the number of (t − λ) factors in the characteristic polynomial fA(t).
1 1 0 Example. Find the algebraic and geometric multiplicities for the eigenvalues of A = 0 0 1. 0 0 1
Solution. The matrix A has characteristic polynomial
1 − λ 1 0 2 fA(λ) = det 0 −λ 1 = −λ(1 − λ) 0 0 1 − λ
so it has two eigenvalues, λ = 0 with algebraic multiplicity 1, and λ = 1 with algebraic multiplic- ity 2.
To compute geometric multiplicities we need to find the dimension of the corresponding eigenspaces. 1 1 0 1 1 0 For λ = 0: A − λI = 0 0 1 row reduces to 0 0 1 which has one free variable. So the 0 0 1 0 0 0 geometric multiplicity of λ = 0 is 1.
0 1 0 0 1 0 For λ = 1: A − λI = 0 −1 1 row reduces to 0 0 1 which has only one free variable. 0 0 0 0 0 0 So the geometric multiplicity of λ = 1 is also 1.
To summarize:
eigenvalue geometric mult. algebraic mult. λ = 0 1 1 λ = 1 2 1 Geometric vs. Algebraic Multiplicities. Let λ be an eigenvalue of a matrix A. The geometric multiplicity of λ is at most the algebraic multiplicity of λ.
Algebraic Multiplicity and Eigenbases. Let A be an n × n matrix. If the sum of the algebraic multiplicities of A is n then there is not an eigenbasis for A.
Similar Matrices. Recall that matrices A and B are similar if there is an invertible matrix S so that B = S−1AS. If A and B are similar, then
1. The characteristic polynomials of A and B are the same. That is, fA(λ) = fB(λ). 2. A and B have the same eigenvalues. 3. The eigenvalues of A and B have the same algebraic multiplicities.
4. The eigenvalues of A and B have the same geometric multiplicities.