Math 217: Eigen Everything Professor Karen E. Smith 1. Eigenvectors And

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1. Eigenvectors and Eigenvalues. Eigenvectors and eigenvalues are objects associated to a ﬁxed linear transformation V →T V. Eigenvectors and eigenvectors always go together: each eigenvector of T has some associated eigenvalue, and each eigenvalue has associated eigenvector(s).

Deﬁnition 1.0.1. An eigenvector of a linear transformation V −→T V is any non-zero vector ~v ∈ V such that T (~v) = λ~v for some scalar λ. The scalar λ is called the eigenvalue of the eigenvector ~v.

Perhaps this is obvious, but it bears repeating: an eigenvector is a vector (in V ) and an eigenvalue is a scalar (in R).

∞ d λx λx Example 1.0.2. Consider the diﬀerentiation map of C . Since dx e = λe for any real number λ, we see that every real number is an eigenvalue of the diﬀerentiation map, with λx corresponding eigenvectors fλ(x) = e . Caution: Not every linear transformation has an eigenvalues!

ρ Example 1.0.3. Consider the map R2 −→ R2 given by rotation counterclockwise through π/2. Since no non-zero vector is taken to a scalar multiple of itself, ρ has no eigenvectors.

Deﬁnition 1.0.4. Let V −→T V be a linear transformation. An eigenbasis is a basis for V consisting of eigenvectors for T .

Example 1.0.5. Consider the map 2 p 2 R −→ R given by projection onto a subspace L. Since every point of L is taken to itself, we have p(~v) = ~v for all ~v ∈ L. That is, every non-zero vector in L is an eigenvector with eigenvalue 1. Since every vector w ∈ L⊥ is send to 0 under p, we have p(~w) = 0~w for all ~w ∈ L⊥. Thus every non-zero vector in L⊥ is an eigenvector with eigenvalue 0. If we let ~v be a basis for L and ~w be a basis for L⊥, then (~v, ~w) is an eigenbasis for p. Caution: Not every linear transformation has an eigenbasis! Example 1.0.3 above is one that does not! Eigenvectors are important geometrically because they help us understand the map better— if we know ~v is an eigenvector with eigenvalue 2, we know that T is scaling vectors by 2 in the direction of ~v. 1 An Eigenbasis for a transformation T is important algebraically because its give us a coordinate system that is especially helpful for dealing with T . For example in Example 1.0.5, the matrix of p in the eigenbasis B = (~v, ~w) is 1 0 [p] = [p(~v)] [p(~w)] = . B B B 0 0 Imagine how much easier it is to work with this B-matrix instead of the standard or some other matrix of p!

Proposition 1.0.6. Let V −→T V be a linear transformation of a ﬁnite dimensional vector space. Then B is an eigenbasis for T if and only if the matrix [T ]B is diagonal. In this case, the elements on the diagonal are the eigenvalues (possibly repeated more than once).

Deﬁnition 1.0.7. A linear transformation V −→T V is diagonalizable if V admits an eigenbasis for T . Equivalently,1 V −→T V is diagonalizable means that there is some basis for V in which the matrix of T is diagonal. Any such basis will be an eigenbasis for T . In terms of matrices, we have

Deﬁnition 1.0.8. An n × n matrix is diagonalizable if it is similar to a diagonal ma- TA trix. This is equivalent to saying that the linear transformation Rn −→ Rn (given by left multiplication by A) is diagonalizable.

Proposition 1.0.9. Let A be an n × n matrix. Then A is similar to a diagonal matrix D if and only if the linear transformation given by left multiplication by A has an eigenbasis. Moreover, if {~v1, . . . ,~vn} is the eigenbasis, with corresponding eigenvalues λ1, . . . , λn, then

λ1 0 ... 0  0 λ ... 0 −1  2  S AS =  . .  ,  ......  0 0 . . . λn where S is the matrix ~v1 . . . ~vn whose columns are the vectors of the eigenbasis. T 2 3 Example 1.0.10. Consider the map 2 −→ 2 given by left multiplication by . R R 0 −1 1 Note that T (~e ) = 2~e . Also, for ~v = , we have T (~v) = −~v. So 1 1 −1 1 1 B = ( , ) 0 −1 is an eigenbasis and the matrix of T in this eigenbasis is 2 0 [T ] = . B 0 −1 Note that the elements on the diagonal are exactly the eigenvalues {2, −1}.

1 assuming V is ﬁnite dimensional so we can model it on Rn 2 2. Eigenspaces

Proposition 2.0.1. Let λ be an eigenvalue of a linear transformation V −→T V . The set

Vλ = {~v ∈ V | T (~v) = λ~v} ⊂ V consisting of all λ-eigenvectors (together with ~0) is a subspace of V. Proof. You should prove this, using the standard technique for checking a subset of V is a subspace.

Deﬁnition 2.0.2. Let V −→T V be a linear transformation with eigenvalue λ. The λ- eigenspace is the subspace Vλ = {~v ∈ V | T (~v) = λ~v}. Its dimension is called the geometric multiplicity of the eigenvalue λ of T . The kernel of T is the 0-eigenspace of T . The geometric multiplicity of 0 is the nullity of T .

T Example 2.0.3. The map R2×2 −→ R2×2 sending A to A + AT is a linear map. Note that if A is symmetric, then T (A) = 2A, so the set of symmetric matrices is contained in the 2-eigenspace. Conversely, if T (A) = 2A, then A + AT = 2A, so A = AT and A is symmetric. Thus the 2-eigenspace of T is precisely the subspace of symmetric matrices. This space is dimension 3, with basis (E11,E22,E12 + E21). Are there any other eigenvalues and vectors? 0 1 Well, the kernel is the zero-eigenspace. The matrix can be easily seen to span the −1 0 kernel, hence it is a basis for the zero-eigenspace. So B = (E11,E22,E12 + E21,E12 − E21) 2 0 0 0 0 2 0 0 is an eigenbasis for T . The corresponding diagonal B-matrix is [T ]B =   . This 0 0 2 0 0 0 0 0 transformation has two eigenspaces. The 2-eigenspace is the space of symmetric matrices. The 0-vector space is the subspace of skew-symmetric matrices. Note that the geometric multiplicity of the eigenvalue 2 is 3 and geometric multiplicity of the eigenvalue 0 is 1.

Remark 2.0.4. We often abuse terminology, referring to "eigenvalues and eigenvectors of a matrix"—this means the eigenvalues and eigenvectors of the corresponding map Rn −→ Rn given by left multiplication by A. Accordingly, the eigenvalues of A are the scalars λ for which there exists a non-zero column vector ~v such that A~v = λ~v. The set of all such column vectors ~v is the λ-eigenspace of A.

3 3. Finding Eigenvalues and Eigenvectors. Deﬁnition 3.0.1. Consider an n × n matrix A. The characteristic polynomial of A is the degree n polynomial χA(x) = det(xIn − A). 0 −1 Example 3.0.2. Let A = . Then the characteristic polynomial of A is the deter- 1 0 x 1 minant of the matrix . Thus χ (x) = x2 + 1. −1 x A Lemma 3.0.3. Similar matrices have the same characteristic polynomial. That is, if A and B are n × n matrices for which there exists an n × n matrix S with B = S−1AS,

then χA(x) = χB(x).

Lemma 3.0.3 ensures that the following deﬁnition makes sense:

Deﬁnition 3.0.4. Let V −→T V be a linear transformation on a vector space V of ﬁnite dimension n. The characteristic polynomial of T is the degree n polynomial

χT (x) = det(xIn − A) where A is the matrix of T in any basis for V .

Theorem 3.0.5. Let V −→T V be a linear transformation on a ﬁnite dimensional vector space. The eigenvalues of T are precisely the roots of the characteristic polynomial of T . Proof. Fix a scalar c. Consider the linear transformation2 V −→Φ V ~v 7→ T (~v) − c~v. Note that c is an eigenvalue if and only if this transformation Φ has a non-zero kernel. By rank-nullity, Φ has a non-zero kernel if and only if it is not invertible, which happens if and only if the determinant of Φ is zero. We can compute the determinant of Φ by computing the determinant of its B for any basis. Fix any basis B for V . We compute the B-basis for Φ by thinking of Φ as T − cIV , where IV is the identity map on V . Let A be the B-matrix of T . Note that the B-matrix of cIV is cIn (where n = dim V ). So

[Φ]B = [T − cIV ]B = [T ]B − [cIV ]B = A − cIn.

Its determinant is precisely the result of plugging c into the polynomial det(A − xIn). Of course, det(A − xIn) is n n det(−1)(xIn − A) = (−1) det(xIn − A) = (−1) χT (x). This means that c is an eigenvalue if and only if c is a root of the characteristic polynomial.

2As a good reader of mathematics, you realize you should check yourself that the given map is really a linear transformation 4 Corollary 3.0.6. A linear transformation of an n-dimensional space has at most n eigenvalues.

Proof. The characteristic polynomial has degree n, so it can have at most n roots.

Example 3.0.7. Suppose that V −→T V is diagonalizable. This means there exists a basis B (an eigenbasis) such that

a11 0 0 ... 0   0 a22 0 ... 0  [T ]B =  . .  .  0 . ... . 0  0 0 ... 0 ann

The diagonal elements aii are the eigenvalues of T . Computing the characteristic polynomial, we see that

χT (x) = (x − a11)(x − a22) ··· (x − ann),

where the aii are the eigenvalues of T (possibly repeated multiple times).

Remark 3.0.8. The characteristic polynomial may or may not have any real roots at all! Remember, however, that over the complex numbers, every polynomial factors completely into linear factors

a1 a2 at (x − λ1) (x − λ2) ··· (x − λt)

where the roots λi ∈ C. Mathematicians (and anyone who uses linear algebra) will call the roots of the characteristic polynomial eigenvalues even when they are complex.3 However, in Math 217, we will be concerned only with real eigenvalues.

The characteristic polynomial gives us a good way to find the eigenvalues of T (at least when V is finite dimensional). But how do we find the associated eigenvectors?

Proposition 3.0.9. Let V −→T V be a linear transformation, and suppose that c is one its eigenvalues. Then the c-eigenspace of T is the kernel of the transformation

V T−→−cIV V ~v 7→ T (~v) − c~v.

[Here IV denotes the identity transformation of V .] Proof. Suppose ~v is an eigenvector of T with eigenvalue c. Then T (~v) = c~v, so T (~v) − c~v = T (~v) − cIV (~v) = (T − cIV )(~v) = 0. So ~v is in the kernel of T − cIV as claimed. For the converse: Suppose ~v is in the kernel of T − cIV . This means (T − cIV )(~v) = 0, so T (v) − cIV (~v) = 0, so T (~v) = c~v.

3If λ is an eigenvalue of a matrix A, then it turns out that there is a column vector with complex entries ~v such that A~v = λ~v. 5 TA Example 3.0.10. Find an eigenbasis for the transformation R2 −→ R2 given by multipli- 5 −7 cation by A = . Find a diagonal matrix similar to A. −6 6 Step 1: find the eigenvalues. The characteristic polynomial is x − 5 7 det[xI − A] = det = (x − 5)(x − 6) − 42 = x2 − 11x − 12 = (x + 1)(x − 12). 2 6 x − 6 This tells us the eigenvalues are 12 and -1. We will find their eigenspaces one at a time. Step 2: For each eigenvalue, compute the eigenspace using Proposition 3.0.9. Make note of its dimension. 6 −7 Here, we first find 12-eigenspace. This is the kernel of A − 12I = . This matrix 2 −6 7 7 has kernel spanned by , so this is a basis for the 12-eigenspace (and in particular, a 6 eigenvector with eigenvalue 12 itself). −7 −7 First next find the −1-eigenspace. This is the kernel of A + I = , which is 2 −6 −6 1 clearly spanned by . −1 Step 3: Find bases for each eigenspace. See if putting them together gives you a basis 1 7 for V . Here, the vectors and span the two-dimensional space 2 and both are −1 6 R 1 7 eigenvectors. So ( , ) is an eigenbasis for T . −1 6 A

Since TA has an eigenbasis B, we know the B-matrix of TA will be diagonal. It is easy to ﬁnd: −1 0 [T ] = . B 0 12 On the other hand, we also know that

[T ]B = SE→B[T ]E SB→E . 1 7 But we compute [T ] = A and S = . So we have E B→E −1 6 −1 0 = S−1AS 0 12 1 7 for S = . −1 6

4. Multiplicities of Eigenvalues

Deﬁnition 4.0.1. Let V −→T V be a linear transformation of a ﬁnite dimensional vector space V . The algebraic multiplicity of an eigenvalue λ is the largest power r such that (x − λ)r divides the characteristic polynomial.

Deﬁnition 4.0.2. The geometric multiplicity of λ is the dimension of the λ-eigenspace, or equivalently, the maximal number of linearly independent eigenvectors with eigenvalue λ. 6 Theorem 4.0.3. Let V −→T V be a linear transformation of a ﬁnite dimensional vector space. Then for each eigenvalue λ, geometric multiplicity of λ ≤ algebraic multiplicity of λ.

Proof. Suppose that m is the geometric multiplicity of λ. Let v1, . . . , vm be a basis for Vλ, the λ-Eigenspace of T . We can extend v1, . . . , vm to a full basis B = (v1, . . . , vm, vm+1, . . . , vn for V . The B-matrix of T has the form   λ 0 0 . . . a1 m+1 a1 m+2 . . . a1 n 0 λ 0 . . . a2 m+1 a2 m+2 . . . a2 n   . .   . .   . . a2 m+1 a2 m+2 . . . a2 n    [T ]B = 0 0 . . . λ am m+1 am m+2 . . . am n  0 0 ... 0 a a . . . a   m+1 m+1 m+1 m+2 m+1 n  ......   ......  0 0 ... 0 an m+1 an m+2 . . . an n

Compute the characteristic polynomial det(xIn − A) by Laplace expansion along the ﬁrst columns. We see that this will produce a polynomial divisible by (x − λ)m. So the algebraic multiplicity of λ is at least m.

Useful Fact 4.0.4. Let V −→T V be a linear transformation, where V has dimension n. Suppose λ is an eigenvalue. A useful trick for ﬁnding the λ-eigenspace is to compute the kernel of the matrix [T ]B − λIn, where B is any basis. Do you see how this helps? The elements of the kernel will the B-coordinates of the λ-eigenvectors.

Example 4.0.5. Consider the three diﬀerent linear transformations of R3 → R3 given by the matrices π 0 0 π 1 0 π 1 0 A1 = 0 π 0 ,A2 = 0 π 0 ,A3 = 0 π 1 . 0 0 π 0 0 π 0 0 π In all three cases, the characteristic polynomial is (x − π)3. So π is the only eigenvalue, and its algebraic multiplicity is 3. The geometric multiplicities of π can be computed by examining the kernels of the matrices (Ai − πI3). These are, respectively, 0 0 0 0 1 0 0 1 0 A1 = 0 0 0 ,A2 = 0 0 0 ,A3 = 0 0 1 . 0 0 0 0 0 0 0 0 0

We conclude that the geometric multiplicity of π is three for A1, two for A2 and one for A3 (remember rank-nullity!). These multiplicities represent the maximal number of linearly independent eigenvectors (of the eigenvalue π, but again, π is the only eigenvalue). Thus only A1 is diagonalizable because it is the only matrix for which there exists an eigenbasis.

7 Proposition 4.0.6. Let V be a vector space of dimension n. Let V −→T V be a linear transformation. The following are equivalent: (1) T is diagonalizable; (2) T has an eigenbasis; (3) The sum of the geometric multiplicities of the (real) eigenvalues of T is n. (4) All eigenvalues are real and the geometric multiplicity of each equals its algebraic multiplicity.

Corollary 4.0.7. Let V −→T V be a linear transformation of an n-dimensional vector space. If T has n distinct eigenvalues, then T is is diagonalizable.

Example 4.0.8. The converse to Corollary 4.0.7 is false: consider the map of R2 that scales every vector by 5.