Eigenvalues and Eigenvectors

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Math 217: Eigenvectors and Eigenvalues

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

We say λ is an eigenvalue of T if there is some non-zero vector in V which is an eigenvector of T with eigenvalue λ.

A. Warm-Up:

1. Find all eigenvectors of the identity map V → V . What are the associated eigenvalues?

2. Find all eigenvectors of the zero map V → V . What are the associated eigenvalues?

2 2 3. Let T : R → R be the map ”scale by 2”. What is the (standard) matrix of T ? What are the eigenvalues of T ? What are the associated eigenvalues?

Solution note: Every non-zero vector is an eigenvector of the identity map, with eigenvalue 1. Every non-zero vector is an eigenvector of the zero map, with eigenvalue 0. Every non-zero vector is an eigenvector of the “scale by k” map, with eigenvalue k.

3 0 B. Let T : 2 → 2 be multiplication by . Explain why any non-zero vector in the span of ~e R R 0 4 1 is an eigenvector. What are the corresponding eigenvalues? Find all eigenvectors with eigenvalue 4. Are there any eigenvectors with eigenvalue zero. Can k 6= 3, 4 be an eigenvalue?

Solution note: Note that T (k~e1) = 3k~e1, so k~e1 is an eigenvector with eigenvalue 3. a a a The eigenvectors with eigenvalue 4 are the vectors such that T ( ) = 4 . To b b b 3a a 4a ﬁnd them, we solve = 4 = . We must have a = 0, but bcan be arbitrary. 4b b 4b

So the eigenvectors with eigenvalues 4 are exactly the span of ~e2 (except for ~0).

C. Consider the diﬀerentiation map d : C∞ → C∞. Why is this linear? Show that the function f(x) = ekx is an eigenvector of d. Find all eigenvectors of the eigenvalue zero. What are all the possible eigenvalues of diﬀerentiation on C∞? d kx kx Solution note: Check: d(f) = dx e = ke = kf, so f is an eigenvector with eigenvalue k. This shows that every real number is an eigenvalue for d.

2 2 D. Explain why the transformation T : R → R which rotates counterclockwise by π/6 has no 3 eigenvectors. Explain why rotation around some axis (through the origin in R ) always has 1 as an eigenvalue. Under what circumstances does it have some other eigenvalue? What is this eigenvalue? 2 Solution note: This rotation in R moves every vector in such a way that no vector is taken to a scaling of itself. So there are no eigenvectors (nor eigenvalues). The 3 rotation in R ﬁxes every vector on the axis of rotation. So each (non-zero) vector in this axis is an eigenvector with eigenvalue 1. This is the only eigenvalue unless we are rotating by π, in which case every vector perpendicular to the axis of rotation is taken to its negative. So if we rotate by π, then −1 is also an eigenvalue.

1 E. Explain why the transformation T : 2 → 2 which projects onto the line spanned by R R 5 1 2 −π −5 −1 has eigenvectors , , , and . What are the eigenvalues of each of these 5 10 −5π 1 1/5 eigenvectors? Are there any other eigenvectors? Find all eigenvector with eigenvalue 1. Solution note: The vectors on the one-dimensional subspace L (a line through the origin) are taken to themselves when we project to L. So all the vectors listed are 1 eigenvectors for the eigenvalue 1, since they are all on the line spanned by . 5 Everything in L⊥ is sent to zero, so zero is also an eigenvalue, and the (non-zero) vectors in L⊥ are eigenvectors.

F. Prove or Disprove:

1. Any non-zero element in the kernel of V −→T V is an eigenvector of T .

2. The linear transformation V −→T V is injective if and only if zero is not an eigenvalue of T .

3. If ~v is an an eigenvector of T , then ~v is an eigenvector of T 2 as well.

Solution note: 1) True: T (~v) = 0 = 0~v so the 0-eigenvectors are exactly the (non- zero) elements in the kernel. 2). True: This basically restates 1, remembering that T is injective if and only if ker T is zero. 3). True: If T (~v) = λ~v, then T 2(~v) = T (T (~v)) = T (λ~v) = λT (~v) = λ2~v. So ~v is an eigenvector for T also (though the corresponding eigenvalue could be diﬀerent.)

Deﬁnition. An eigenbasis for a linear transformation T is a basis for V consisting of eigenvectors of T .

G. Prove or disprove: every linear transformation has an eigenbasis. 2 Solution note: FALSE! Rotation by π/6 in R has no eigenvectors! So it can’t have an eigenbasis.

1 H. Suppose that 2 −→T 2 is a linear transformation with 3-eigenvector and 2-eigenvector R R 1 1 1 1 . Does T have an eigenbasis? Find the matrix of T in the basis B = { , }. −1 1 −1 2 Draw the square in the source R determined by the elements of B (meaning whose vertices are 0,~v1,~v2,~v1 + ~v2). Now draw its image under T . Describe T geometrically in words. Solution note: You should see a square in the source, with vertices [0, 0]T , [1, 1]T , [1, −1]T and [2, 0]T . You should see a rectangle in the target, also with vertex [0, 0]T but whose sides are parallel to the original squares, but with lengths scaled by 2 and 3 from the original square.

I. Suppose B is an eigenbasis for T , a linear transformation of P3, the vector space of polynomials of degree at most 3. Suppose that the corresponding eigenvalues are a, b, c, d. Find the B-matrix for T . a 0 0 0 0 b 0 0 Solution note: [T ]B =  . 0 0 c 0 0 0 0 d

J. Deﬁne an orthogonal transformation V → V on an inner product space. Now explain why the only eigenvalues of an orthogonal transformation are ±1. Solution note: An orthogonal transformation satisﬁes ||T (~v)|| = ||~v|| for all ~v. Sup- pose ~v has eigenvalue λ. So T (~v) = λ~v. Taking magnitudes, we have ||T (~v)|| = ||λ~v|| = |λ|||~v||. So in order for T to be orthogonal, we must have |lambda| = 1. K. Let A be the matrix 5 −6 A = . 4 −5 In this problem we will compute A1000.

2 2 1. Let L : R → R be the linear transformation given by left multiplication by A. Find the eigenvalues of L.

2. Find the corresponding eigenvectors. You have to make some choice since you can always scale eigenvectors. Call these ~v1 and ~v2.

3. Compute the matrix B = [L]B of L in the basis B = {~v1,~v2}. 4. Find an invertible matrix P such that P −1AP = B.

5. Now compute A1000. Solution note: The char poly is x2 − 1, so the eigenvalues are ±1. To ﬁnd the x x 3 1 eigenvectors, we solve A = . We get eigenvectors and . [You can y y 2 1 also have scalar multiples of these, or have them in the other order]. The matrix 1 0 of L in this basis is . [Note: this is not the only correct answer–you could 0 −1 have the roles of −1 and 1 reversed.] The change of basis matrix from the eigenbasis 3 1 to the standard is P = [or whatever your eigenvectors are....there is more 2 1 1 −1 than one correct answer here.] Its inverse is . Since B = P −1AP , we have −2 3 3 1 11000 0 1 −1 A1000 = PB1000P −1. So A1000 = = PP −1 = I . 2 1 0 (−1)1000 −2 3 2

B. Let T : V → V be a linear transformation. Prove or disprove each statement:

1. If ~v is a eigenvector of T , then ~v is also an eigenvector of T 2.

2. If T has no real eigenvalues, then also T 2 has no real eigenvalues.

3. If λ is an eigenvalue of some linear transformation T : V → V , then λn is a eigenvalue of T n : V → V .

4. Then T is not injective if and only if 0 is an eigenvalue.

Solution note: 1. True. Suppose T (~v) = λ~v. Then applying T to both sides we have T 2(~v) = T (λ~v) = λT (~v) = λ2~v. So ~v is also an eigenvector of T (with eigenvalue λ2). 2. False. Let T be rotation through π/2. There are no real eigenvalues because no vector is taken to a parallel vector by T . However T 2 is rotation by π. This takes 2 ~e1 to −~e1, so −1 is a real eigenvalue of T . 3. True. Proof by induction. We did the case n = 2 in (1). Inductive hypothesis: T n−1(~v) = λn−1~v. Apply T to both sides to get T n(~v) = λn−1T (~v) = λn~v. 4. True. Say T is not injective. Then kernel T is not zero. If ~v ∈ kerT , then ~v is an eigenvector with eigenvalue 0, so 0 is an eigenvalue. Conversely, say 0 is an eigenvalue. Then there is some non-zero vector ~v such that T (~v) = 0~v = 0. This means ~v is in the kernel, so T is not injective.