Worksheet 10/14. Math 110, Fall 2015. Some Solutions

Home , Invariant subspace

Worksheet 10/14. Math 110, Fall 2015. Some solutions These problems are intended as supplementary material to the homework exercises and will hopefully give you some more practice with actual examples. In particular, they may be easier/harder than homework. Remember that F ∈ {R, C}. Send me an email if you have any questions! Eigenstuﬀ/Invariant subspaces 1. Consider the operator 0 −1 T : F 2 → F 2 ; v 7→ v. 1 0

a) Find the eigenvalues/eigenvectors for T when F = C.

b) Are there any eigenvectors/eigenvalues when F = R?

v Solution: An eigenvector is a nonzero vector 1 such that T (v) = λv, for some λ. So, we see v2 that v1, v2 must satisfy v1 = λv2, −v2 = λv1. Hence, we combine these equations to obtain 2 2 2 v2 = −λ v2 and v1 = −λ v1. Since at least one of v1, v2 are nonzero, we ﬁnd that λ = −1, and the allowed eigenvalues are λ = ±i. The above equations show that v1 6= 0 ⇔ v2 6= 0, so 1 1 we ﬁnd the eigenvector for λ = i, and the eigenvector for λ = −i. There no real −i i solutions to λ2 = −1 so that there are no real eigenvalues; hence, no eigenvectors.

2. Let (v1, ... , vk ) be a list of eigenvectors for T : V → V . Is U = span(v1, ... , vk ) T -invariant? Prove or provide a counterexample.

Solution: Let v = a1v1 +...+ak vk ∈ span(v1, ... , vk ). Then, since T (vj ) = λj vj (here λ1, ... , λk are the eigenvalues of v1, ... , vk )), then we have

T (v) = a1T (v1) + ... + ak T (vk ) = a1λ1v1 + ... + ak λk vk ∈ span(v1, ... , vk ).

Hence, span(v1, ... , vk ) is T -invariant. 3. True/False: Let V be a vector space, T , S ∈ L(V ).

a) If v ∈ V is an eigenvector for T and S, then it is an eigenvector for T + S.

b) If λ is an eigenvalue for T and S, then it is an eigenvalue for T + S.

c) If v, u are eigenvectors for T then v + u is an eigenvector for T .

Solution:

a) True: if T (v) = av, S(v) = bv, for some a, b, then (T + S)(v) = T (v) + S(v) = av + bv = (a + b)v.

b) False: let S = T = idR2 , both having eigenvalue 1. Then, S + T = 2idR2 . 2 2 c) False: consider the linear map T : R → R ; x 7→ Ax, where A is diagonal with 1, 2 on the diagonal.

1 4. Consider the linear map

2 1 −1 3 3 T : R → R ; v 7→ 1 2 0  v 0 0 −1

Find a 1-dimensional T -invariant subspace U and a 2-dimensional subspace T -invariant subspace W .

Solution: It is straightforward to see (by looking at the matrix deﬁning T ) that U = span(e1, e2) is T -invariant: indeed T (e1) = 2e1 + e2 ∈ U and T (e2) = e1 + 2e2 ∈ U. Since U is T - invariant, we can consider the new operator S ∈ L(U), where S : U → U ; u 7→ T (u) (ie S is the restriction of T to U). So, if we can ﬁnd an S-invariant line L ⊂ U, then L will also be T -invariant: if x ∈ L then T (x) = S(x) ∈ L. Thus, we require L = span(v), where v = ae1 + be2 6= 0, and so that S(v) = λv, for some λ. As S(v) = aS(e1) + bS(e2) = (2a + b)e1 + (a + 2b)e2, we must have

(2a + b) = λa,(a + 2b) = λb, for some λ.

Since we require that v 6= 0 ∈ U, we must have that either a 6= 0 or b 6= 0 - the equations show that a = 0 if and only if b = 0. Hence, we can assume, without loss of generality, that a 6= 0. Hence, using b = (λ − 2)a (from the ﬁrst equation), the second equation gives a + 2(λ − 2)a = λ(λ − 2)a =⇒ a(λ2 − 4λ + 3) = 0 =⇒ (λ2 − 4λ + 3) = 0, since a 6= 0. So, the allowed λ are λ = 1, 3. When λ = 1 we solve the equations above to obtain a = −b, so that v = e1 − e2 is an eigenvector of S (and also of T !) with eigenvalue λ = 1. Thus, take L = span(v). 5. Consider the operator

1 0 1 3 3 T : C → C ; v 7→ 0 −2 0 v. 1 0 1

3 3 a) Find a 1-dim T -invariant subspace U ⊂ C , where U 6= {0}, C . 3 3 b) Find a 2-dim T -invariant subspace V ⊂ C , where V 6= {0}, C , U.

Solution:

a) The second column of the matrix implies that span(e2) is T -invariant.

b) It can be seen that span(e1, e3) is T -invariant.

6. Let T ∈ L(V ) and suppose that U, W ⊂ V are T -invariant subspaces. Is U∩WT -invariant? Prove or give a counterexample. What about U + W ? Prove or give a counterexample. Solution: U ∩ W is T -invariant. Let x ∈ U ∩ W . Then, T (x) ∈ U (since U is T -invariant) and T (x) ∈ W (since W is T -invariant). Hence, T (x) ∈ U ∩ W . Also, U + W is T -invariant: if x = u + w then T (x) = T (u) + T (w) ∈ U + W . 7. Consider the linear operator

0 T : P(R) → P(R); p(x) 7→ (xp(x))

2 Show that T has an inﬁnite number of eigenvalues. Solution: Observe that T (xn) = (n + 1)xn, for every n. Hence, every positive integer is an eigenvalue. 8. Recall that an operator T ∈ L(V ) is diagonalisable if there exists a basis B of V consisting of eigenvectors of T . Assume that F = C. Give examples of the following types of operators or explain why such an operator can’t exist: (look for T ∈ L(F 2), deﬁned by a 2 × 2 matrix A)

i) diagonalisable, invertible,

ii) diagonalisable, not invertible,

iii) not diagonalisable, invertible,

iv) not diagonalisable, not invertible.

Do you think there is a relationship between invertibility and diagonalisability? 9. (Harder) Consider the following operators

0 1 0 0 1 0 0 0 4 4 1 0 0 0 4 4 0 1 0 0 S1 : C → C ; x 7→   x, S3 : C → C ; x 7→   x 0 0 1 0 0 0 0 1 0 0 0 1 0 0 1 0

2 2 i) Verify that S = idC2 and R = idC2 and that S1S3 = S3S1. ii) What are the allowed eigenvalues of S and R? Show that both S and R must admit two distinct eigenvalues 1, −1.

iii) Suppose that S1(w) = −w. Show that S3(w) is also an eigenvector of S1 with eigenvalue

−1. Show that dim null(S1 + idC2 ) = 1. Show that S3(w) = w.

iv) Determine a basis B of null(S1 − idC4 ) (Hint: dim null(S1 − idC3 ) = 3)

v) Explain why null(S1 − idC4 ) is S3-invariant. vi) Consider the restriction operator

0 S3 : null(S1 − idC4 ) → null(S1 − idC4 ); x 7→ S3(x).

0 0 Find the matrix [S3]B of S3 with respect to B. (Hint: it should be 3 × 3)

vii) Find two distinct eigenvalues and three linearly independent eigenvectors (v1, v2, v3) of 0 S3. viii) Show that C = (w, v1, v2, v3) is linearly independent, and determine [S1]C ,[S3]C . What do you notice?