2/28/13 Linear Algebra II Test 1 Solutions 1. Complete the Following Definition: a Vector Space V Is Called the Direct Sum of Su

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2/28/13 Linear Algebra II Test 1 Solutions 1. Complete the following deﬁnition: A vector space V is called the direct sum of subspaces

W1,...,Wk if every vector v in V ...

can be written uniquely as a sum v = w1 + ···+ wk with wj in Wj for each j.

2. Suppose A is a matrix whose characteristic polynomial is ∆(t)=(t − 2)5 and whose minimum polynomial is m(t)=(t − 2)2. List all possible Jordan normal forms for A (up to block-rearrangement).

From the information given, we know that the matrix is 5×5, with only 2s on the diagonal, and that the largest block appearing is 2×2. Thus the options for A are

21000 21000 02000 02000 00200 , 00210 .     00020 00020     00002 00002    

3. (a) Complete the following deﬁnition: A subspace W of a vector space V is invariant under a linear operator T : V → V if . . .

for each w in W , T (w) is again in W .

(b) Show that the kernel of a linear operator is always an invariant subspace.

Suppose v is in the kernel of T . We need to show that T (v) is again in the kernel of T . In other words, we need to show that T (T (v)) = 0. But this is true, because T (v)= 0, so T (T (v)) = T (0)= 0.

2 −4 (c) Determine any invariant subspaces of the matrix thought of as a linear map 5 −2 R2 → R2. We know that {0} and R2 are automatically invariant subspaces. Thus we only need to check for 1--dimensional invariant subspaces. But these are exactly eigenspaces, so we check for eigenvalues/vectors:

2 − λ −4 det(A − λI)= det = λ2 + 16. 5 −2 − λ

This has no real roots, so there are no real eigenvalues, so no real eigenvectors, so no 1--dimensional invariant subspaces. Thus the only invariant subspaces are the trivial ones: {0} and R2.

(d) Explain how your answer to part (c) above be diﬀerent if we thought of the matrix as a linear map C2 → C2? (No computation necessary.)

The characteristic polynomial λ2+16 factors over C with two distinct roots. Thus there would be two 1--dimensional invariant subspaces as well as the two trivial ones.

21 4 2 4. Consider B = 0 2 −1 which has characteristic polynomial ∆(t)=(t − 2) (t − 3). 00 3   (a) Find the minimum polynomial of B.

The characteristic polynomial has roots 2 and 3, each of which must also be roots of the minimum polynomial. We know the matrix satisfies the characteristic polynomial, so the only thing we have to check is whether it solves the more ‘minimal’ polynomial (t−2)(t−3). For this we compute:

01 4 −1 1 4 0 −1 ∗ (B − 2I)(B − 3I)= 0 0 −1  0 −1 −1 = ∗ ∗ ∗ . 00 1 0 0 0 ∗ ∗ ∗       This is not the zero matrix. It follows that the minimum polynomial is ∆(t)= m(t)=(t − 2)2(t − 3).

(b) What is the Jordan normal form for B? There must be a 2×2 block with 2s on the diagonal and a 1×1 block with a 3 on the diagonal. Since the matrix is 3×3, this is all there is: 2 1 0 0 2 0 . 0 0 3  

Since the eigenvalue 2 has multiplicity 2, we compute

00 3 2 (A − 2I) = 0 0 −1 , 00 1   the kernel of which has basis 1 0 0 , 1 .  0 0        1 1 5. Consider the basis , for R2. Find the dual basis for (R2)∗. 2 −1

The dual basis is {ϕ1,ϕ2}, where

1 1 ϕ1 =1, ϕ1 =0 2 −1 and 1 1 ϕ2 =0, ϕ2 =1. 2 −1 On the other hand, linearity ensures that

x x ϕ1 = ax + by, ϕ2 = cx + dy y y

for some a,b,c,d. Setting corresponding outputs equal, we find that 1 a = 1/3, b = 1/3, c = 2/3, and d = −1/3. Thus we have ϕ1(x, y)= 3 x + 1 2 1 3 y and ϕ2(x, y)= 3 x − 3 y. If we express these maps as row matrices, we have {ϕ1,ϕ2} = { 1/2 1/3 , 2/3 −1/3 }. 6. (a) Complete the following deﬁnition. Suppose T : V → W is a linear map. The transpose of T is the map T t : W ∗ → V ∗ deﬁned as follows: for any ϕ : W → K in W ∗ we have . . .

(T t(ϕ))(v)= ϕ(T (v))

(b) Find T t(ϕ)(x, y, z), where ϕ : R2 → R is deﬁned by ϕ(x, y)=3x − 2y and T : R3 → R2 is deﬁned by T (x, y, z)=(x + y,y + z).

Using the formula above, we see that

T t(ϕ)(x, y, z)= ϕ(T (x, y, z)) = ϕ(x+y,y +z)=3(x+y)−2(y +z)=3x+y −2z.

(c) Note that the standard matrix representation for ϕ is 3 −2 while that for T is 1 1 0 . Use the matrices, as well as the fact that the standard matrix representation 0 1 1 for T t is the transpose of that for T , to verify your answer to part (b) above.

We compute T t(ϕ), where ϕ is thought of as a vector in the dual, and 3 hence is represented by the column . We find that the column vector −2 representing T t(ϕ) in the dual space is

1 0 3 3 1 1 =  1  . −2 0 1 −2     This corresponds to the map sending (x, y, z) to 3x+y−2z, as we found in part (b).