Homework 6 Solutions

1. Let V be the real vector space spanned by the functions et, tet, t2et, e2t. Find a Jordan canonical basisand a Jordan canonical form of T on V de�ned by T (f) = f ′.

Solution. Note {et, tet, t2et, e2t} is linearly independent. If β = {et, tet, t2et, e2t}, then

[T ]β = A =

1 1 0 00 1 2 00 0 1 00 0 0 2

Thus, pT (x) = (x− 1)3(x− 2). We look at N((A− I)3). We have

A− I =

0 1 0 00 0 2 00 0 0 00 0 0 1

which has rank 3, so V1 = N((A− I)) has dimension d1 = 1. From this we must have d2 = 2 and d3 = 3 (asthere is only one Jordan block). We see

(A− I)2 =

0 0 2 00 0 0 00 0 0 00 0 0 1

and

(A− I)3 =

0 0 0 00 0 0 00 0 0 00 0 0 1

Hence, N((A − I)2) = {(a, b, c, d)|c = d = 0} and N((A − I)3) = {(a, b, c, d)|d = 0}, so {(0, 0, 1, 0), (A −I)(0, 0, 1, 0), (A − I)2(0, 0, 1, 0)} works as a basis for N((A − I)). This is {(0, 0, 1, 0), (0, 2, 0, 0), (2, 0, 0, 0)},or {t2et, 2tet, 2et}. For N((A − 2I)), we see e2t works, so our Jordan canonical basis is {t2et, 2tet, 2et, e2t}with corresponding matrix

1 0 0 01 1 0 00 1 1 00 0 0 2

2. Find a Jordan canonical basis and a Jordan canonical form of the operator T on P3(R) given byT (f(x)) = xf ′′(x).

Solution. In the basis β = {1, x, x2, x3}, we have

A = [T ]β =

0 0 0 00 0 2 00 0 0 60 0 0 0

Thus, pT (x) = x4. We see

A2 =

0 0 0 00 0 0 120 0 0 00 0 0 0

1

and A3 = 0, so we have d1 = dimN(A) = 2, d2 = 3, and d4 = 4. We see N(A) = {(a, b, c, d)|c = d = 0},N(A2) = {(a, b, c, d)|d = 0}. Hence, we may choose (0, 0, 0, 1) to start our �rst chain. We get A(0, 0, 0, 1) =(0, 0, 6, 0) and A2(0, 0, 0, 1) = (0, 12, 0, 0). Now, in N(A) we already have (0, 12, 0, 0). Hence, to complete abasis ofN(A) we take (1, 0, 0, 0). Thus, our Jordan canonical basis is {(0, 0, 0, 1), (0, 0, 6, 0), (0, 12, 0, 0), (1, 0, 0, 0)}which is {x3, 6x2, 12x, 1} with corresponding matrix

0 0 0 01 0 0 00 1 0 00 0 0 0

3. Find a Jordan canonical basis and a Jordan canonical form of the operator T on P3(R) given byT (f) = f ′′ + 2f .

Solution. In the basis β = {1, x, x2, x3}, we have

A = [T ]β =

2 0 2 00 2 0 60 0 2 00 0 0 2

so pT (x) = (x− 2)4. We have

(A− 2I) =

0 0 2 00 0 0 60 0 0 00 0 0 0

and (A − 2I)2 = 0. We see N((A − 2I)) = {(a, b, c, d)|c = d = 0} and so a basis of N((A − 2I)2) relativeto N((A− 2I)) is given by {(0, 0, 1, 0), (0, 0, 0, 1)}. Applying A− 2, we get the chains {(0, 0, 1, 0), (2, 0, 0, 0)}and {(0, 0, 0, 1), (0, 6, 0, 0)}. Hence, our Jordan basis is {(0, 0, 1, 0), (2, 0, 0, 0), (0, 0, 0, 1), (0, 6, 0, 0)} which is

{x2, 2, x3, 6x} and which has corresponding matrix

2 0 0 01 2 0 00 0 2 00 0 1 2

.4. Find a Jordan canonical basis and a Jordan canonical form of the operator T on M2×2(R) given by

T (A) =

[3 10 3

]A−At.

Solution. We'll use B as the matrix of T in the basis β = {e1, e2, e3, e4} where e1 =

[1 00 0

], e2 =[

0 10 0

], e3 =

[0 01 0

], and e4 =

[0 00 1

]. We see

B = [T ]β =

2 0 1 00 3 −1 10 −1 3 00 0 0 2

so pT (x) = (x− 2)((x− 3)(x− 3)(x− 2)− 1(x− 2)) = (x− 2)2((x− 3)2− 1). This is (x− 2)2(x2− 6x+8) =(x− 2)3(x− 4).

For λ = 2, we have

(B − 2I) =

0 0 1 00 1 −1 10 −1 1 00 0 0 0

2

which has rank 3. We see

(B − 2I)2 =

0 −1 1 00 2 −2 10 −2 2 −10 0 0 0

which has rank 2. Lastly

(B − 2I)3 =

0 −2 2 −10 4 −4 20 −4 4 −20 0 0 0

which has rank 1. Thus we have N((B − 2I)2) = {(a, b, c, d)| − b + c = 0 and 2b − 2c + d = 0} andN((B − 2I)3) = {(a, b, c, d)| − 2b+ 2c− d = 0}. Hence, a basis for N((B − 2I)3) relative to N((B − 2I)2) is{(0, 0, 1, 2)}. We have (B − 2I)(0, 0, 1, 2) = (1, 1, 1, 0) and (B − 2I)2(0, 0, 1, 2) = (1, 0, 0, 0).

We have

(B − 4I) =

−2 0 1 00 −1 −1 10 −1 −1 00 0 0 −2

so N((B−4I)) = {(a, b, c, d)|−2a+ c = −b− c+d = −b− c = d = 0} which is {(a, b, c, d)|−2a+ c = b+ c =d = 0}. Hence, a nonzero vector in N((B − 4I)) is (1,−2, 2, 0). Adding this to our basis for N((B − 2I)3),we get {(0, 0, 1, 2), (1, 1, 1, 0), (1, 0, 0, 0), (1,−2, 2, 0)} as a Jordan basis with corresponding matrix

2 0 0 01 2 0 00 1 2 00 0 0 4

5. Find a Jordan canonical basis and a Jordan canonical form of the operator T on M2×2(R) given by

T (A) =

[3 10 3

](A−At).

Solution. With β as in 4 we have

B = [T ]β =

0 −1 1 00 3 −3 00 −3 3 00 0 0 0

so pT (x) = x2((x − 3)2 − 9) = x2(x2 − 6x) = x3(x − 6). We see B has rank 1, so dimN(B) = 3. We seeN(B) = {(a, b, c, d)| − b+ c = 0}, so a basis for this space is {(0, 1, 1, 0), (1, 0, 0, 0), (0, 0, 0, 1)}. Now,

B − 6I =

−6 −1 1 00 −3 −3 00 −3 −3 00 0 0 −6

so N((B−6I) = {(a, b, c, d)|−6a− b+c = −3b−3c = −6d = 0}, or {(a, b, c, d)|−6a− b+c = b+c = d = 0}.Hence, a nonzero vector in this space is (1,−3, 3, 0). With the other vectors, our Jordan basis is thus{(0, 1, 1, 0), (1, 0, 0, 0), (0, 0, 0, 1), (1,−3, 3, 0)} with corresponding matrix

0 0 0 00 0 0 00 0 0 00 0 0 6

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6. Let V be the real vector space of polynomials in 2 variables of degree at most 2. Find a Jordanbasis and a Jordan canonical form for the operator T on V given by T (f) = fx + fy (here subscripts denotedi�erentiation with respect to those variables).

Proof. Let β = {1, x, y, x2, xy, y2} be a basis for V . Then,

A = [T ]β =

0 1 1 0 0 00 0 0 2 1 00 0 0 0 1 20 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

so pT (x) = x6. We see

A2 =

0 0 0 2 2 20 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

and so A3 = 0. Thus, d1 = 3, d2 = 5, and d3 = 6. For a basis of N(A3)/N(A2), we may take anyvector (a, b, c, d, e, f) with d + e + f 6= 0, such as (0, 0, 0, 0, 1, 1). Then, A(0, 0, 0, 0, 1, 1) = (0, 1, 3, 0, 0, 0)and A2(0, 0, 0, 0, 1, 1) = (4, 0, 0, 0, 0, 0). Now, in N(A2)/N(A), we have (0, 1, 3, 0, 0, 0) + N(A). We wish to�nd a basis relative to N(A), so �rst we �nd a basis of N(A). We have N(A) = {(a, b, c, d, e, f)|b + c =2d+e = e+2f = 0}, so a basis for this space is {(1, 0, 0, 0, 0, 0), (0, 1,−1, 0, 0, 0), (0, 0, 0, 1,−2, 1)}. We extendthis to a basis of N(A2) by including (0, 1, 3, 0, 0, 0) from our chain and some other vector (a, b, c, d, e, f)with d + e + f = 0. We see (0, 0, 0, 0,−1, 1) works, and we use it to get a chain with A(0, 0, 0, 0,−1, 1) =(0,−1, 1, 0, 0, 0). Now, in N(A), we have (4, 0, 0, 0, 0, 0) and (0,−1, 1, 0, 0, 0). To extend this to a basis ofN(A), we include (0, 0, 0, 1,−2, 1). Hence, our Jordan basis is

{(0, 0, 0, 0, 1, 1), (0, 1, 3, 0, 0, 0), (4, 0, 0, 0, 0, 0), (0, 0, 0, 0,−1, 1), (0,−1, 1, 0, 0, 0), (0, 0, 0, 1,−2, 1)}

which is {xy + y2, x+ 3y, 4,−xy + y2,−x+ y, x2 − 2xy + y2} with corresponding matrix0 0 0 0 0 01 0 0 0 0 00 1 0 0 0 00 0 0 0 0 00 0 0 1 0 00 0 0 0 0 0

7. Let A ∈Mn×n(F ) be such that pA splits. Prove that A and At are similar.

Proof 1. Let JA be a Jordan canonical form for A, which exists as pA splits. We note A ∼ JA. We claimJA ∼ J tA. Indeed, suppose β = ∪mk=1βk is a Jordan canonical basis for which each block in JA correspondsto some βk. For βk = {x1, x2, . . . , x`}, let γk = {x`, . . . , x1} Then, for Wk = span(βk), we see that Wk isT -invariant with

[T ]βk=

λ 0 . . . 01 λ 0...

. . . 00 0 0 λ

4

and so

[T ]γk =

λ 1 . . . 00 λ 0...

. . . 10 0 0 λ

Thus, JA ∼ J tA. We claim JAt ∼ J tA. Indeed, if JA = C−1AC, then J tA = CtAt(C−1)t and (C−1)t = (Ct)−1,so J tA ∼ At.

Proof 2. We note that for arbitrary λ ∈ F and k ∈ N0, dimN((At − λI)k) = dimN(((A − λI)k)t) =dimV − columnrank(((A−λI)k)t) which, as column rank = row rank, is dimV − rowrank(((A−λI)k)t) =dimV − columnrank((A−λI)k) = dimN((A−λI)k). As JA is determined by the values dimN((A−λI)k),we conclude JA = JAt up to rearrangement. Hence, A ∼ JA = JAt ∼ At.

8. Let T be a linear operator on a �nite dimensional vector space such that pT splits and let λ1, . . . , λk bethe eigenvalues of T . De�ne S : V → V by S(v) =

∑kj=1 λjvj where v =

∑kj=1 vj is the (unique) expansion

of v as a sum of generalized eigenvectors.

Claim. S is diagonalizable and T − S is nilpotent.

Proof. Let β be a Jordan basis for T , so [T ]β = J where J is a Jordan block matrix. Then, if v ∈ β,we have v ∈ Wλj

(the generalized eigenspace corresponding to λj) for some 1 ≤ j ≤ k. Hence, S(v) = λjv,so [S]β is a diagonal matrix. Moreover, [S]β − [T ]β is a matrix consisting of all zeros except possiblysome ones immediately below the diagonal. Thus, [S − T ]β is strictly lower triangular and hence nilpotent(eg by (S−T )t being strinctly upper triangular and hence nilpotent and ((S−T )t)m = ((S−T )m)t for allm).

9. Let P1, P2 be two projectors in the vector space V such that P1P2 = P2P1. Show that P1P2 is also aprojector and that R(P1P2) = R(P1) ∩R(P2).

Proof. We see (P1P2)2 = P1P2P1P2 = P1P1P2P2 = P1P2, so P1P2 is a projector. Now, suppose

x ∈ R(P1P2), say x = P1P2(v). Then, x = P1(P2(v)), so x ∈ R(P1). Moreover, x = P2(P1(v)) by hy-pothesis, so x ∈ R(P2). Thus, x ∈ R(P1) ∩ R(P2). Now, let x ∈ R(P1) ∩ R(P2), so x = P1(v) = P2(u)for some v, u ∈ V . Then, P1(x) = P1(P1(v)) = P1(v) = x and P2(x) = P2(P2(u)) = P2(u) = x. Thus,P1(P2(x)) = P1(x) = x, so x ∈ R(P1P2).

10. Let P : V → V be a projector in a vector space V of dimension n. Prove there is a basis β of V such

that [P ]β =

[Ik 00 0n−k

].

Proof. P 2 − P = 0 so if f(x) = x(x− 1), we see mp|f . Thus, mP is a product of distinct linear factors,so P is diagonalizable. As 0 and 1 are the only zeros of f , they are the only eigenvalues of P , so by choosinga basis β of eigenvectors in which those with eigenvalue 1 come �rst, we get the desired result.

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