Math 406, Linear Algebra 3, Winter 2005

Problem Set 5, Solutions

1. Let T : F3 → F3 be the map given by M(T ) =

3 9/7 −13/72 1 23 1 1

.

(a) Find a basis of generalized eigenvectors of T .The eigenvalues of T are 1 and 2, with multiplicities 1 and 2. One can find these bycomputing the characteristic polynomial det(λI −M(T )) = (λ − 1)(λ − 2)2, or by find-ing a basis which makes M(T ) upper triangular, and reading the eigenvalues and theirmultiplicities off the diagonal. In any case, we can find a basis of generalized eigenvaluesby finding bases for null (T − 1I)3 and null (T − 2I)3.Now

M((T − I)3) =

0 17 − 3

718 64

7 − 667

13 477 − 50

7

,

while

M(T − 2I)3) =

2 137 − 18

7−6 − 39

7547

−2 − 137

187

.

Then null (T − I)3 consists of all xyz

such that 0 1

7 − 37

18 647 − 66

713 47

7 − 507

xyz

=

000

.

This amounts to augmenting 0 17 − 3

718 64

7 − 667

13 477 − 50

7

,

and row reducing. We get the matrix1 0 1 00 1 −3 00 0 0 0

,

so that the solutions to our equation consist of the vectors in the set−c

3cc

| c ∈ F

.

Thus {(−1, 3, 1)} is a basis for null (T − I)3. We can do the same for null (T − 2I)3. Theaugmented matrix in this situation is1 13

14 − 97 0

0 0 0 00 0 0 0

so that null (T − 2I)3 is the set

− 1314b + 9

7cbc

| b, c ∈ F

,

and {(−13, 14, 0), (9, 0, 7)} is a basis.We conclude that {(−1, 3, 1), (−13, 14, 0), (9, 0, 7)} is a basis of generalized eigenvectorsof F3.

(b) There is a basis of V such that the matrix of T is of the form described by theorem 8.28.Find such a basis and give the resulting matrix.We can find a basis which behaves in this fashion by following the construction in lemma8.26. First we choose a basis for null (T − I), extend to a basis for null (T − I)2, thenextend to a basis for null (T − I)3. Repeating this process for (T − 2I) gives the basisof the form we are looking for (according to theorem 8.28). Since dim null (T − I)3 = 1,there is no work to do for (T − I) (we can take {(−1, 3, 1)} as computed in part (a)).Without too much trouble we find that a basis for null (T − 2I) is (−1, 8, 5) (computethis as in part (a) by row reducing M(T − 2I)), and we can extend this to a basis fornull (T − 2I)2 by adding any linear independent vector from null (T − 2I)3 (here we usethe fact that dim null (T −2I)3 = 2). So we choose (9, 0, 7) to be the second basis vector,as it is clearly independent from (−1, 8, 5).Thus the basis we are looking for is {(−1, 3, 1), (−1, 8, 5), (9, 0, 7)}. Just to check wecompute the matrix for T with respect to this basis. Note that

T (−1, 3, 1) = (−1, 3, 1)T (−1, 8, 5) = (−2, 16, 10) = 2(−1, 8, 5)

T (9, 0, 7) = (14, 32, 34) = 4(−1, 8, 5) + 2(9, 0, 7)

(I just computed these by augmenting the matrix whose columns are our basis vectorswith the image vector, and then row reducing). Thus

M(T ) =

1 0 00 2 40 0 2

,

which is in the required form.

8.19 Prove that if V is a complex vector space, then every invertible operator on V has a cube root.

First we show that I + N has a cube root for any nilpotent map N ∈ L(V ). Because N isnilpotent, there is m ∈ N such that Nm = 0. Then it is enough to show that we can findai ∈ C such that (I + a1N + a2N

2 + · · ·+ amNm−1)3 = I + N . Note that that

(I + a1N + a2N2 + · · ·+ amNm−1)3 = I + 3a1N + (3a2 + a2

1)N2 + (3a3 + 3a1a2 + a3

1)N3

+ · · ·+ (3am−1 + terms involving a1, . . . , am−1)Nm−1.

It will be the case that this sum equals I + N if we let a1 = 1/3, a2 = −1/27, a3 =(−3a1a2 − a3

1)/3 = −2/81, and so on. More precisely, we see that the coefficient of N i is3ai + terms involving a1, . . . , ai−1, which we may write as 3ai + fi(a1, . . . , ai−1) where fi issome polynomial in a1, . . . , ai−1. Then I +a1N + · · ·+am−1N

m−1 is the cube root of I +N ifwe let a1 = 1/3, a2 = −f1(a1)/3, a3 = −f2(a1, a2)/3, . . . , ai = −fi(a1, . . . , ai−1)/3, . . . , am =−fm(a1, . . . , am−1)/3.

Now suppose that T ∈ L(V ) is invertible, let λ1, . . . , λm be the distinct eigenvalues of T , andlet U1, . . . , Um be the corresponding generalized eigenspaces. By theorem 8.23, (T −λjI)|Uj isnilpotent, so let Nj = (T − λjI)|Uj . Note that λj 6= 0 (see theorems 5.18 and 5.16), and thuswe may write T |Uj

= λj(I + Nj

λj) for each j. It is the case that Njλj is nilpotent (because if

Npj = 0 then (Nj/λj)p = Np

j /λpj = 0 as well), and thus (by the result above), (I + Nj/λj) has

a cube root, which we call Sj . Then the map Sj : Uj → Uj given by Sj(−→u ) = 3√

λjSj(−→u ) is acube root of T |Uj

(because S3j = ( 3

√λjSj)3 = λjS

3j = λj(I + Nj

λj) = T |Uj

).

Now V = U1 ⊕ · · · ⊕ Um, so a typical vector −→v ∈ V is −→v = −→u 1 + · · · + −→u m where −→u i ∈ Ui

for i = 1, . . . ,m. Thus we can define S : V → V by S(−→v ) = S1(−→u 1) + · · · + Sm(−→u m). Notethat S is a well defined linear map. It also follows that S is a cube root of T because S3(−→v ) =S2S(−→u 1 + · · · + −→u m) = S2(S1(−→u 1) + · · · + Sm(−→u m)) = S(S(S1(−→u 1) + · · · + Sm(−→u m))) =S(S2

1(−→u 1) + · · · + S2m(−→u m)) = S3

1(−→u 1) + · · · + S3m(−→u m) = T |U1(

−→u 1) + · · · + T |Um(−→u m) =T (−→u 1 + · · ·+−→u m) = T (−→v ) as required (we have used that the Ui are Si invariant).

8.27 Give an example of an operator on C4 whose characteristic polynomial equals z(z− 1)2(z− 3)and whose minimal polynomial equals z(z − 1)(z − 3).

The first thing to try is, of course, the map T such that

M(T ) =

0 0 0 00 1 0 00 0 1 00 0 0 3

.

This in fact works, that is, the characteristic polynomial of T is clearly p(z) = z(z−1)2(z−3),but it is also true that T (T − 1I)(T − 3I) = 0. As the minimal polynomial is monic, dividesany polynomial for which T vanishes, and has roots exactly the eigenvalues, we know thatq(z) = z(z − 1)(z − 3) must be it.

More interesting is what happens if we introduce other entries into the matrix. Consider, for

example, the map T such that M(T ) =

0 a b c0 1 d e0 0 1 f0 0 0 3

. The characteristic polynomial is still

the same, but it turns out (you can work this out by multiplying) that the minimal polynomialis q(z) if and only if d = 0. Here the minimal polynomial is helping us discern between thetwo possible Jordan form (for maps with eigenvalues 0, 1, and 3, and multiplicities 1, 2, and1).

8.30 Suppose V is a complex vector space and T ∈ L(V ). Prove that there does not exist a directsum decomposition of V into two proper subspaces invariant under T if and only if the minimalpolynomial of T is of the form (z − λ)dim V for some λ ∈ C.

Suppose that M(T ) is a matrix in Jordan form. Then M(T ) is block upper triangular. Letthe vectors corresponding to the columns in the ith block generate a subspace of V which wecall Ui for i = 1, . . . ,m. Then certainly V = U1⊕ · · · ⊕Um, and it is also true (because M(T )is block diagonal) that each Ui is T -invariant. Note that the direct sum of two T -invariantsubspaces is T -invariant. Thus there does not exist a direct sum decomposition of V into twoproper subspaces invariant under T if and only if M(T ) has Jordan form with only one block.So to finish the problem we need to show that M(T ) has Jordan form with only one block ifand only if (z − λ)dim V is the minimal polynomial.

Question 29 in the text is relevant here. It claims that the dimension of the minimal polynomialof any nilpotent map N is the longest consecutive strings of ones that appear on the diagonaldirectly above the main diagonal when N is is Jordan form. We will prove this below, but inthe meantime, note that, if the minimal polynomial of T is (z − λ)dim V , then T has only oneeigenvalue, and because T has a basis of generalized eigenvectors, T − λI is nilpotent. Thusby question 29, the the length of the longest consecutive strings of ones on the off diagonalof (T − λI) in Jordan form is dim V − 1. This implies that when we consider the matrix of(T − λI) in Jordan form, it can have only one block, and hence the matrix of T in Jordanform has only one block. Similarly, suppose that M(T ) has Jordan form with only one blockand hence only one eigenvalue, λ. Thus the minimal polynomial of T is (z − λ)q for someq ≤ dim V . This implies that (T − λI)q = 0. But M(T − λI) is an n × n nilpotent matrix,so again by question number 29, the minimal polynomial of M(T − λI) is zp+1 where p is thelongest consecutive string of ones that appears above the main diagonal when the matrix isin Jordan form. Thus, the minimal polynomial of (T − λI) is zdim V , and we conclude thatq = dim V , as required.

I will sketch the proof of problem 29 (it is really a 206 problem). We know that any nilpotentmatrix is upper triangular with zeros on the main diagonal (we may assume that we areover C, because they ask us to use Jordan form). So if A is a nilpotent matrix in Jordanform, its only nonzero entries are ones along the off diagonal. Note that if A is n × n andthe whole off diagonal is full of ones, then Aj 6= 0 for 0 < j ≤ n − 1 and An = 0. Thisis because if B is is the matrix

[−→b 1 . . .

−→b n

]where

−→b 1, . . . ,

−→b n are the columns, then

AB =[−→

0−→b 1 . . .

−→b n−1

](then some quick thought shows us that AAj = 0 if and only if

j = n− 1). Now if A is a matrix in Jordan form, let A1, . . . , Am be the blocks along the main

diagonal. So A =

A1

A2

. . .Am

, and it is easy to see that Aj =

Aj

1

Aj2

. . .Aj

m

.

We conclude that Aj = 0 for the largest j such that At is j× j, and that Ap 6= 0 for p < j. Asthe largest integer such that At is j × j is exactly equal to the largest consecutive strings ofones along the off diagonal minus one (each Aj contains an unbroken string of ones), we arefinished.