7.2 The Jordan Canonical Form II – Linear Algebra, 5th Edition

7.2 The Jordan Canonical Form II

For the purposes of this section, we fix a linear operator T on an n-dimensional vector space V such that the characteristic polynomial of T splits. Let λ1, λ2, , λk be the distinct eigenvalues of T.

By Theorem 7.7 (p. 484), each generalized eigenspace Kλi contains an ordered basis βi consisting of a union of disjoint cycles of generalized eigenvectors corresponding to λi. So by Theorems 7.4(b) (p. 480) and 7.5 (p. 482), the union β=i=1kβi is a Jordan canonical basis for T. For each i, let Ti be the restriction of T to Kλi, and let Ai=[Ti]βi. Then Ai is the Jordan canonical form of Ti, and

J = [ T ] β = ( A 1 O O O A 2 O O O A k )

is the Jordan canonical form of T. In this matrix, each O is a zero matrix of appropriate size.

In this section, we compute the matrices Ai and the bases βi, thereby computing J and β as well. While developing a method for finding J, it becomes evident that in some sense the matrices Ai are unique.

To aid in formulating the uniqueness theorem for J, we adopt the following convention: The basis βi for Kλi will henceforth be ordered in such a way that the cycles appear in order of decreasing length. That is, if βi is a disjoint union of cycles γ1, γ2, , γni and if the length of the cycle γj is pj, we index the cycles so that p1p2pni. This ordering of the cycles limits the possible orderings of vectors in βi, which in turn determines the matrix Ai. It is in this sense that Ai is unique. It then follows that the Jordan canonical form for T is unique up to an ordering of the eigenvalues of T. As we will see, there is no uniqueness theorem for the bases βi or for β. However, we show that for each i, the number ni of cycles that form βi, and the length pj(j=1, 2, , ni) of each cycle, is completely determined by T.

Example 1

To illustrate the discussion above, suppose that, for some i, the ordered basis βi for Kλi is the union of four cycles βi=γ1γ2γ3γ4 with respective lengths p1=3, p2=3, p3=2, and p4=1. Then

To help us visualize each of the matrices Ai and ordered bases βi, we use an array of dots called a dot diagram of Ti, where Ti is the restriction of T to Kλi. Suppose that βi is a disjoint union of cycles of generalized eigenvectors γ1, γ2, , γni with lengths p1p2pni, respectively. The dot diagram of Ti contains one dot for each vector in βi, and the dots are configured according to the following rules.

  1. The array consists of ni columns (one column for each cycle).

  2. Counting from left to right, the jth column consists of the pj dots that correspond to the vectors of γj starting with the initial vector at the top and continuing down to the end vector.

Denote the end vectors of the cycles by v1, v2, , vni. In the following dot diagram of Ti, each dot is labeled with the name of the vector in βi to which it corresponds.

Notice that the dot diagram of Ti has ni columns (one for each cycle) and p1 rows. Since p1p2pni, the columns of the dot diagram become shorter (or at least not longer) as we move from left to right.

Now let rj denote the number of dots in the jth row of the dot diagram. Observe that r1r2rp1. Furthermore, the diagram can be reconstructed from the values of the ri’s. The proofs of these facts, which are combinatorial in nature, are treated in Exercise 9.

In Example 1, with ni=4, p1=p2=3, p3=2, and p4=1, the dot diagram of Ti is as follows:

Here r1=4, r2=3, and r3=2.

We now devise a method for computing the dot diagram of Ti using the ranks of linear operators determined by T and λi. It will follow that the dot diagram is completely determined by T, from which it will follow that it is unique. On the other hand, βi is not unique. For example, see Exercise 8. (It is for this reason that we associate the dot diagram with Ti rather than with βi.)

To determine the dot diagram of Ti, we devise a method for computing each rj, the number of dots in the jth row of the dot diagram, using only T and λi. The next three results give us the required method. To facilitate our arguments, we fix a basis βi for Kλi so that βi is a disjoint union of ni cycles of generalized eigenvectors with lengths p1p2pni.

Theorem 7.9.

For any positive integer r, the vectors in βi that are associated with the dots in the first r rows of the dot diagram of Ti constitute a basis for N((TλiI)r). Hence the number of dots in the first r rows of the dot diagram equals nullity((TλiI)r).

Proof.

Clearly, N((TλiI)r)Kλi, and Kλi is invariant under (TλiI)r. Let U denote the restriction of (TλiI)r to Kλi. By the preceding remarks, N((TλiI)r)=N(U), and hence it suffices to establish the theorem for U. Now define

S 1 = { x β i :  U ( x ) = 0 } and S 2 = { x β i :  U ( x ) 0 } .

Let a and b denote the number of vectors in S1 and S2, respectively, and let mi=dim(Kλi). Then a+b=mi. For any xβi, xS1 if and only if x is one of the first r vectors of a cycle, and this is true if and only if x corresponds to a dot in the first r rows of the dot diagram. Hence a is the number of dots in the first r rows of the dot diagram. For any xS2, the effect of applying U to x is to move the dot corresponding to x exactly r places up its column to another dot. It follows that U maps S2 in a one-to-one fashion into βi. Thus {U(x): xS2} is a basis for R(U) consisting of b vectors. Hence rank(U)=b, and so nullity(U)=mib=a. But S1 is a linearly independent subset of N(U) consisting of a vectors; therefore S1 is a basis for N(U).

In the case that r=1, Theorem 7.9 yields the following corollary.

Corollary.

The dimension of Eλi is ni. Hence in a Jordan canonical form of T, the number of Jordan blocks corresponding to λi equals the dimension of Eλi.

Proof.

Exercise.

We are now able to devise a method for describing the dot diagram in terms of the ranks of operators.

Theorem 7.10.

Let rj denote the number of dots in the jth row of the dot diagram of Ti, the restriction of T to Kλi. Then the following statements are true.

  1. r1=dim(V)rank(TλiI).

  2. rj=rank((TλiI)j1)rank((TλiI)j)

    if j>1.

Proof.

By Theorem 7.9, for j=1, 2, , p1, we have

r 1 + r 2 + + r j = nullity ( ( T λ i I ) j ) = dim ( V ) rank ( ( T λ i I ) j ) .

Hence

r 1 = dim ( V ) rank ( T λ i I )

and for j>1,

r j = ( r 1 + r 2 + + r j ) ( r 1 + r 2 + + r j 1 ) = [ dim ( V ) rank ( T λ i I ) j ] [ dim ( V ) rank( ( T λ i I ) j 1 ) ] = rank ( ( T λ i I ) j 1 ) rank ( ( T λ i I ) j ) .

Theorem 7.10 shows that the dot diagram of Ti is completely determined by T and λi. Hence we have proved the following result.

Corollary.

For any eigenvalue λi of T, the dot diagram of Ti is unique. Thus, subject to the convention that the cycles of generalized eigenvectors for the bases of each generalized eigenspace are listed in order of decreasing length, the Jordan canonical form of a linear operator or a matrix is unique up to the ordering of the eigenvalues.

We apply these results to find the Jordan canonical forms of two matrices and a linear operator.

Example 2

Let

A = ( 2 1 0 1 0 3 1 0 0 1 1 0 0 1 0 3 ) .

We find the Jordan canonical form of A and a Jordan canonical basis for the linear operator T=LA. The characteristic polynomial of A is

det ( A t I ) = ( t 2 ) 3 ( t 3 ) .

Thus A has two distinct eigenvalues, λ1=2 and λ2=3, with multiplicities 3 and 1, respectively. Let T1 and T2 be the restrictions of LA to the generalized eigenspaces Kλ1 and Kλ2, respectively.

Suppose that β1 is a Jordan canonical basis for T1. Since λ1 has multiplicity 3, it follows that dim(Kλ1)=3 by Theorem 7.4(c) (p. 480); hence the dot diagram of T1 has three dots. As we did earlier, let rj denote the number of dots in the jth row of this dot diagram. Then, by Theorem 7.10,

r 1 = 4 rank ( A 2 I ) = 4 rank ( 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 ) = 4 2 = 2 ,

and

r 2 = rank ( A 2 I ) rank ( ( A 2 I ) 2 ) = 2 1 = 1.

(Actually, the computation of r2 is unnecessary in this case because r1=2 and the dot diagram only contains three dots.) Hence the dot diagram associated with β1 is

So

A 1 = [ T 1 ] β 1 = ( 2 1 0 0 2 0 0 0 2 ) .

Since λ2=3 has multiplicity 1, it follows that dim(Kλ2)=1, and consequently any basis β2 for Kλ2 consists of a single eigenvector corresponding to λ2=3. Therefore

A 2 = [ T 2 ] β 2 = ( 3 ) .

Setting β=β1β2, we have

J = [ L A ] β = ( 2 1 0 0 0 2 0 0 0 0 2 0 0 0 0 3 ) ,

and so J is the Jordan canonical form of A.

We now find a Jordan canonical basis for T=LA. We begin by determining a Jordan canonical basis β1 for T1. Since the dot diagram of T1 has two columns, each corresponding to a cycle of generalized eigenvectors, there are two such cycles. Let v1 and v2 denote the end vectors of the first and second cycles, respectively. We reprint below the dot diagram with the dots labeled with the names of the vectors to which they correspond.

From this diagram we see that v1N((T2I)2) but v1N(T2I). Now

A 2 I = ( 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 ) and ( A 2 I ) 2 = ( 0 2 1 1 0 0 0 0 0 0 0 0 0 2 1 1 ) .

It is easily seen that

{ ( 1 0 0 0 ) ,   ( 0 1 2 0 ) ,   ( 0 1 0 2 ) }

is a basis for N((T2I)2)=Kλ1. Of these three basis vectors, the last two do not belong to N(T2I), and hence we select one of these for v1. Suppose that we choose

v 1 = ( 0 1 2 0 ) .

Then

( T 2 I ) ( v 1 ) = ( A 2 I ) ( v 1 ) = ( 0 1 0 1 0 1 1 0 0 1 1 0 0 1 0 1 ) ( 0 1 2 0 ) = ( 1 1 1 1 ) .

Now simply choose v2 to be a vector in Eλ1 that is linearly independent of (T2I)(v1); for example, select

v 2 = ( 1 0 0 0 ) .

Thus we have associated the Jordan canonical basis

β 1 = { ( 1 1 1 1 ) ,   ( 0 1 2 0 ) ,   ( 1 0 0 0 ) }

with the dot diagram in the following manner.

By Theorem 7.6 (p. 483), the linear independence of β1 is guaranteed since v2 was chosen to be linearly independent of (T2I)(v1).

Since λ2=3 has multiplicity 1, dim(Kλ2)=dim(Eλ2)=1. Hence any eigenvector of LA corresponding to λ2=3 constitutes an appropriate basis β2. For example,

β 2 = { ( 1 0 0 1 ) } .

Thus

β = β 1 β 2 = { ( 1 1 1 1 ) ,   ( 0 1 2 0 ) ,   ( 1 0 0 0 ) ,   ( 1 0 0 1 ) }

is a Jordan canonical basis for LA.

Notice that if

Q = ( 1 0 1 1 1 1 0 0 1 2 0 0 1 0 0 1 ) ,

then J=Q1AQ.

Example 3

Let

A = ( 2 4 2 2 2 0 1 3 2 2 3 3 2 6 3 7 ) .

We find the Jordan canonical form J of A, a Jordan canonical basis for LA, and a matrix Q such that J=Q1AQ.

The characteristic polynomial of A is det(AtI)=(t2)2(t4)2. Let T=LA, λ1=2, and λ2=4, and let Ti be the restriction of LA to Kλi for i=1, 2.

We begin by computing the dot diagram of T1. Let r1 denote the number of dots in the first row of this diagram. Then

r 1 = 4 rank ( A 2 I ) = 4 2 = 2 ;

hence the dot diagram of T1 is as follows.

 

Therefore

A 1 = [ T 1 ] β 1 = ( 2 0 0 2 ) ,

where β1 is any basis corresponding to the dots. In this case, β1 is an arbitrary basis for Eλ1=N(T2I), for example,

β 1 = { ( 2 1 0 2 ) ,   ( 0 1 2 0 ) } .

Next we compute the dot diagram of T2. Since rank(A4I)=3, there is only 43=1 dot in the first row of the diagram. Since λ2=4 has multiplicity 2, we have dim(Kλ2)=2, and hence this dot diagram has the following form:

Thus

A 2 = [ T 2 ] β 2 = ( 4 1 0 4 ) ,

where β2 is any basis for Kλ2 corresponding to the dots. In this case, β2 is a cycle of length 2. The end vector of this cycle is a vector vKλ2=N((T4I)2) such that vN(T4I). One way of finding such a vector was used to select the vector v1 in Example 2. In this example, we illustrate another method. A simple calculation shows that a basis for the null space of LA4I is

{ ( 0 1 1 1 ) } .

Choose v to be any solution to the system of linear equations

( A 4 I ) x = ( 0 1 1 1 ) ,

for example,

v = ( 1 1 1 0 ) .

Thus

β 2 = { ( L A 4 I ) ( v ) ,   v } = { ( 0 1 1 1 ) ,   ( 1 1 1 0 ) } .

Therefore

β = β 1 β 2 = { ( 2 1 0 2 ) ,   ( 0 1 2 0 ) ,   ( 0 1 1 1 ) ,   ( 1 1 1 0 ) }

is a Jordan canonical basis for LA. The corresponding Jordan canonical form is given by

J = [ L A ] β = ( A 1 O O A 2 ) = ( 2 0 0 0 0 2 0 0 0 0 4 1 0 0 0 4 ) .

Finally, we define Q to be the matrix whose columns are the vectors of β listed in the same order, namely,

Q = ( 2 0 0 1 1 1 1 1 0 2 1 1 2 0 1 0 ) .

Then J=Q1AQ.

Example 4

Let V be the vector space of polynomial functions in two real variables x and y of degree at most 2. Then V is a vector space over R and α={1, x, y, x2, y2, xy} is an ordered basis for V. Let T be the linear operator on V defined by

T ( f ( x ,   y ) ) = x f ( x ,   y ) .

For example, if f(x, y)=x+2x23xy+y, then

T ( f ( x ,   y ) ) = x ( x + 2 x 2 3 x y + y ) = 1 + 4 x 3 y .

We find the Jordan canonical form and a Jordan canonical basis for T.

Let A=[T]α. Then

A = ( 0 1 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ,

and hence the characteristic polynomial of T is

det ( A t I ) = det ( t 1 0 0 0 0 0 t 0 2 0 0 0 0 t 0 0 1 0 0 0 t 0 0 0 0 0 0 t 0 0 0 0 0 0 t ) = t 6 .

Thus λ=0 is the only eigenvalue of T, and Kλ=V. For each j, let rj denote the number of dots in the jth row of the dot diagram of T. By Theorem 7.10,

r 1 = 6 rank ( A ) = 6 3 = 3 ,

and since

A 2 = ( 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ,

r2=rank(A)rank(A2)=31=2.

Because there are a total of six dots in the dot diagram and r1=3 and r2=2, it follows that r3=1. So the dot diagram of T is

We conclude that the Jordan canonical form of T is

We now find a Jordan canonical basis for T. Since the first column of the dot diagram of T consists of three dots, we must find a polynomial f1(x, y) such that 2x2f1(x, y)0. Examining the basis α={1, x, y, x2, y2, xy} for Kλ=V, we see that x2 is a suitable candidate. Setting f1(x, y)=x2, we see that

( T λ I ) ( f 1 ( x ,   y ) ) = T ( f 1 ( x ,   y ) ) = x ( x 2 ) = 2 x

and

( T λ I ) 2 ( f 1 ( x ,   y ) ) = T 2 ( f 1 ( x ,   y ) ) = 2 x 2 ( x 2 ) = 2.

Likewise, since the second column of the dot diagram consists of two dots, we must find a polynomial f2(x, y) such that

x ( f 2 ( x ,   y ) ) 0 , but 2 x 2 ( f 2 ( x ,   y ) ) = 0.

Since our choice must be linearly independent of the polynomials already chosen for the first cycle, the only choice in α that satisfies these constraints is xy. So we set f2(x, y)=xy. Thus

( T λ I ) ( f 2 ( x ,   y ) ) = T ( f 2 ( x ,   y ) ) = x ( x y ) = y .

Finally, the third column of the dot diagram consists of a single polynomial that lies in the null space of T. The only remaining polynomial in α is y2, and it is suitable here. So set f3(x, y)=y2. Therefore we have identified polynomials with the dots in the dot diagram as follows.

Thus β={2, 2x, x2, y, xy, y2} is a Jordan canonical basis for T.

In the three preceding examples, we relied on our ingenuity and the context of the problem to find Jordan canonical bases. The reader can do the same in the exercises. We are successful in these cases because the dimensions of the generalized eigenspaces under consideration are small. We do not attempt, however, to develop a general algorithm for computing Jordan canonical bases, although one could be devised by following the steps in the proof of the existence of such a basis (Theorem 7.7 p. 484).

The following result may be thought of as a corollary to Theorem 7.10.

Theorem 7.11.

Let A and B be n×n matrices, each having Jordan canonical forms computed according to the conventions of this section. Then A and B are similar if and only if they have (up to an ordering of their eigenvalues) the same Jordan canonical form.

Proof.

If A and B have the same Jordan canonical form J, then A and B are each similar to J and hence are similar to each other.

Conversely, suppose that A and B are similar. Then A and B have the same eigenvalues. Let JA and JB denote the Jordan canonical forms of A and B, respectively, with the same ordering of their eigenvalues. Then A is similar to both JA and JB, and therefore, by the corollary to Theorem 2.23 (p. 115), JA and JB are matrix representations of LA. Hence JA and JB are Jordan canonical forms of LA. Thus JA=JB by the corollary to Theorem 7.10.

Example 5

We determine which of the matrices

A = ( 3 3 2 7 6 3 1 1 2 ) , B = ( 0 1 1 4 4 2 2 1 1 ) , C = ( 0 1 1 3 1 2 7 5 6 ) , and D = ( 0 1 2 0 1 1 0 0 2 )

are similar. Observe that A, B, and C have the same characteristic polynomial (t1)(t2)2, whereas D has (t1)(t2) as its characteristic polynomial. Because similar matrices have the same characteristic polynomials, D cannot be similar to A, B, or C. Let JA, JB, and JC be the Jordan canonical forms of A, B, and C, respectively, using the ordering 1, 2 for their common eigenvalues. Then (see Exercise 4)

J A = ( 1 0 0 0 2 1 0 0 2 ) ,   J B = ( 1 0 0 0 2 0 0 0 2 ) , and J C = ( 1 0 0 0 2 1 0 0 2 ) .

Since JA=JC, A is similar to C. Since JB is different from JA and JC, B is similar to neither A nor C.

The reader should observe that any diagonal matrix is a Jordan canonical form. Thus a linear operator T on a finite-dimensional vector space V is diagonalizable if and only if its Jordan canonical form is a diagonal matrix. Hence T is diagonalizable if and only if the Jordan canonical basis for T consists of eigenvectors of T. Similar statements can be made about matrices. Thus, of the matrices A, B, and C in Example 5, A and C are not diagonalizable because their Jordan canonical forms are not diagonal matrices.

Exercises

  1. Label the following statements as true or false. Assume that the characteristic polynomial of the matrix or linear operator splits.

    1. The Jordan canonical form of a diagonal matrix is the matrix itself.

    2. Let T be a linear operator on a finite-dimensional vector space V that has a Jordan canonical form J. If β is any basis for V, then the Jordan canonical form of [T]β is J.

    3. Linear operators having the same characteristic polynomial are similar.

    4. Matrices having the same Jordan canonical form are similar.

    5. Every matrix is similar to its Jordan canonical form.

    6. Every linear operator with the characteristic polynomial (1)n(tλ)n has the same Jordan canonical form.

    7. Every linear operator on a finite-dimensional vector space has a unique Jordan canonical basis.

    8. The dot diagrams of a linear operator on a finite-dimensional vector space are unique.

  2. Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Suppose that λ1=2, λ2=4, and λ3=3 are the distinct eigenvalues of T and that the dot diagrams for the restriction of T to Kλi(i=1, 2, 3) are as follows:

    λ1=2 λ2=4 λ3=3

    Find the Jordan canonical form J of T.

  3. Let T be a linear operator on a finite-dimensional vector space V with Jordan canonical form

    1. Find the characteristic polynomial of T.

    2. Find the dot diagram corresponding to each eigenvalue of T.

    3. For which eigenvalues λi, if any, does Eλi=Kλi?

    4. For each eigenvalue λi, find the smallest positive integer pi for which Kλi=N((TλiI)pi).

    5. Compute the following numbers for each i, where Ui denotes the restriction of TλiI to Kλi.

      1. rank(Ui)

      2. rank(Ui2)

      3. nullity(Ui)

      4. nullity(Ui2)

  4. For each of the matrices A that follow, find a Jordan canonical form J and an invertible matrix Q such that J=Q1AQ. Notice that the matrices in (a), (b), and (c) are those used in Example 5.

    1. A=(332763112)

    2. A=(011442211)

    3. A=(011312756)

    4. A=(0312211221122314)

  5. For each linear operator T, find a Jordan canonical form J of T and a Jordan canonical basis β for T.

    1. (a) V is the real vector space of functions spanned by the set of real-valued functions {et, tet, t2et, e2t}, and T is the linear operator on V defined by T(f)=f.

    2. (b) T is the linear operator on P3(R) defined by T(f(x))=xf(x).

    3. (c) T is the linear operator on P3(R) defined by T(f(x))=f(x)+2f(x)

    4. (d) T is the linear operator on M2×2(R) defined by

      T ( A ) = ( 3 1 0 3 ) A A t .
    5. (e) T is the linear operator on M2×2(R) defined by

      T ( A ) = ( 3 1 0 3 ) ( A A t ) .
    6. (f) V is the vector space of polynomial functions in two real variables x and y of degree at most 2, as defined in Example 4, and T is the linear operator on V defined by

      T ( f ( x ,   y ) ) = x f ( x ,   y ) + y f ( x ,   y ) .
  6. Let A be an n×n matrix whose characteristic polynomial splits. Prove that A and At have the same Jordan canonical form, and conclude that A and At are similar. Hint: For any eigenvalue λ of A and At and any positive integer r, show that rank((AλI)r)=rank((AtλI)r).

  7. Let A be an n×n matrix whose characteristic polynomial splits, γ be a cycle of generalized eigenvectors corresponding to an eigenvalue λ, and W be the subspace spanned by γ. Define γ to be the ordered set obtained from γ by reversing the order of the vectors in γ.

    1. (a) Prove that [TW]γ=([TW]γ)t.

    2. (b) Let J be the Jordan canonical form of A. Use (a) to prove that J and Jt are similar.

    3. (c) Use (b) to prove that A and At are similar.

  8. Let T be a linear operator on a finite-dimensional vector space, and suppose that the characteristic polynomial of T splits. Let β be a Jordan canonical basis for T.

    1. (a) Prove that for any nonzero scalar c, {cx: xβ} is a Jordan canonical basis for T.

    2. (b) Suppose that γ is one of the cycles of generalized eigenvectors that forms β, and suppose that γ corresponds to the eigenvalue λ and has length greater than 1. Let x be the end vector of γ, and let y be a nonzero vector in Eλ. Let γ be the ordered set obtained from γ by replacing x by x+y. Prove that γ is a cycle of generalized eigenvectors corresponding to λ, and that if γ replaces γ in the union that defines β, then the new union is also a Jordan canonical basis for T.

    3. (c) Apply (b) to obtain a Jordan canonical basis for LA, where A is the matrix given in Example 2, that is different from the basis given in the example.

  9. Suppose that a dot diagram has k columns and m rows with pj dots in column j and ri dots in row i. Prove the following results.

    1. (a) m=p1 and k=r1.

    2. (b) pj=max{i: rij} for 1jk and ri=max{j: pji} for 1im. Hint: Use mathematical induction on m.

    3. (c) r1r2rm.

    4. (d) Deduce that the number of dots in each column of a dot diagram is completely determined by the number of dots in the rows.

  10. Let T be a linear operator whose characteristic polynomial splits, and let λ be an eigenvalue of T.

    1. (a) Prove that dim(Kλ) is the sum of the lengths of all the blocks corresponding to λ in the Jordan canonical form of T.

    2. (b) Deduce that Eλ=Kλ if and only if all the Jordan blocks corresponding to λ are 1×1 matrices.

The following definitions are used in Exercises 11–19.

Definitions.

A linear operator T on a vector space V is called nilpotent if Tp=T0 for some positive integer p. An n×n matrix A is called nilpotent if Ap=O for some positive integer p.

  1. Let T be a linear operator on a finite-dimensional vector space V, and let β be an ordered basis for V. Prove that T is nilpotent if and only if [T]β is nilpotent.

  2. Prove that any square upper triangular matrix with each diagonal entry equal to zero is nilpotent.

  3. Let T be a nilpotent operator on an n-dimensional vector space V, and suppose that p is the smallest positive integer for which Tp=T0. Prove the following results.

    1. (a) N(Ti)N(Ti+1) for every positive integer i.

    2. (b) There is a sequence of ordered bases β1, β2, , βp such that βi is a basis for N(Ti) and βi+1 contains βi for 1ip1.

    3. (c) Let ββp be the ordered basis for N(Tp)=V V in (b). Then [T]β is an upper triangular matrix with each diagonal entry equal to zero.

    4. (d) The characteristic polynomial of T is (1)ntn. Hence the characteristic polynomial of T splits, and 0 is the only eigenvalue of T.

  4. Prove the converse of Exercise 13(d): If T is a linear operator on an n-dimensional vector space V and (1)ntn is the characteristic polynomial of T, then T is nilpotent.

  5. Give an example of a linear operator T on a finite-dimensional vector space over the field of real numbers such that T is not nilpotent, but zero is the only eigenvalue of T. Characterize all such operators. Visit goo.gl/nDjsWm for a solution.

  6. Let T be a nilpotent linear operator on a finite-dimensional vector space V. Recall from Exercise 13 that λ=0 is the only eigenvalue of T, and hence V=Kλ. Let β be a Jordan canonical basis for T. Prove that for any positive integer i, if we delete from β the vectors corresponding to the last i dots in each column of a dot diagram of β, the resulting set is a basis for R(Ti). (If a column of the dot diagram contains fewer than i dots, all the vectors associated with that column are removed from β.)

  7. Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits, and let λ1, λ2, , λk be the distinct eigenvalues of T. For each i, let vi denote the unique vector in Kλi such that x=v1+v2++vk. (This unique representation is guaranteed by Theorem 7.3 (p. 479).) Define a mapping S: VV by

    S ( x ) = λ 1 v 1 + λ 2 v 2 + + λ k v k .
    1. (a) Prove that S is a diagonalizable linear operator on V.

    2. (b) Let U=TS. Prove that U is nilpotent and commutes with S, that is, SU=US.

  8. Let T be a linear operator on a finite-dimensional vector space V over C, and let J be the Jordan canonical form of T. Let D be the diagonal matrix whose diagonal entries are the diagonal entries of J, and let M=JD. Prove the following results.

    1. (a) M is nilpotent.

    2. (b) MD=DM.

    3. (c) If p is the smallest positive integer for which Mp=O, then, for any positive integer r<p,

      J r = D r + r D r 1 M + r ( r 1 ) 2 ! D r 2 M 2 + + r D M r 1 + M r ,

      and, for any positive integer rp,

      J r = D r + r D r 1 M + r ( r 1 ) 2 ! D r 2 M 2 + + r ! ( r p + 1 ) ! ( p 1 ) ! D r p + 1 M p 1 .
  9. Let F=C and

    J = ( λ 1 0 0 0 λ 1 0 0 0 λ 0 0 0 0 1 0 0 0 λ )

    be the m×m Jordan block corresponding to λ, and let N=TλIm. Prove the following results:

    1. (a) Nm=O, and for 1r<m,

      N i j r = { 1 if  j = i + r 0 otherwise .
    2. (b) For any integer rm,

      J r = ( λ r r λ r 1 r ( r 1 ) 2 ! λ r 2 r ( r 1 ) ( r m + 2 ) ( m 1 ) ! λ r m + 1 0 λ r r λ r 1 r ( r 1 ) ( r m + 3 ) ( m 2 ) ! λ r m + 2   0 0 0 λ r ) .
    3. (c) limrJr exists if and only if one of the following holds:

      1. |λ|<1.

      2. λ=1 and m=1.

        (Note that limrλr exists under these conditions. See the discussion preceding Theorem 5.12 on page 284.) Furthermore, limrJr is the zero matrix if condition (i) holds and is the 1×1 matrix (1) if condition (ii) holds.

    4. (d) Prove Theorem 5.12 on page 284.

The following definition is used in Exercises 20 and 21.

Definition.

For any AMn×n(C), define the norm of A by

| | A | | m = max { | A i j | :   1 i , j n } .
  1. Let A, BMn×n(C). Prove the following results.

    1. (a) ||A||m0.

    2. (b) ||A||m=0 if and only if A=O.

    3. (c) ||cA||m=|c|||A||m for any scalar c.

    4. (d) ||A+B||m||A||m+||B||m.

    5. (e) ||AB||mn||A||m||B||m.

  2. Let AMn×n(C) be a transition matrix. (See Section 5.3.) Since C is an algebraically closed field, A has a Jordan canonical form J to which A is similar. Let P be an invertible matrix such that P1AP=J. Prove the following results.

    1. (a) ||Ak||m1 for every positive integer k.

    2. (b) There exists a positive number c such that ||Jk||mc for every positive integer k.

    3. (c) Each Jordan block of J corresponding to the eigenvalue λ=1 is a 1×1 matrix.

    4. (d) limkAk exists if and only if 1 is the only eigenvalue of A with absolute value 1.

    5. (e) Theorem 5.19(a), using (c) and Theorem 5.18.

The next exercise requires knowledge of absolutely convergent series as well as the definition of eA for a matrix A. (See page 310.)

  1. Use Exercise 20(d) to prove that eA exists for every AMn×n(C).

  2. Let x=Ax be a system of n linear differential equations, where x is an n-tuple of differentiable functions x1(t), x2(t), , xn(t) of the real variable t, and A is an n×n coefficient matrix as in Exercise 16 of Section 5.2. In contrast to that exercise, however, do not assume that A is diagonalizable, but assume that the characteristic polynomial of A splits. Let λ1, λ2, , λk be the distinct eigenvalues of A.

    1. (a) Prove that if u is the end vector of a cycle of generalized eigenvectors of LA of length p and u corresponds to the eigenvalue λi, then for any polynomial f(t) of degree less than p, the function

      e λ i t [ f ( t ) ( A λ i I ) p 1 + f ( t ) ( A λ i I ) p 2 + + f ( p 1 ) ( t ) ] u

      is a solution to the system x=Ax.

    2. (b) Prove that the general solution to x=Ax is a sum of the functions of the form given in (a), where the vectors u are the end vectors of the distinct cycles that constitute a fixed Jordan canonical basis for LA.

  3. Use Exercise 23 to find the general solution to each of the following systems of linear equations, where x, y, and z are real-valued differentiable functions of the real variable t.

    1. x=2x+yy=2yzz=3z

    2. x=2x+yy=2y+zz=2z