7.2 The Jordan Canonical Form II
For the purposes of this section, we fix a linear operator T on an ndimensional vector space V such that the characteristic polynomial of T splits. Let ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ be the distinct eigenvalues of T.
By Theorem 7.7 (p. 484), each generalized eigenspace ${\text{K}}_{{\lambda}_{i}}$ contains an ordered basis ${\beta}_{i}$ consisting of a union of disjoint cycles of generalized eigenvectors corresponding to ${\lambda}_{i}$. So by Theorems 7.4(b) (p. 480) and 7.5 (p. 482), the union $\beta ={\displaystyle \underset{i=1}{\overset{k}{\cup}}{\beta}_{i}}$ is a Jordan canonical basis for T. For each i, let ${\text{T}}_{i}$ be the restriction of T to ${\text{K}}_{{\lambda}_{i}}$, and let ${A}_{i}={[{\text{T}}_{i}]}_{{\beta}_{i}}$. Then ${A}_{i}$ is the Jordan canonical form of ${\text{T}}_{i}$, and
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 ${A}_{i}$ and the bases ${\beta}_{i}$, thereby computing J and $\beta $ as well. While developing a method for finding J, it becomes evident that in some sense the matrices ${A}_{i}$ are unique.
To aid in formulating the uniqueness theorem for J, we adopt the following convention: The basis ${\beta}_{i}$ for ${\text{K}}_{{\lambda}_{i}}$ will henceforth be ordered in such a way that the cycles appear in order of decreasing length. That is, if ${\beta}_{i}$ is a disjoint union of cycles ${\gamma}_{1},\text{}{\gamma}_{2},\text{}\dots ,\text{}{\gamma}_{{n}_{i}}$ and if the length of the cycle ${\gamma}_{j}$ is ${p}_{j}$, we index the cycles so that ${p}_{1}\ge {p}_{2}\ge \cdots \ge {p}_{{n}_{i}}$. This ordering of the cycles limits the possible orderings of vectors in ${\beta}_{i}$, which in turn determines the matrix ${A}_{i}$. It is in this sense that ${A}_{i}$ 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 ${\beta}_{i}$ or for $\beta $. However, we show that for each i, the number ${n}_{i}$ of cycles that form ${\beta}_{i}$, and the length ${p}_{j}(j=1,\text{}2,\text{}\dots ,\text{}{n}_{i})$ of each cycle, is completely determined by T.
Example 1
To illustrate the discussion above, suppose that, for some i, the ordered basis ${\beta}_{i}$ for ${\text{K}}_{{\lambda}_{i}}$ is the union of four cycles ${\beta}_{i}={\gamma}_{1}\cup {\gamma}_{2}\cup {\gamma}_{3}\cup {\gamma}_{4}$ with respective lengths ${p}_{1}=3,\text{}{p}_{2}=3,\text{}{p}_{3}=2$, and ${p}_{4}=1$. Then
To help us visualize each of the matrices ${A}_{i}$ and ordered bases ${\beta}_{i}$, we use an array of dots called a dot diagram of ${\text{T}}_{i}$, where ${\text{T}}_{i}$ is the restriction of T to ${\text{K}}_{{\lambda}_{i}}$. Suppose that ${\beta}_{i}$ is a disjoint union of cycles of generalized eigenvectors ${\gamma}_{1},\text{}{\gamma}_{2},\text{}\dots ,\text{}{\gamma}_{{n}_{i}}$ with lengths ${p}_{1}\ge {p}_{2}\ge \cdots \ge {p}_{{n}_{i}}$, respectively. The dot diagram of ${\text{T}}_{i}$ contains one dot for each vector in ${\beta}_{i}$, and the dots are configured according to the following rules.

The array consists of ${n}_{i}$ columns (one column for each cycle).

Counting from left to right, the jth column consists of the ${p}_{j}$ dots that correspond to the vectors of ${\gamma}_{j}$ starting with the initial vector at the top and continuing down to the end vector.
Denote the end vectors of the cycles by ${v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{{n}_{i}}$. In the following dot diagram of ${\text{T}}_{i}$, each dot is labeled with the name of the vector in ${\beta}_{i}$ to which it corresponds.
Notice that the dot diagram of ${\text{T}}_{i}$ has ${n}_{i}$ columns (one for each cycle) and ${p}_{1}$ rows. Since ${p}_{1}\ge {p}_{2}\ge \cdots \ge {p}_{{n}_{i}}$, the columns of the dot diagram become shorter (or at least not longer) as we move from left to right.
Now let ${r}_{j}$ denote the number of dots in the jth row of the dot diagram. Observe that ${r}_{1}\ge {r}_{2}\ge \cdots \ge {r}_{{p}_{1}}$. Furthermore, the diagram can be reconstructed from the values of the ${r}_{i}$’s. The proofs of these facts, which are combinatorial in nature, are treated in Exercise 9.
In Example 1, with ${n}_{i}=4,\text{}{p}_{1}={p}_{2}=3,\text{}{p}_{3}=2$, and ${p}_{4}=1$, the dot diagram of ${\text{T}}_{i}$ is as follows:
Here ${r}_{1}=4,\text{}{r}_{2}=3$, and ${r}_{3}=2$.
We now devise a method for computing the dot diagram of ${\text{T}}_{i}$ using the ranks of linear operators determined by T and ${\lambda}_{i}$. 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, ${\beta}_{i}$ is not unique. For example, see Exercise 8. (It is for this reason that we associate the dot diagram with ${\text{T}}_{i}$ rather than with ${\beta}_{i}$.)
To determine the dot diagram of ${\text{T}}_{i}$, we devise a method for computing each ${r}_{j}$, the number of dots in the jth row of the dot diagram, using only T and ${\lambda}_{i}$. The next three results give us the required method. To facilitate our arguments, we fix a basis ${\beta}_{i}$ for ${\text{K}}_{{\lambda}_{i}}$ so that ${\beta}_{i}$ is a disjoint union of ${n}_{i}$ cycles of generalized eigenvectors with lengths ${p}_{1}\ge {p}_{2}\ge \cdots \ge {p}_{{n}_{i}}$.
Theorem 7.9.
For any positive integer r, the vectors in ${\beta}_{i}$ that are associated with the dots in the first r rows of the dot diagram of ${\text{T}}_{i}$ constitute a basis for $\text{N}({(\text{T}{\lambda}_{i}\text{I})}^{r})$. Hence the number of dots in the first r rows of the dot diagram equals $\text{nullity}({(\text{T}{\lambda}_{i}\text{I})}^{r})$.
Proof.
Clearly, $\text{N}({(\text{T}{\lambda}_{i}\text{I})}^{r})\subseteq {\text{K}}_{{\lambda}_{i}}$, and ${\text{K}}_{{\lambda}_{i}}$ is invariant under ${(\text{T}{\lambda}_{i}\text{I})}^{r}$. Let U denote the restriction of ${(\text{T}{\lambda}_{i}\text{I})}^{r}$ to ${\text{K}}_{{\lambda}_{i}}$. By the preceding remarks, $\text{N}({(\text{T}{\lambda}_{i}\text{I})}^{r})=\text{N}(\text{U})$, and hence it suffices to establish the theorem for U. Now define
Let a and b denote the number of vectors in ${S}_{1}$ and ${S}_{2}$, respectively, and let ${m}_{i}=\mathrm{dim}({\text{K}}_{{\lambda}_{i}})$. Then $a+b={m}_{i}$. For any $x\in {\beta}_{i},\text{}x\in {S}_{1}$ 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 $x\in {S}_{2}$, 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 ${S}_{2}$ in a onetoone fashion into ${\beta}_{i}$. Thus $\{\text{U}(x):\text{}x\in {S}_{2}\}$ is a basis for R(U) consisting of b vectors. Hence $\text{rank}(\text{U})=b$, and so $\text{nullity}(\text{U})={m}_{i}b=a$. But ${S}_{1}$ is a linearly independent subset of N(U) consisting of a vectors; therefore ${S}_{1}$ is a basis for N(U).
In the case that $r=1$, Theorem 7.9 yields the following corollary.
Corollary.
The dimension of ${\text{E}}_{{\lambda}_{i}}$ is ${n}_{i}$. Hence in a Jordan canonical form of T, the number of Jordan blocks corresponding to ${\lambda}_{i}$ equals the dimension of ${\text{E}}_{{\lambda}_{i}}$.
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 ${r}_{j}$ denote the number of dots in the jth row of the dot diagram of ${\text{T}}_{i}$, the restriction of T to ${\text{K}}_{{\lambda}_{i}}$. Then the following statements are true.

${r}_{1}=\mathrm{dim}(\text{V})\text{rank}(\text{T}{\lambda}_{i}\text{I}).$

${r}_{j}=\text{rank}({(\text{T}{\lambda}_{i}\text{I})}^{j1})\text{rank}({(\text{T}{\lambda}_{i}\text{I})}^{j})$
if $j>1$.
Proof.
By Theorem 7.9, for $j=1,\text{}2,\text{}\dots ,\text{}p1$, we have
Hence
and for $j>1$,
Theorem 7.10 shows that the dot diagram of ${\text{T}}_{i}$ is completely determined by T and ${\lambda}_{i}$. Hence we have proved the following result.
Corollary.
For any eigenvalue ${\lambda}_{i}$ of T, the dot diagram of ${\text{T}}_{i}$ 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
We find the Jordan canonical form of A and a Jordan canonical basis for the linear operator $\text{T}={\text{L}}_{A}$. The characteristic polynomial of A is
Thus A has two distinct eigenvalues, ${\lambda}_{1}=2$ and ${\lambda}_{2}=3$, with multiplicities 3 and 1, respectively. Let ${\text{T}}_{1}$ and ${\text{T}}_{2}$ be the restrictions of ${\text{L}}_{A}$ to the generalized eigenspaces ${\text{K}}_{{\lambda}_{1}}$ and ${\text{K}}_{{\lambda}_{2}}$, respectively.
Suppose that ${\beta}_{1}$ is a Jordan canonical basis for ${\text{T}}_{1}$. Since ${\lambda}_{1}$ has multiplicity 3, it follows that $\mathrm{dim}({\text{K}}_{{\lambda}_{1}})=3$ by Theorem 7.4(c) (p. 480); hence the dot diagram of ${\text{T}}_{1}$ has three dots. As we did earlier, let ${r}_{j}$ denote the number of dots in the jth row of this dot diagram. Then, by Theorem 7.10,
and
(Actually, the computation of ${r}_{2}$ is unnecessary in this case because ${r}_{1}=2$ and the dot diagram only contains three dots.) Hence the dot diagram associated with ${\beta}_{1}$ is
So
Since ${\lambda}_{2}=3$ has multiplicity 1, it follows that $\mathrm{dim}({\text{K}}_{{\lambda}_{2}})=1$, and consequently any basis ${\beta}_{2}$ for ${\text{K}}_{{\lambda}_{2}}$ consists of a single eigenvector corresponding to ${\lambda}_{2}=3$. Therefore
Setting $\beta ={\beta}_{1}\cup {\beta}_{2}$, we have
and so J is the Jordan canonical form of A.
We now find a Jordan canonical basis for $\text{T}={\text{L}}_{A}$. We begin by determining a Jordan canonical basis ${\beta}_{1}$ for ${\text{T}}_{1}$. Since the dot diagram of ${\text{T}}_{1}$ has two columns, each corresponding to a cycle of generalized eigenvectors, there are two such cycles. Let ${v}_{1}$ and ${v}_{2}$ 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 ${v}_{1}\in \text{N}({(\text{T}2\text{I})}^{2})$ but ${v}_{1}\notin \text{N}(\text{T}2\text{I})$. Now
It is easily seen that
is a basis for $\text{N}({(\text{T}2\text{I})}^{2})={\text{K}}_{{\lambda}_{1}}$. Of these three basis vectors, the last two do not belong to $\text{N}(\text{T}2\text{I})$, and hence we select one of these for ${v}_{1}$. Suppose that we choose
Then
Now simply choose ${v}_{2}$ to be a vector in ${\text{E}}_{{\lambda}_{1}}$ that is linearly independent of $(\text{T}2\text{I})({v}_{1})$; for example, select
Thus we have associated the Jordan canonical basis
with the dot diagram in the following manner.
By Theorem 7.6 (p. 483), the linear independence of ${\beta}_{1}$ is guaranteed since ${v}_{2}$ was chosen to be linearly independent of $(\text{T}2\text{I})({v}_{1})$.
Since ${\lambda}_{2}=3$ has multiplicity 1, $\mathrm{dim}({\text{K}}_{{\lambda}_{2}})=\mathrm{dim}({\text{E}}_{{\lambda}_{2}})=1$. Hence any eigenvector of ${\text{L}}_{A}$ corresponding to ${\lambda}_{2}=3$ constitutes an appropriate basis ${\beta}_{2}$. For example,
Thus
is a Jordan canonical basis for ${\text{L}}_{A}$.
Notice that if
then $J={Q}^{1}AQ$.
Example 3
Let
We find the Jordan canonical form J of A, a Jordan canonical basis for ${\text{L}}_{A}$, and a matrix Q such that $J={Q}^{1}AQ$.
The characteristic polynomial of A is $\mathrm{det}(AtI)={(t2)}^{2}{(t4)}^{2}$. Let $\text{T}={\text{L}}_{A},\text{}{\lambda}_{1}=2$, and ${\lambda}_{2}=4$, and let ${\text{T}}_{i}$ be the restriction of ${\text{L}}_{A}$ to ${\text{K}}_{{\lambda}_{i}}$ for $i=1,\text{}2$.
We begin by computing the dot diagram of ${\text{T}}_{1}$. Let ${r}_{1}$ denote the number of dots in the first row of this diagram. Then
hence the dot diagram of ${\text{T}}_{1}$ is as follows.
Therefore
where ${\beta}_{1}$ is any basis corresponding to the dots. In this case, ${\beta}_{1}$ is an arbitrary basis for ${\text{E}}_{{\lambda}_{1}}=\text{N}(\text{T}2\text{I})$, for example,
Next we compute the dot diagram of ${\text{T}}_{2}$. Since $\text{rank}(A4I)=3$, there is only $43=1$ dot in the first row of the diagram. Since ${\lambda}_{2}=4$ has multiplicity 2, we have $\mathrm{dim}({\text{K}}_{{\lambda}_{2}})=2$, and hence this dot diagram has the following form:
Thus
where ${\beta}_{2}$ is any basis for ${\text{K}}_{{\lambda}_{2}}$ corresponding to the dots. In this case, ${\beta}_{2}$ is a cycle of length 2. The end vector of this cycle is a vector $v\in {\text{K}}_{{\lambda}_{2}}=\text{N}({(\text{T}4\text{I})}^{2})$ such that $v\notin \text{N}(\text{T}4\text{I})$. One way of finding such a vector was used to select the vector ${v}_{1}$ in Example 2. In this example, we illustrate another method. A simple calculation shows that a basis for the null space of ${\text{L}}_{A}4\text{I}$ is
Choose v to be any solution to the system of linear equations
for example,
Thus
Therefore
is a Jordan canonical basis for ${\text{L}}_{A}$. The corresponding Jordan canonical form is given by
Finally, we define Q to be the matrix whose columns are the vectors of $\beta $ listed in the same order, namely,
Then $J={Q}^{1}AQ$.
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 $\alpha =\{1,\text{}x,\text{}y,\text{}{x}^{2},\text{}{y}^{2},\text{}xy\}$ is an ordered basis for V. Let T be the linear operator on V defined by
For example, if $f(x,\text{}y)=x+2{x}^{2}3xy+y$, then
We find the Jordan canonical form and a Jordan canonical basis for T.
Let $A={[\text{T}]}_{\alpha}$. Then
and hence the characteristic polynomial of T is
Thus $\lambda =0$ is the only eigenvalue of T, and ${\text{K}}_{\lambda}=\text{V}$. For each j, let ${r}_{j}$ denote the number of dots in the jth row of the dot diagram of T. By Theorem 7.10,
and since
${r}_{2}=\text{rank}(A)\text{rank}({A}^{2})=31=2$.
Because there are a total of six dots in the dot diagram and ${r}_{1}=3$ and ${r}_{2}=2$, it follows that ${r}_{3}=1$. 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 ${f}_{1}(x,\text{}y)$ such that $\frac{{\partial}^{2}}{\partial {x}^{2}}}{f}_{1}(x,\text{}y)\ne \mathit{0$. Examining the basis $\alpha =\{1,\text{}x,\text{}y,\text{}{x}^{2},\text{}{y}^{2},\text{}xy\}$ for ${\text{K}}_{\lambda}=\text{V}$, we see that ${x}^{2}$ is a suitable candidate. Setting ${f}_{1}(x,\text{}y)={x}^{2}$, we see that
and
Likewise, since the second column of the dot diagram consists of two dots, we must find a polynomial ${f}_{2}(x,\text{}y)$ such that
Since our choice must be linearly independent of the polynomials already chosen for the first cycle, the only choice in $\alpha $ that satisfies these constraints is xy. So we set ${f}_{2}(x,\text{}y)=xy$. Thus
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 $\alpha $ is ${y}^{2}$, and it is suitable here. So set ${f}_{3}(x,\text{}y)={y}^{2}$. Therefore we have identified polynomials with the dots in the dot diagram as follows.
Thus $\beta =\{2,\text{}2x,\text{}{x}^{2},\text{}y,\text{}xy,\text{}{y}^{2}\}$ 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 $n\times n$ 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 ${J}_{A}$ and ${J}_{B}$ denote the Jordan canonical forms of A and B, respectively, with the same ordering of their eigenvalues. Then A is similar to both ${J}_{A}$ and ${J}_{B}$, and therefore, by the corollary to Theorem 2.23 (p. 115), ${J}_{A}$ and ${J}_{B}$ are matrix representations of ${\text{L}}_{A}$. Hence ${J}_{A}$ and ${J}_{B}$ are Jordan canonical forms of ${\text{L}}_{A}$. Thus ${J}_{A}={J}_{B}$ by the corollary to Theorem 7.10.
Example 5
We determine which of the matrices
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 ${J}_{A},\text{}{J}_{B}$, and ${J}_{C}$ be the Jordan canonical forms of A, B, and C, respectively, using the ordering 1, 2 for their common eigenvalues. Then (see Exercise 4)
Since ${J}_{A}={J}_{C}$, A is similar to C. Since ${J}_{B}$ is different from ${J}_{A}$ and ${J}_{C}$, 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 finitedimensional 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

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

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

Let T be a linear operator on a finitedimensional vector space V that has a Jordan canonical form J. If $\beta $ is any basis for V, then the Jordan canonical form of ${[\text{T}]}_{\beta}$ is J.

Linear operators having the same characteristic polynomial are similar.

Matrices having the same Jordan canonical form are similar.

Every matrix is similar to its Jordan canonical form.

Every linear operator with the characteristic polynomial ${(1)}^{n}{(t\lambda )}^{n}$ has the same Jordan canonical form.

Every linear operator on a finitedimensional vector space has a unique Jordan canonical basis.

The dot diagrams of a linear operator on a finitedimensional vector space are unique.


Let T be a linear operator on a finitedimensional vector space V such that the characteristic polynomial of T splits. Suppose that ${\lambda}_{1}=2,\text{}{\lambda}_{2}=4$, and ${\lambda}_{3}=3$ are the distinct eigenvalues of T and that the dot diagrams for the restriction of T to ${\text{K}}_{{\lambda}_{i}}(i=1,\text{}2,\text{}3)$ are as follows:
${\lambda}_{1}=2$ ${\lambda}_{2}=4$ ${\lambda}_{3}=3$ Find the Jordan canonical form J of T.

Let T be a linear operator on a finitedimensional vector space V with Jordan canonical form

Find the characteristic polynomial of T.

Find the dot diagram corresponding to each eigenvalue of T.

For which eigenvalues ${\lambda}_{i}$, if any, does ${\text{E}}_{{\lambda}_{i}}={\text{K}}_{{\lambda}_{i}}$?

For each eigenvalue ${\lambda}_{i}$, find the smallest positive integer ${p}_{i}$ for which ${\text{K}}_{{\lambda}_{i}}=\text{N}({(\text{T}{\lambda}_{i}\text{I})}^{{p}_{i}})$.

Compute the following numbers for each i, where ${\text{U}}_{i}$ denotes the restriction of $\text{T}{\lambda}_{i}\text{I}$ to ${\text{K}}_{{\lambda}_{i}}$.

$\text{rank}({\text{U}}_{\text{i}})$

$\text{rank}({\text{U}}_{i}^{2})$

$\text{nullity}({\text{U}}_{i})$

$\text{nullity}({\text{U}}_{i}^{2})$



For each of the matrices A that follow, find a Jordan canonical form J and an invertible matrix Q such that $J={Q}^{1}AQ$. Notice that the matrices in (a), (b), and (c) are those used in Example 5.

$A=\left(\begin{array}{rrr}3& 3& 2\\ 7& 6& 3\\ 1& 1& 2\end{array}\right)$

$A=\left(\begin{array}{rrr}0& 1& 1\\ 4& 4& 2\\ 2& 1& 1\end{array}\right)$

$A=\left(\begin{array}{rrr}0& 1& 1\\ 3& 1& 2\\ 7& 5& 6\end{array}\right)$

$A=\left(\begin{array}{rrrr}0& 3& 1& 2\\ 2& 1& 1& 2\\ 2& 1& 1& 2\\ 2& 3& 1& 4\end{array}\right)$


For each linear operator T, find a Jordan canonical form J of T and a Jordan canonical basis $\beta $ for T.

(a) V is the real vector space of functions spanned by the set of realvalued functions $\{{e}^{t},\text{}t{e}^{t},\text{}{t}^{2}{e}^{t},\text{}{e}^{2t}\}$, and T is the linear operator on V defined by $\text{T}(f)={f}^{\prime}$.

(b) T is the linear operator on ${\text{P}}_{3}(R)$ defined by $\text{T}(f(x))=x{f}^{\u2033}(x)$.

(c) T is the linear operator on ${\text{P}}_{3}(R)$ defined by $\text{T}(f(x))={f}^{\u2033}(x)+2f(x)$

(d) T is the linear operator on ${\text{M}}_{2\times 2}(R)$ defined by
$$\text{T}(A)=\left(\begin{array}{rr}3& 1\\ 0& 3\end{array}\right)\cdot A{A}^{t}.$$ 
(e) T is the linear operator on ${\text{M}}_{2\times 2}(R)$ defined by
$$\text{T}(A)=\left(\begin{array}{rr}3& 1\\ 0& 3\end{array}\right)\cdot (A{A}^{t}).$$ 
(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
$$\text{T}(f(x,\text{}y))={\displaystyle \frac{\partial}{\partial x}}f(x,\text{}y)+{\displaystyle \frac{\partial}{\partial y}}f(x,\text{}y).$$


Let A be an $n\times n$ matrix whose characteristic polynomial splits. Prove that A and ${A}^{t}$ have the same Jordan canonical form, and conclude that A and ${A}^{t}$ are similar. Hint: For any eigenvalue $\lambda $ of A and ${A}^{t}$ and any positive integer r, show that $\text{rank}({(A\lambda I)}^{r})=\text{rank}({({A}^{t}\lambda I)}^{r})$.

Let A be an $n\times n$ matrix whose characteristic polynomial splits, $\gamma $ be a cycle of generalized eigenvectors corresponding to an eigenvalue $\lambda $, and W be the subspace spanned by $\gamma $. Define ${\gamma}^{\prime}$ to be the ordered set obtained from $\gamma $ by reversing the order of the vectors in $\gamma $.

(a) Prove that ${[{\text{T}}_{\text{W}}]}_{{\gamma}^{\prime}}={({[{\text{T}}_{\text{W}}]}_{\gamma})}^{t}$.

(b) Let J be the Jordan canonical form of A. Use (a) to prove that J and ${J}^{t}$ are similar.

(c) Use (b) to prove that A and ${A}^{t}$ are similar.


Let T be a linear operator on a finitedimensional vector space, and suppose that the characteristic polynomial of T splits. Let $\beta $ be a Jordan canonical basis for T.

(a) Prove that for any nonzero scalar c, $\{cx:\text{}x\in \beta \}$ is a Jordan canonical basis for T.

(b) Suppose that $\gamma $ is one of the cycles of generalized eigenvectors that forms $\beta $, and suppose that $\gamma $ corresponds to the eigenvalue $\lambda $ and has length greater than 1. Let x be the end vector of $\gamma $, and let y be a nonzero vector in ${\text{E}}_{\lambda}$. Let ${\gamma}^{\prime}$ be the ordered set obtained from $\gamma $ by replacing x by $x+y$. Prove that ${\gamma}^{\prime}$ is a cycle of generalized eigenvectors corresponding to $\lambda $, and that if ${\gamma}^{\prime}$ replaces $\gamma $ in the union that defines $\beta $, then the new union is also a Jordan canonical basis for T.

(c) Apply (b) to obtain a Jordan canonical basis for ${\text{L}}_{A}$, where A is the matrix given in Example 2, that is different from the basis given in the example.


Suppose that a dot diagram has k columns and m rows with ${p}_{j}$ dots in column j and ${r}_{i}$ dots in row i. Prove the following results.

(a) $m={p}_{1}$ and $k={r}_{1}$.

(b) ${p}_{j}=\mathrm{max}\{i:\text{}{r}_{i}\ge j\}$ for $1\le j\le k$ and ${r}_{i}=\mathrm{max}\{j:\text{}{p}_{j}\ge i\}$ for $1\le i\le m$. Hint: Use mathematical induction on m.

(c) ${r}_{1}\ge {r}_{2}\ge \cdots \ge {r}_{m}.$

(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.


Let T be a linear operator whose characteristic polynomial splits, and let $\lambda $ be an eigenvalue of T.

(a) Prove that $\mathrm{dim}({\text{K}}_{\lambda})$ is the sum of the lengths of all the blocks corresponding to $\lambda $ in the Jordan canonical form of T.

(b) Deduce that ${\text{E}}_{\lambda}={\text{K}}_{\lambda}$ if and only if all the Jordan blocks corresponding to $\lambda $ are $1\times 1$ matrices.

The following definitions are used in Exercises 11–19.
Definitions.
A linear operator T on a vector space V is called nilpotent if ${\text{T}}^{p}={\text{T}}_{0}$ for some positive integer p. An $n\times n$ matrix A is called nilpotent if ${A}^{p}=O$ for some positive integer p.

Let T be a linear operator on a finitedimensional vector space V, and let $\beta $ be an ordered basis for V. Prove that T is nilpotent if and only if ${[\text{T}]}_{\beta}$ is nilpotent.

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

Let T be a nilpotent operator on an ndimensional vector space V, and suppose that p is the smallest positive integer for which ${\text{T}}^{p}={\text{T}}_{0}$. Prove the following results.

(a) $\text{N}({\text{T}}^{i})\subseteq \text{N}({\text{T}}^{i+1})$ for every positive integer i.

(b) There is a sequence of ordered bases ${\beta}_{1},\text{}{\beta}_{2},\text{}\dots ,\text{}{\beta}_{p}$ such that ${\beta}_{i}$ is a basis for $\text{N}({\text{T}}^{i})$ and ${\beta}_{i+1}$ contains ${\beta}_{i}$ for $1\le i\le p1$.

(c) Let $\beta {\beta}_{p}$ be the ordered basis for $\text{N}({\text{T}}^{p})=\text{V}$ V in (b). Then ${[\text{T}]}_{\beta}$ is an upper triangular matrix with each diagonal entry equal to zero.

(d) The characteristic polynomial of T is ${(1)}^{n}{t}^{n}$. Hence the characteristic polynomial of T splits, and 0 is the only eigenvalue of T.


Prove the converse of Exercise 13(d): If T is a linear operator on an ndimensional vector space V and ${(1)}^{n}{t}^{n}$ is the characteristic polynomial of T, then T is nilpotent.

Give an example of a linear operator T on a finitedimensional 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. 
Let T be a nilpotent linear operator on a finitedimensional vector space V. Recall from Exercise 13 that $\lambda =0$ is the only eigenvalue of T, and hence $\text{V}={\text{K}}_{\lambda}$. Let $\beta $ be a Jordan canonical basis for T. Prove that for any positive integer i, if we delete from $\beta $ the vectors corresponding to the last i dots in each column of a dot diagram of $\beta $, the resulting set is a basis for $\text{R}({\text{T}}^{i})$. (If a column of the dot diagram contains fewer than i dots, all the vectors associated with that column are removed from $\beta $.)

Let T be a linear operator on a finitedimensional vector space V such that the characteristic polynomial of T splits, and let ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ be the distinct eigenvalues of T. For each i, let ${v}_{i}$ denote the unique vector in ${\text{K}}_{{\lambda}_{i}}$ such that $x={v}_{1}+{v}_{2}+\cdots +{v}_{k}$. (This unique representation is guaranteed by Theorem 7.3 (p. 479).) Define a mapping $\text{S}:\text{V}\to \text{V}$ by
$$\text{S}(x)={\lambda}_{1}{v}_{1}+{\lambda}_{2}{v}_{2}+\cdots +{\lambda}_{k}{v}_{k}.$$
(a) Prove that S is a diagonalizable linear operator on V.

(b) Let $\text{U}=\text{T}\text{S}$. Prove that U is nilpotent and commutes with S, that is, $\text{SU}=\text{US}$.


Let T be a linear operator on a finitedimensional 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.

(a) M is nilpotent.

(b) $MD=DM.$

(c) If p is the smallest positive integer for which ${M}^{p}=O$, then, for any positive integer $r<p$,
$${J}^{r}={D}^{r}+r{D}^{r1}M+{\displaystyle \frac{r(r1)}{2!}}{D}^{r2}{M}^{2}+\cdots +rD{M}^{r1}+{M}^{r},$$and, for any positive integer $r\ge p$,
$$\begin{array}{l}{J}^{r}={D}^{r}+r{D}^{r1}M+{\displaystyle \frac{r(r1)}{2!}}{D}^{r2}{M}^{2}+\cdots \\ \phantom{\rule{8.7em}{0ex}}+{\displaystyle \frac{r!}{(rp+1)!(p1)!}}{D}^{rp+1}{M}^{p1}.\end{array}$$


Let $F=C$ and
$$J=\left(\begin{array}{ccccc}\lambda & 1& 0& \cdots & 0\\ 0& \lambda & 1& \cdots & 0\\ 0& 0& \lambda & \cdots & 0\\ \vdots & \vdots & \vdots & & \vdots \\ 0& 0& 0& \cdots & 1\\ 0& 0& 0& \cdots & \lambda \end{array}\right)$$be the $m\times m$ Jordan block corresponding to $\lambda $, and let $N=T\lambda {I}_{m}$. Prove the following results:

(a) ${N}^{m}=O$, and for $1\le r<m$,
$${N}_{ij}^{r}=\{\begin{array}{ll}1& \text{if}j=i+r\\ 0& \text{otherwise}\text{.}\end{array}$$ 
(b) For any integer $r\ge m$,
$${J}^{r}=\left(\begin{array}{ccccc}{\lambda}^{r}& r{\lambda}^{r1}& {\displaystyle \frac{r(r1)}{2!}}{\lambda}^{r2}& \cdots & {\displaystyle \frac{r(r1)\cdots (rm+2)}{(m1)!}}{\lambda}^{rm+1}\\ 0& {\lambda}^{r}& r{\lambda}^{r1}& \cdots & {\displaystyle \frac{r(r1)\cdots (rm+3)}{(m2)!}}{\lambda}^{rm+2}\\ \vdots & \vdots & \vdots \text{}& & \vdots \\ 0& 0& 0& \cdots & {\lambda}^{r}\end{array}\right).$$ 
(c) $\underset{r\to \infty}{\mathrm{lim}}{J}^{r}$ exists if and only if one of the following holds:

$\left\lambda \right<1$.

$\lambda =1$ and $m=1$.
(Note that $\underset{r\to \infty}{\mathrm{lim}}{\lambda}^{r}$ exists under these conditions. See the discussion preceding Theorem 5.12 on page 284.) Furthermore, $\underset{r\to \infty}{\mathrm{lim}}{J}^{r}$ is the zero matrix if condition (i) holds and is the $1\times 1$ matrix (1) if condition (ii) holds.


(d) Prove Theorem 5.12 on page 284.

The following definition is used in Exercises 20 and 21.
Definition.
For any $A\in {\text{M}}_{n\times n}(C)$, define the norm of A by

Let $A,\text{}B\in {\text{M}}_{n\times n}(C)$. Prove the following results.

(a) $\left\rightA{}_{m}\ge 0$.

(b) $\left\rightA{}_{m}=0$ if and only if $A=O$.

(c) $\left\rightcA{}_{m}=c\cdot \leftA\right{}_{m}$ for any scalar c.

(d) $\left\rightA+B{}_{m}\le \leftA\right{}_{m}+\left\rightB{}_{m}$.

(e) $\left\rightAB{}_{m}\le n\leftA\right{}_{m}\left\rightB{}_{m}.$


Let $A\in {\text{M}}_{n\times n}(C)$ 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 ${P}^{1}AP=J$. Prove the following results.

(a) $\left\right{A}^{k}{}_{m}\le 1$ for every positive integer k.

(b) There exists a positive number c such that $\left\right{J}^{k}{}_{m}\le c$ for every positive integer k.

(c) Each Jordan block of J corresponding to the eigenvalue $\lambda =1$ is a $1\times 1$ matrix.

(d) $\underset{k\to \infty}{\mathrm{lim}}{A}^{k}$ exists if and only if 1 is the only eigenvalue of A with absolute value 1.

(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 ${e}^{A}$ for a matrix A. (See page 310.)

Use Exercise 20(d) to prove that ${e}^{A}$ exists for every $A\in {\text{M}}_{n\times n}(C)$.

Let ${x}^{\prime}=Ax$ be a system of n linear differential equations, where x is an ntuple of differentiable functions ${x}_{1}(t),\text{}{x}_{2}(t),\text{}\dots ,\text{}{x}_{n}(t)$ of the real variable t, and A is an $n\times n$ 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 ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ be the distinct eigenvalues of A.

(a) Prove that if u is the end vector of a cycle of generalized eigenvectors of ${\text{L}}_{A}$ of length p and u corresponds to the eigenvalue ${\lambda}_{i}$, then for any polynomial f(t) of degree less than p, the function
$${e}^{{\lambda}_{i}t}[f(t){(A{\lambda}_{i}I)}^{p1}+{f}^{\prime}(t){(A{\lambda}_{i}I)}^{p2}+\cdots +{f}^{(p1)}(t)]u$$is a solution to the system ${x}^{\prime}=Ax$.

(b) Prove that the general solution to ${x}^{\prime}=Ax$ 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 ${\text{L}}_{A}$.


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

$\begin{array}{rcrcrcr}{x}^{\prime}& =& 2x& +& y& & \\ {y}^{\prime}& =& & & 2y& & z\\ {z}^{\prime}& =& & & & & 3z\end{array}$

$\begin{array}{rcrcrcr}{x}^{\prime}& =& 2x& +& y& & \\ {y}^{\prime}& =& & & 2y& +& z\\ {z}^{\prime}& =& & & & & 2z\end{array}$
