# 5.2 Diagonalizability

In Section 5.1, we presented the diagonalization problem and observed that not all linear operators or matrices are diagonalizable. Although we are able to diagonalize operators and matrices and even obtain a necessary and sufficient condition for diagonalizability (Theorem 5.1 p. 247), we have not yet solved the diagonalization problem. What is still needed is a simple test to determine whether an operator or a matrix can be diagonalized, as well as a method for actually finding a basis of eigenvectors. In this section, we develop such a test and method.

In Example 6 of Section 5.1, we obtained a basis of eigenvectors by choosing one eigenvector corresponding to each eigenvalue. In general, such a procedure does not yield a basis, but the following theorem shows that any set constructed in this manner is linearly independent.

# Theorem 5.5.

*Let* T *be a linear operator on a vector space, and let* ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ *be distinct eigenvalues of* T. *For each* $i=1,\text{}2,\text{}\mathrm{...},\text{}k$, *let* ${S}_{i}$ *be a finite set of eigenvectors of* T *corresponding to* ${\lambda}_{i}$. *If each* ${S}_{i}(i=1,\text{}2,\text{}\mathrm{...},\text{}k)$, *is linearly independent, then* ${S}_{1}\cup {S}_{2}\cup \cdots \cup {S}_{k}$ *is linearly independent.*

# Proof.

The proof is by mathematical induction on `k`. If $k=1$, there is nothing to prove. So assume that the theorem holds for $k-1$ distinct eigenvalues, where $k-1\ge 1$, and that we have `k` distinct eigenvalues ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\mathrm{...},\text{}{\lambda}_{k}$ of T. For each $i=1,\text{}2,\text{}\mathrm{...},\text{}k$, let ${S}_{i}=\{{v}_{i1},\text{}{v}_{i2},\text{}\mathrm{...},\text{}{v}_{i{n}_{i}}\}$ be a linearly independent set of eigenvectors of T corresponding to ${\lambda}_{i}$. We wish to show that $S={S}_{1}\cup {S}_{2}\cup \cdots \cup {S}_{k}$ is linearly independent.

Consider any scalars $\left\{{a}_{ij}\right\}$, where $i=1,\text{}2,\text{}\mathrm{...},\text{}k$ and $j=1,\text{}2,\text{}\mathrm{...},\text{}{n}_{i}$, such that

Because ${v}_{ij}$ is an eigenvector of T corresponding to ${\lambda}_{i}$, applying $\text{T}-{\lambda}_{k}\text{I}$ to both sides of (1) yields

But ${S}_{1}\cup {S}_{2}\cup \cdots \cup {S}_{k-1}$ is linearly independent by the induction hypothesis, so that (2) implies ${a}_{ij}({\lambda}_{i}-{\lambda}_{k})=0$ for $i=1,\text{}2,\text{}\dots ,\text{}k-1$ and $j=1,\text{}2,\text{}\dots ,\text{}{n}_{i}$. Since ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ are distinct, it follows that ${\lambda}_{i}-{\lambda}_{k}\ne 0$ for $1\le i\le k-1$. Hence ${a}_{ij}=0$ for $i=1,\text{}2,\text{}\dots ,\text{}k-1$ and $j=1,\text{}2,\text{}\mathrm{...}\text{}{n}_{i}$, and therefore (1) reduces to ${\sum}_{j=1}^{{n}_{k}}{a}_{kj}}{v}_{kj}=0$. But ${S}_{k}$ is also linearly independent, and so ${a}_{kj}=0$ for $j=1,\text{}2,\text{}\mathrm{...}\text{}{n}_{k}$. Consequently ${a}_{ij}=0$ for $i=1,\text{}2,\text{}\mathrm{...},\text{}k$ and $j=1,\text{}2,\text{}\mathrm{...}\text{}{n}_{i}$, proving that `S` is linearly independent.

# Corollary.

*Let* T *be a linear operator on an n-dimensional vector space* V

*. If*T

*has*T

`n`distinct eigenvalues, then*is diagonalizable.*

# Proof.

Suppose that T has `n` distinct eigenvalues ${\lambda}_{1},\text{}\mathrm{...},\text{}{\lambda}_{n}$. For each `i` choose an eigenvector ${v}_{i}$ corresponding to ${\lambda}_{i}$. By Theorem 5.5, $\{{v}_{1},\text{}\mathrm{...},\text{}{v}_{n}\}$ is linearly independent, and since $\text{dim}(\text{V})=n$, this set is a basis for V. Thus, by Theorem 5.1 (p. 247), T is diagonalizable.

# Example 1

Let

The characteristic polynomial of A (and hence of ${\text{L}}_{A}$) is

and thus the eigenvalues of ${\text{L}}_{A}$ are 0 and 2. Since ${\text{L}}_{A}$ is a linear operator on the two-dimensional vector space ${\text{R}}^{2}$, we conclude from the preceding corollary that ${\text{L}}_{A}$ (and hence `A`) is diagonalizable.

The converse of the corollary to Theorem 5.5 is false. That is, it is not true that if T is diagonalizable, then it has `n` distinct eigenvalues. For example, the identity operator is diagonalizable even though it has only one eigenvalue, namely, $\lambda =1$.

We have seen that diagonalizability requires the existence of eigenvalues. Actually, diagonalizability imposes a stronger condition on the characteristic polynomial.

# Definition.

*A polynomial f*(`t`) *in* P(`F`) *splits over F if there are scalars* $c,\text{}{a}_{1},\text{}\mathrm{...},\text{}{a}_{n}$ *(not necessarily distinct) in F such that*

For example, ${t}^{2}-1=(t+1)(t-1)$ splits over `R`, but $({t}^{2}+1)(t-2)$ does not split over `R` because ${t}^{2}+1$ cannot be factored into a product of linear factors. However, $({t}^{2}+1)(t-2)$ does split over `C` because it factors into the product $(t+i)(t-i)(t-2)$. If `f`(`t`) is the characteristic polynomial of a linear operator or a matrix over a field `F`, then the statement that `f`(`t`) splits is understood to mean that it splits over `F`.

# Theorem 5.6.

*The characteristic polynomial of any diagonalizable linear operator on a vector space* V *over a field F splits over F*.

# Proof.

Let T be a diagonalizable linear operator on the `n`-dimensional vector space V, and let $\beta $ be an ordered basis for V such that ${[\text{T}]}_{\beta}=D$ is a diagonal matrix. Suppose that

and let `f` (`t`) be the characteristic polynomial of T. Then

From this theorem, it is clear that if T is a diagonalizable linear operator on an `n`-dimensional vector space that fails to have `n` distinct eigenvalues, then the characteristic polynomial of T must have repeated zeros.

The converse of Theorem 5.6 is false; that is, the characteristic polynomial of T may split, but T need not be diagonalizable. (See Example 3, which follows.) The following concept helps us determine when an operator whose characteristic polynomial splits is diagonalizable.

# Definition.

*Let* $\lambda $ *be an eigenvalue of a linear operator or matrix with characteristic polynomial f*(`t`). *The multiplicity (sometimes called the algebraic multiplicity) of* $\lambda $ *is the largest positive integer k for which* ${(t-\lambda )}^{k}$

*is a factor of f*(

`t`).

# Example 2

Let

which has characteristic polynomial $f(t)=-{(t-3)}^{2}(t-4)$. Hence $\lambda =3$ is an eigenvalue of `A` with multiplicity 2, and $\lambda =4$ is an eigenvalue of `A` with multiplicity 1.

If T is a diagonalizable linear operator on a finite-dimensional vector space V, then there is an ordered basis $\beta $ for V consisting of eigenvectors of T. We know from Theorem 5.1 (p. 247) that ${[\text{T}]}_{\beta}$ is a diagonal matrix in which the diagonal entries are the eigenvalues of T. Since the characteristic polynomial of T is $\text{det}({[\text{T}]}_{\beta}-tI)$, it is easily seen that each eigenvalue of T must occur as a diagonal entry of ${[\text{T}]}_{\beta}$ exactly as many times as its multiplicity. Hence $\beta $ contains as many (linearly independent) eigenvectors corresponding to an eigenvalue as the multiplicity of that eigenvalue. So the number of linearly independent eigenvectors corresponding to a given eigenvalue is of interest in determining whether an operator can be diagonalized. Recalling from Theorem 5.4 (p. 250) that the eigenvectors of T corresponding to the eigenvalue $\lambda $ are the nonzero vectors in the null space of $\text{T}-\lambda \text{I}$, we are led naturally to the study of this set.

# Definition.

*Let* T *be a linear operator on a vector space* V, *and let* $\lambda $ *be an eigenvalue of* T. *Define* ${\text{E}}_{\lambda}=\{x\in \text{V}:\text{}\text{T}(x)=\lambda x\}=\text{N}(\text{T}-\lambda {\text{I}}_{\text{V}})$. *The set* ${\text{E}}_{\lambda}$ *is called the eigenspace of* T *corresponding to the eigenvalue $\lambda $. Analogously, we define the eigenspace of a square matrix A corresponding to the eigenvalue* $\lambda $ *to be the eigenspace of* ${\text{L}}_{A}$ *corresponding to* $\lambda $*.*

Clearly, ${\text{E}}_{\lambda}$ is a subspace of V consisting of the zero vector and the eigenvectors of T corresponding to the eigenvalue $\lambda $. The maximum number of linearly independent eigenvectors of T corresponding to the eigenvalue $\lambda $ is therefore the dimension of ${\text{E}}_{\lambda}$. Our next result relates this dimension to the multiplicity of $\lambda $.

# Theorem 5.7.

*Let* T *be a linear operator on a finite-dimensional vector space* V, *and let* $\lambda $ *be an eigenvalue of* T *having multiplicity m. Then* $1\le \text{dim}({\text{E}}_{\lambda})\le m$.

# Proof.

Choose an ordered basis $\{{v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{p}\}$ for ${\text{E}}_{\lambda}$, extend it to an ordered basis $\beta =\left\{{v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{p},\text{}{v}_{p+1},\text{}\dots ,\text{}{v}_{n}\right\}$ for V, and let $A={[\text{T}]}_{\beta}$. Observe that ${v}_{i}(1\le i\le p)$ is an eigenvector of T corresponding to $\lambda $, and therefore

By Exercise 21 of Section 4.3, the characteristic polynomial of T is

where `g`(`t`) is a polynomial. Thus ${(\lambda -t)}^{p}$ is a factor of `f`(`t`), and hence the multiplicity of $\lambda $ is at least `p`. But $\text{dim}({\text{E}}_{\lambda})=p$, and so $\text{dim}({\text{E}}_{\lambda})\le m$.

# Example 3

Let T be the linear operator on ${\text{P}}_{2}(R)$ defined by $\text{T}(f(x))={f}^{\prime}(x)$. The matrix representation of T with respect to the standard ordered basis $\beta $ for ${\text{P}}_{2}(R)$ is

Consequently, the characteristic polynomial of T is

Thus T has only one eigenvalue $(\lambda =0)$ with multiplicity 3. Solving $\text{T}(f(x))={f}^{\prime}(x)=0$ shows that ${\text{E}}_{\lambda}=\text{N}(\text{T}-\lambda \text{I})=\text{N}(\text{T})$ is the subspace of ${\text{P}}_{2}(R)$ consisting of the constant polynomials. So {1} is a basis for ${\text{E}}_{\lambda}$, and therefore $\text{dim}({\text{E}}_{\lambda})=1$. Consequently, there is no basis for ${\text{P}}_{2}(R)$ consisting of eigenvectors of T, and therefore T is not diagonalizable. Even though T is not diagonalizable, we will see in Chapter 7 that its eigenvalue and eigenvectors are still useful for describing the behavior of T.

# Example 4

Let T be the linear operator on ${\text{R}}^{3}$ defined by

We determine the eigenspace of T corresponding to each eigenvalue. Let $\beta $ be the standard ordered basis for ${\text{R}}^{3}$. Then

and hence the characteristic polynomial of T is

So the eigenvalues of T are ${\lambda}_{1}=5$ and ${\lambda}_{2}=3$ with multiplicities 1 and 2, respectively.

Since

${\text{E}}_{{\lambda}_{1}}$ is the solution space of the system of linear equations

It is easily seen (using the techniques of Chapter 3) that

is a basis for ${\text{E}}_{{\lambda}_{1}}$. Hence $\text{dim}({\text{E}}_{{\lambda}_{1}})=1$.

Similarly, ${\text{E}}_{{\lambda}_{2}}=\text{N}(\text{T}-{\lambda}_{2}\text{I})$ is the solution space of the system

Since the unknown ${x}_{2}$ does not appear in this system, we assign it a parametric value, say, ${x}_{2}=s$, and solve the system for ${x}_{1}$ and ${x}_{3}$, introducing another parameter `t`. The result is the general solution to the system

It follows that

is a basis for ${\text{E}}_{{\lambda}_{2}}$, and $\text{dim}({\text{E}}_{{\lambda}_{2}})=2$.

In this case, the multiplicity of each eigenvalue ${\lambda}_{i}$ is equal to the dimension of the corresponding eigenspace ${\text{E}}_{{\lambda}_{i}}$. Observe that the union of the two bases just derived, namely,

is linearly independent by Theorem 5.5 and hence is a basis for ${\text{R}}^{3}$ consisting of eigenvectors of T. Consequently, T is diagonalizable.

Examples 3 and 4 suggest that an operator on V whose characteristic polynomial splits is diagonalizable if and only if the dimension of each eigenspace is equal to the multiplicity of the corresponding eigenvalue. This is indeed true, as our next theorem shows. Moreover, when the operator is diagonalizable, we can use Theorem 5.5 to construct a basis for V consisting of eigenvectors of the operator by collecting bases for the individual eigenspaces.

# Theorem 5.8.

*Let* T *be a linear operator on a finite-dimensional 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. *Then*

(a) T

*is diagonalizable if and only if the multiplicity of*${\lambda}_{i}$*is equal to*$\text{dim}({\text{E}}_{{\lambda}_{i}})$*for all i.*(b)

*If*T*is diagonalizable and*${\beta}_{i}$*is an ordered basis for*${\text{E}}_{{\lambda}_{i}}$*for each i, then*$\beta ={\beta}_{1}\cup {\beta}_{2}\cup \cdots \cup {\beta}_{k}$*is an ordered basis*^{2}*for*V*consisting of eigenvectors of*T.

# Proof.

For each `i`, let ${m}_{i}$ denote the multiplicity of ${\lambda}_{i},\text{}{d}_{i}=\text{dim}({\text{E}}_{{\lambda}_{i}})$, and $n=\text{dim}(\text{V})$.

First, suppose that T is diagonalizable. Let $\beta $ be a basis for V consisting of eigenvectors of T. For each `i`, let ${\beta}_{i}=\beta \cap {\text{E}}_{{\lambda}_{i}}$, the set of vectors in $\beta $ that are eigenvectors corresponding to ${\lambda}_{i}$, and let ${n}_{i}$ denote the number of vectors in ${\beta}_{i}$. Then ${n}_{i}\le {d}_{i}$ for each `i` because ${\beta}_{i}$ is a linearly independent subset of a subspace of dimension ${d}_{i}$, and ${d}_{i}\le {m}_{i}$ by Theorem 5.7. The ${n}_{i}$’s sum to `n` because $\beta $ contains `n` vectors. The ${m}_{i}$’s also sum to `n` because the degree of the characteristic polynomial of T is equal to the sum of the multiplicities of the eigenvalues. Thus

It follows that

Since $({m}_{i}-{d}_{i})\ge 0$ for all `i`, we conclude that ${m}_{i}={d}_{i}$ for all `i`.

Conversely, suppose that ${m}_{i}={d}_{i}$ for all `i`. We simultaneously show that T is diagonalizable and prove (b). For each `i`, let ${\beta}_{i}$ be an ordered basis for ${\text{E}}_{{\lambda}_{i}}$, and let $\beta ={\beta}_{1}\cup {\beta}_{2}\cup \cdots \cup {\beta}_{k}$. By Theorem 5.5, $\beta $ is linearly independent. Furthermore, since ${d}_{i}={m}_{i}$ for all `i`, $\beta $ contains

vectors. Therefore $\beta $ is an ordered basis for V consisting of eigenvectors of V, and we conclude that T is diagonalizable.

This theorem completes our study of the diagonalization problem. We summarize our results.

# Test for Diagonalizability

Let T be a linear operator on an `n`-dimensional vector space V. Then T is diagonalizable if and only if both of the following conditions hold.

The characteristic polynomial of T splits.

For each eigenvalue $\lambda $ of T, the multiplicity of $\lambda $ equals $\text{nullity}(\text{T}-\lambda \text{I})$, that is, the multiplicity of $\lambda $ equals $n-\text{rank}(\text{T}-\lambda \text{I})$.

These same conditions can be used to test if a square matrix `A` is diagonalizable because diagonalizability of `A` is equivalent to diagonalizability of the operator ${\text{L}}_{A}$.

If T is a diagonalizable operator and ${\beta}_{1},\text{}{\beta}_{2},\text{}\dots ,\text{}{\beta}_{k}$ are ordered bases for the eigenspaces of T, then the union $\beta ={\beta}_{1}\cup {\beta}_{2}\cup \cdots \cup {\beta}_{k}$ is an ordered basis for V consisting of eigenvectors of T, and hence ${[\text{T}]}_{\beta}$ is a diagonal matrix.

When testing T for diagonalizability, it is usually easiest to choose a convenient basis $\alpha $ for V and work with $B={[\text{T}]}_{\alpha}$. If the characteristic polynomial of `B` splits, then use condition 2 above to check if the multiplicity of each *repeated* eigenvalue of `B` equals $n-\text{rank}(B-\lambda I)$. (By Theorem 5.7, condition 2 is automatically satisfied for eigenvalues of multiplicity 1.) If so, then `B`, and hence T, is diagonalizable.

If T is diagonalizable and a basis $\beta $ for V consisting of eigenvectors of T is desired, then we first find a basis for each eigenspace of `B`. The union of these bases is a basis $\gamma $ for ${\text{F}}^{n}$ consisting of eigenvectors of `B`. Each vector in $\gamma $ is the coordinate vector relative to $\alpha $ of an eigenvector of T. The set consisting of these `n` eigenvectors of T is the desired basis $\beta $.

Furthermore, if `A` is an $n\times n$ diagonalizable matrix, we can use the corollary to Theorem 2.23 (p. 115) to find an invertible $n\times n$ matrix `Q` and a diagonal $n\times n$ matrix `D` such that ${Q}^{-1}AQ=D$. The matrix `Q` has as its columns the vectors in a basis of eigenvectors of `A`, and `D` has as its `j` th diagonal entry the eigenvalue of `A` corresponding to the `j` th column of `Q`.

We now consider some examples illustrating the preceding ideas.

# Example 5

We test the matrix

for diagonalizability.

The characteristic polynomial of `A` is $\text{det}(A-tI)=-(t-4){(t-3)}^{2}$, which splits, and so condition 1 of the test for diagonalization is satisfied. Also `A` has eigenvalues ${\lambda}_{1}=4$ and ${\lambda}_{2}=3$ with multiplicities 1 and 2, respectively. Since ${\lambda}_{1}$ has multiplicity 1, condition 2 is satisfied for ${\lambda}_{1}$. Thus we need only test condition 2 for ${\lambda}_{2}$. Because

has rank 2, we see that $3-\text{rank}(A-{\lambda}_{2}I)=1$, which is not the multiplicity of ${\lambda}_{2}$. Thus condition 2 fails for ${\lambda}_{2}$, and `A` is therefore not diagonalizable.

# Example 6

Let T be the linear operator on ${\text{P}}_{2}(R)$ defined by

We first test T for diagonalizability. Let *α* denote the standard ordered basis for ${\text{P}}_{2}(R)$ and $B={[\text{T}]}_{\alpha}$. Then

The characteristic polynomial of `B`, and hence of T, is $-{(t-1)}^{2}(t-2)$, which splits. Hence condition 1 of the test for diagonalization is satisfied. Also `B` has the eigenvalues ${\lambda}_{1}=1$ and ${\lambda}_{2}=2$ with multiplicities 2 and 1, respectively. Condition 2 is satisfied for ${\lambda}_{2}$ because it has multiplicity 1. So we need only verify condition 2 for ${\lambda}_{1}=1$. For this case,

which is equal to the multiplicity of ${\lambda}_{1}$. Therefore T is diagonalizable.

We now find an ordered basis $\gamma $ for ${\text{R}}^{3}$ consisting of eigenvectors of `B`. We consider each eigenvalue separately.

The eigenspace corresponding to ${\lambda}_{1}=1$ is

which is the solution space for the system

and has

as a basis.

The eigenspace corresponding to ${\lambda}_{2}=2$ is

which is the solution space for the system

and has

as a basis.

Let

Then $\gamma $ is an ordered basis for ${\text{R}}^{3}$ consisting of eigenvectors of `B`.

Finally, observe that the vectors in $\gamma $ are the coordinate vectors relative to $\alpha $ of the vectors in the set

which is an ordered basis for ${\text{P}}_{2}(R)$ consisting of eigenvectors of T. Thus

Our next example is an application of diagonalization that is of interest in Section 5.3.

# Example 7

Let

We show that `A` is diagonalizable and find a $2\times 2$ matrix `Q` such that ${Q}^{-1}AQ$ is a diagonal matrix. We then show how to use this result to compute ${A}^{n}$ for any positive integer `n`.

First observe that the characteristic polynomial of `A` is $(t-1)(t-2)$, and hence `A` has two distinct eigenvalues, ${\lambda}_{1}=1$ and ${\lambda}_{2}=2$. By applying the corollary to Theorem 5.5 to the operator ${\text{L}}_{A}$, we see that `A` is diagonalizable. Moreover,

are bases for the eigenspaces ${\text{E}}_{{\lambda}_{1}}$ and ${\text{E}}_{{\lambda}_{2}}$, respectively. Therefore

is an ordered basis for ${\text{R}}^{2}$ consisting of eigenvectors of `A`. Let

the matrix whose columns are the vectors in $\gamma $. Then, by the corollary to Theorem 2.23 (p. 115),

To find ${A}^{n}$ for any positive integer `n`, observe that $A=QD{Q}^{-1}$. Therefore

We now consider an application that uses diagonalization to solve a system of differential equations.

# Systems of Differential Equations

Consider the system of differential equations

where, for each `i`, ${x}_{i}={x}_{i}(t)$ is a differentiable real-valued function of the real variable `t`. Clearly, this system has a solution, namely, the solution in which each ${x}_{i}(t)$ is the zero function. We determine all of the solutions to this system.

Let $x:\text{}R\to {\text{R}}^{3}$ be the function defined by

The derivative of `x`, denoted ${x}^{\prime}$, is defined by

Let

be the coefficient matrix of the given system, so that we can rewrite the system as the matrix equation ${x}^{\prime}=Ax$.

It can be verified that for

we have ${Q}^{-1}AQ=D$. Substitute $A=QD{Q}^{-1}$ into ${x}^{\prime}=Ax$ to obtain ${x}^{\prime}=QD{Q}^{-1}x$ or, equivalently, ${Q}^{-1}{x}^{\prime}=D{Q}^{-1}x$. The function $y:\text{}R\to {\text{R}}^{3}$ defined by $y(t)={Q}^{-1}x(t)$ can be shown to be differentiable, and ${y}^{\prime}={Q}^{-1}{x}^{\prime}$ (see Exercise 17). Hence the original system can be written as ${y}^{\prime}=Dy$.

Since `D` is a diagonal matrix, the system ${y}^{\prime}=Dy$ is easy to solve. Setting

we can rewrite ${y}^{\prime}=Dy$ as

The three equations

are independent of each other, and thus can be solved individually. It is easily seen (as in Example 3 of Section 5.1) that the general solution to these equations is ${y}_{1}(t)={c}_{1}{e}^{2t},\text{}{y}_{2}(t)={c}_{2}{e}^{2t}$, and ${y}_{3}(t)={c}_{3}{e}^{4t}$, where ${c}_{1},\text{}{c}_{2}$, and ${c}_{3}$ are arbitrary constants. Finally,

yields the general solution of the original system. Note that this solution can be written as

The expressions in brackets are arbitrary vectors in ${\text{E}}_{{\lambda}_{1}}$ and ${\text{E}}_{{\lambda}_{2}}$, respectively, where ${\lambda}_{1}=2$ and ${\lambda}_{2}=4$. Thus the general solution of the original system is $x(t)={e}^{2t}{z}_{1}+{e}^{4t}{z}_{2}$, where ${z}_{1}\in {\text{E}}_{{\lambda}_{1}}$ and ${z}_{2}\in {\text{E}}_{{\lambda}_{2}}$. This result is generalized in Exercise 16.

# Direct Sums*

Let T be a linear operator on a finite-dimensional vector space V. There is a way of decomposing V into simpler subspaces that offers insight into the behavior of T. This approach is especially useful in Chapter 7, where we study nondiagonalizable linear operators. In the case of diagonalizable operators, the simpler subspaces are the eigenspaces of the operator.

# Definition.

*Let* ${\text{W}}_{1},\text{}{\text{W}}_{2},\text{}\dots ,\text{}{\text{W}}_{k}$ *be subspaces of a vector space V. We define the sum of these subspaces to be the set*

*which we denote by* ${\text{W}}_{1}+{\text{W}}_{2}+\cdots +{\text{W}}_{k}$ *or* $\sum _{i=1}^{k}{W}_{i}$.

It is a simple exercise to show that the sum of subspaces of a vector space is also a subspace.

# Example 8

Let $\text{V}={\text{R}}^{3}$, let ${\text{W}}_{1}$ denote the *xy*-plane, and let ${\text{W}}_{2}$ denote the *yz*-plane. Then ${\text{R}}^{3}={\text{W}}_{1}+{\text{W}}_{2}$ because, for any vector $(a,\text{}b,\text{}c)\in {\text{R}}^{3}$, we have

where $(a,\text{}0,\text{}0)\in {\text{W}}_{1}$ and $(0,\text{}b,\text{}c)\in {\text{W}}_{2}$.

Notice that in Example 8 the representation of (`a`, `b`, `c`) as a sum of vectors in ${\text{W}}_{1}$ and ${\text{W}}_{2}$ is not unique. For example, $(a,\text{}b,\text{}c)=(a,\text{}b,\text{}0)+(0,\text{}0,\text{}c)$ is another representation. Because we are often interested in sums for which representations are unique, we introduce a condition that assures this outcome. The definition of *direct sum* that follows is a generalization of the definition given in the exercises of Section 1.3.

# Definition.

*Let* $\text{W},\text{}{\text{W}}_{1},\text{}{\text{W}}_{2},\text{}\dots ,\text{}{\text{W}}_{k}$ *be subspaces of a vector space* V *such that* ${\text{W}}_{i}\subseteq \text{W}$ *for* $i=1,\text{}2,\text{}\dots ,\text{}k$. *We call* W *the direct sum of the subspaces* ${\text{W}}_{1},\text{}{\text{W}}_{2},\text{}\dots ,\text{}{\text{W}}_{k}$,

*and write*$\text{V}={\text{W}}_{1}\oplus {\text{W}}_{2}\oplus \cdots \oplus {\text{W}}_{k}$

*, if*

# Example 9

In ${\text{R}}^{5},\text{}\text{letW}=\{({x}_{1},\text{}{x}_{2},\text{}{x}_{3},\text{}{x}_{4},\text{}{x}_{5}):\text{}{x}_{5}=0\},\text{}{\text{W}}_{1}=\{(a,\text{}b,\text{}0,\text{}0,\text{}0):\text{}a,\text{}b\in R\},\text{}{\text{W}}_{2}=\{(0,\text{}0,\text{}c,\text{}0,\text{}0):\text{}c\in R\}$, and ${\text{W}}_{3}=\{(0,\text{}0,\text{}0,\text{}d,\text{}0):\text{}d\in R\}$. For any $(a,\text{}b,\text{}c,\text{}d,\text{}0)\in \text{W,}$

Thus

To show that W is the direct sum of ${\text{W}}_{1},\text{}{\text{W}}_{2}$, and ${\text{W}}_{3}$, we must prove that ${\text{W}}_{1}\cap ({\text{W}}_{2}+{\text{W}}_{3})={\text{W}}_{2}\cap ({\text{W}}_{1}+{\text{W}}_{3})={\text{W}}_{3}\cap ({\text{W}}_{1}+{\text{W}}_{2})=\left\{0\right\}$. But these equalities are obvious, and so $\text{W}={\text{W}}_{1}\oplus {\text{W}}_{2}\oplus {\text{W}}_{3}$.

Our next result contains several conditions that are equivalent to the definition of a direct sum.

# Theorem 5.9.

*Let* ${\text{W}}_{1},\text{}{\text{W}}_{2},\text{}\dots ,\text{}{\text{W}}_{k}$ *be subspaces of a finite-dimensional vector space* V*. The following conditions are equivalent.*

(a) $\text{V}={\text{W}}_{1}\oplus {\text{W}}_{2}\oplus \cdots \oplus {\text{W}}_{k}.$

(b) $\text{V}={\displaystyle \sum _{i=1}^{k}{\text{W}}_{i}}$

*and, for any vectors*${v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{k}$*such that*${v}_{i}\in {\text{W}}_{i}$ $(1\le i\le k)$,*if*${v}_{1}+{v}_{2}+\cdots +{v}_{k}=0$,*then*${v}_{i}=0$*for all i.*(c)

*Each vector*$v\in \text{V}$*can be uniquely written as*$v={v}_{1}+{v}_{2}+\cdots +{v}_{k}$,*where*${v}_{i}\in {\text{W}}_{i}$.(d)

*If*${\gamma}_{i}$*is an ordered basis for*${\text{W}}_{i}(1\le i\le k)$*, then*${\gamma}_{i}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$*is an ordered basis for*V.(e)

*For each*$i=1,\text{}2,\text{}\dots ,\text{}k$*, there exists an ordered basis*${\gamma}_{i}$*for*${\text{W}}_{i}$*such that*${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$*is an ordered basis for*V.

# Proof.

Assume (a). We prove (b). Clearly

Now suppose that ${v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{k}$ are vectors such that ${v}_{i}\in {\text{W}}_{i}$ for all `i` and ${v}_{1}+{v}_{2}+\cdots +{v}_{k}=0$. Then for any `j`

But $-{v}_{j}\in {\text{W}}_{j}$ and hence

So ${v}_{j}=0$, proving (b).

Now assume (b). We prove (c). Let $v\in \text{V}$. By (b), there exist vectors ${v}_{1},\text{}{v}_{2},\text{}\dots ,{v}_{k}$ such that ${v}_{i}\in {\text{W}}_{i}$ and $v={v}_{1}+{v}_{2}+\cdots +{v}_{k}$. We must show that this representation is unique. Suppose also that $v={w}_{1}+{w}_{2}+\cdots +{w}_{k}$, where ${w}_{i}\in {\text{W}}_{i}$ for all `i`. Then

But ${v}_{i}-{w}_{i}\in {\text{W}}_{i}$ for all `i`, and therefore ${v}_{i}-{w}_{i}=0$ for all `i` by (b). Thus ${v}_{i}={w}_{i}$ for all `i`, proving the uniqueness of the representation.

Now assume (c). We prove (d). For each `i`, let ${\gamma}_{i}$ be an ordered basis for ${\text{W}}_{i}$. Since

by (c), it follows that ${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ generates V. To show that this set is linearly independent, consider vectors ${v}_{ij}\in {\gamma}_{i}\text{}(j=1,\text{}2,\text{}\dots ,\text{}{m}_{i}$ and $i=1,\text{}2,\text{}\dots ,\text{}k)$ and scalars ${a}_{ij}$ such that

For each `i`, set

Then for each `i`, ${w}_{i}\in \text{span}({\gamma}_{i})={\text{W}}_{i}$ and

Since $0\in {\text{W}}_{i}$ for each `i` and $0+0+\cdots +0={w}_{1}+{w}_{2}+\cdots +{w}_{k}$, (c) implies that ${w}_{i}=0$ for all `i`. Thus

for each `i`. But each ${\gamma}_{i}$ is linearly independent, and hence ${a}_{ij}=0$ for all `i` and `j`. Consequently ${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ is linearly independent and therefore is a basis for V.

Clearly (e) follows immediately from (d).

Finally, we assume (e) and prove (a). For each `i`, let ${\gamma}_{i}$ be an ordered basis for ${\text{W}}_{i}$ such that ${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ is an ordered basis for V. Then

by repeated applications of Exercise 14 of Section 1.4. Fix $j(1\le j\le k)$, and suppose that, for some nonzero vector $v\in \text{V}$,

Then

Hence `v` is a nontrivial linear combination of both ${\gamma}_{j}$ and $\underset{i\ne j}{\cup}{\gamma}_{i}$, so that `v` can be expressed as a linear combination of ${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ in more than one way. But these representations contradict Theorem 1.8 (p. 44), and so we conclude that

proving (a).

With the aid of Theorem 5.9, we are able to characterize diagonalizability in terms of direct sums.

# Theorem 5.10.

*A linear operator* T *on a finite-dimensional vector space* V *is diagonalizable if and only if* V *is the direct sum of the eigenspaces of* T.

# Proof.

Let ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ be the distinct eigenvalues of T.

First suppose that T is diagonalizable, and for each `i` choose an ordered basis ${\gamma}_{i}$ for the eigenspace ${\text{E}}_{{\lambda}_{i}}$. By Theorem 5.8, ${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ is a basis for V, and hence V is a direct sum of the ${\text{E}}_{{\lambda}_{i}}\text{'}\text{s}$ by Theorem 5.9.

Conversely, suppose that V is a direct sum of the eigenspaces of T. For each `i`, choose an ordered basis ${\gamma}_{i}$ of ${\text{E}}_{{\lambda}_{i}}$. By Theorem 5.9, the union ${\gamma}_{1}\cup {\gamma}_{2}\cup \cdots \cup {\gamma}_{k}$ is a basis for V. Since this basis consists of eigenvectors of T, we conclude that T is diagonalizable.

# Example 10

Let T be the linear operator on ${\text{R}}^{4}$ defined by

It is easily seen that T is diagonalizable with eigenvalues ${\lambda}_{1}=1,\text{}{\lambda}_{2}=2$, and ${\lambda}_{3}=3$. Furthermore, the corresponding eigenspaces coincide with the subspaces ${\text{W}}_{1},\text{}{\text{W}}_{2}$, and ${\text{W}}_{3}$ of Example 9. Thus Theorem 5.10 provides us with another proof that ${\text{R}}^{4}={\text{W}}_{1}\oplus {\text{W}}_{2}\oplus {\text{W}}_{3}$.

# Exercises

Label the following statements as true or false.

(a) Any linear operator on an

`n`-dimensional vector space that has fewer than`n`distinct eigenvalues is not diagonalizable.(b) Two distinct eigenvectors corresponding to the same eigenvalue are always linearly dependent.

(c) If $\lambda $ is an eigenvalue of a linear operator T, then each vector in ${\text{E}}_{\lambda}$ is an eigenvector of T.

(d) If ${\lambda}_{1}$ and ${\lambda}_{2}$ are distinct eigenvalues of a linear operator T, then ${\text{E}}_{{\lambda}_{1}}\cap {\text{E}}_{{\lambda}_{2}}=\left\{0\right\}$.

(e) Let $A\in {\text{M}}_{n\times n}(F)$ and $\beta =\{{v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{n}\}$ be an ordered basis for ${\text{F}}^{n}$ consisting of eigenvectors of

`A`. If`Q`is the $n\times n$ matrix whose`j`th column is ${v}_{j}(1\le j\le n)$, then ${Q}^{-1}AQ$ is a diagonal matrix.(f) A linear operator T on a finite-dimensional vector space is diagonalizable if and only if the multiplicity of each eigenvalue $\lambda $ equals the dimension of ${\text{E}}_{\lambda}$.

(g) Every diagonalizable linear operator on a nonzero vector space has at least one eigenvalue.

The following two items relate to the optional subsection on direct sums.

(h) If a vector space is the direct sum of subspaces ${\text{W}}_{1},\text{}{\text{W}}_{2},\text{}\dots ,\text{}{\text{W}}_{k}$, then ${\text{W}}_{i}\cap {\text{W}}_{j}=\left\{0\right\}$ for $i\ne j$.

(i) If

$$\text{V}={\displaystyle \sum _{i=1}^{k}{\text{W}}_{i}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\text{W}}_{i}\cap {\text{W}}_{j}={\displaystyle \left\{0\right\}}\phantom{\rule{1em}{0ex}}\text{for}i\ne j,\text{}$$then $\text{V}={\text{W}}_{1}\oplus {\text{W}}_{2}\oplus \cdots \oplus {\text{W}}_{k}$.

For each of the following matrices $A\in {\text{M}}_{n\times n}(R)$, test

`A`for diagonalizability, and if`A`is diagonalizable, find an invertible matrix`Q`and a diagonal matrix`D`such that ${Q}^{-1}AQ=D$.(a) $\left(\begin{array}{rr}1& 2\\ 0& 1\end{array}\right)$

(b) $\left(\begin{array}{rr}1& 3\\ 3& 1\end{array}\right)$

(c) $\left(\begin{array}{rr}1& 4\\ 3& 2\end{array}\right)$

(d) $\left(\begin{array}{rrr}7& -4& 0\\ 8& -5& 0\\ 6& -6& 3\end{array}\right)$

(e) $\left(\begin{array}{rrr}0& 0& 1\\ 1& 0& -1\\ 0& 1& 1\end{array}\right)$

(f) $\left(\begin{array}{rrr}1& 1& 0\\ 0& 1& 2\\ 0& 0& 3\end{array}\right)$

(g) $\left(\begin{array}{rrr}3& 1& 1\\ 2& 4& 2\\ -1& -1& 1\end{array}\right)$

For each of the following linear operators T on a vector space V, test T for diagonalizability, and if T is diagonalizable, find a basis $\beta $ for V such that ${[\text{T}]}_{\beta}$ is a diagonal matrix.

(a) $\text{V}={\text{P}}_{3}(R)$ and T is defined by $\text{T}(f(x))={f}^{\prime}(x)+{f}^{\u2033}(x)$.

(b) $\text{V}={\text{P}}_{2}(R)$ and T is defined by $\text{T}(a{x}^{2}+bx+c)=c{x}^{2}+bx+a$.

(c) $\text{V}={\text{R}}^{3}$ and T is defined by

$$\text{T}\left(\begin{array}{r}{a}_{1}\\ {a}_{2}\\ {a}_{3}\end{array}\right)=\left(\begin{array}{r}\hfill {a}_{2}\\ \hfill -{a}_{1}\\ \hfill 2{a}_{3}\end{array}\right).$$(d) $\text{V}={\text{P}}_{2}\left(R\right)$ and T is defined by $\text{T}(f(x))=f(0)+f(1)(x+{x}^{2})$.

(e) $\text{V}={\text{C}}^{2}$ and T is defined by $\text{T}(z,\text{}w)=(z+iw,\text{}iz+w)$.

(f) $\text{V}={\text{M}}_{2\times 2}(R)$ and T is defined by $\text{T}(A)={A}^{t}$.

Prove the matrix version of the corollary to Theorem 5.5: If $A\in {\text{M}}_{n\times n}(F)$ has

`n`distinct eigenvalues, then`A`is diagonalizable.State and prove the matrix version of Theorem 5.6.

(a) Justify the test for diagonalizability and the method for diagonalization stated in this section.

(b) Formulate the results in (a) for matrices.

For

$$A=\left(\begin{array}{rr}1& 4\\ 2& 3\end{array}\right)\in {\text{M}}_{2\times 2}(R),\text{}$$find an expression for ${A}^{n}$, where

`n`is an arbitrary positive integer.Suppose that $A\in {\text{M}}_{n\times n}(F)$ has two distinct eigenvalues, ${\lambda}_{1}$ and ${\lambda}_{2}$, and that $\text{dim}({\text{E}}_{{\lambda}_{1}})=n-1$. Prove that

`A`is diagonalizable.Let T be a linear operator on a finite-dimensional vector space V, and suppose there exists an ordered basis $\beta $ for V such that ${[\text{T}]}_{\beta}$ is an upper triangular matrix.

(a) Prove that the characteristic polynomial for T splits.

(b) State and prove an analogous result for matrices.

The converse of (a) is treated in Exercise 12(b).

Let T be a linear operator on a finite-dimensional vector space V with the distinct eigenvalues ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,{\lambda}_{k}$ and corresponding multiplicities ${m}_{1},\text{}{m}_{2},\text{}\dots ,\text{}{m}_{k}$. Suppose that $\beta $ is a basis for V such that ${[\text{T}]}_{\beta}$ is an upper triangular matrix. Prove that the diagonal entries of ${[\text{T}]}_{\beta}$ are ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ and that each ${\lambda}_{i}$ occurs ${m}_{i}$ times $(1\le i\le k)$.

Let

`A`be an $n\times n$ matrix that is similar to an upper triangular matrix and has the distinct eigenvalues ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$ with corresponding multiplicities ${m}_{1},\text{}{m}_{2},\text{}\dots ,\text{}{m}_{k}$. Prove the following statements.(a) $\text{tr}(A)={\displaystyle \sum _{i=1}^{k}{m}_{i}{\lambda}_{i}}$

(b) $\text{det}(A)={({\lambda}_{1})}^{{m}_{1}}{({\lambda}_{2})}^{{m}_{2}}\cdots {({\lambda}_{k})}^{{m}_{k}}$.

(a) Prove that if $A\in {\text{M}}_{n\times n}(F)$ and the characteristic polynomial of

`A`splits, then`A`is similar to an upper triangular matrix. (This proves the converse of Exercise 9(b).)*Hint:*Use mathematical induction on`n`. For the general case, let ${v}_{1}$ be an eigenvector of`A`, and extend $\left\{{v}_{1}\right\}$ to a basis $\{{v}_{1},\text{}{v}_{2},\text{}\dots ,\text{}{v}_{n}\}$ for ${\text{F}}^{n}$. Let`P`be the $n\times n$ matrix whose`j`th column is ${v}_{j}$, and consider ${P}^{-1}AP$. Exercise 13(a) in Section 5.1 and Exercise 21 in Section 4.3 can be helpful.(b) Prove the converse of Exercise 9(a).

Visit goo.gl/

gJSjRU for a solution.

Let T be an invertible linear operator on a finite-dimensional vector space V.

(a) Recall that for any eigenvalue $\lambda $ of T, ${\lambda}^{-1}$ is an eigenvalue of ${\text{T}}^{-1}$ (Exercise 9 of Section 5.1). Prove that the eigenspace of T corresponding to $\lambda $ is the same as the eigenspace of ${\text{T}}^{-1}$ corresponding to ${\lambda}^{-1}$.

(b) Prove that if T is diagonalizable, then ${\text{T}}^{-1}$ is diagonalizable.

Let $A\in {\text{M}}_{n\times n}(F)$. Recall from Exercise 15 of Section 5.1 that

`A`and ${A}^{t}$ have the same characteristic polynomial and hence share the same eigenvalues with the same multiplicities. For any eigenvalue $\lambda $ of`A`and ${A}^{t}$, let ${\text{E}}_{\lambda}$ and ${{\text{E}}^{\prime}}_{\lambda}$ denote the corresponding eigenspaces for`A`and ${A}^{t}$, respectively.(a) Show by way of example that for a given common eigenvalue, these two eigenspaces need not be the same.

(b) Prove that for any eigenvalue $\lambda $, $\text{dim}({\text{E}}_{\lambda})=\text{dim}({{\text{E}}^{\prime}}_{\lambda})$.

(c) Prove that if

`A`is diagonalizable, then ${A}^{t}$ is also diagonalizable.

Find the general solution to each system of differential equations.

(a) $\begin{array}{rrr}{x}^{\prime}& =& x+y\\ {y}^{\prime}& =& 3x-y\end{array}$

(b) $\begin{array}{rrrrr}{{x}^{\prime}}_{1}& =& 8{x}_{1}& +& 10{x}_{2}\\ {{x}^{\prime}}_{2}& =& -5{x}_{1}& -& 7{x}_{2}\end{array}$

(c) $\begin{array}{rrrrrr}{{x}^{\prime}}_{1}& =& {x}_{1}& & +& {x}_{3}\\ {{x}^{\prime}}_{2}& =& & {x}_{2}& +& {x}_{3}\\ {{x}^{\prime}}_{3}& =& & & & 2{x}_{3}\end{array}$

Let

$$A=\left(\begin{array}{cccc}{a}_{11}& {a}_{12}& \cdots & {a}_{1n}\\ {a}_{21}& {a}_{22}& \cdots & {a}_{2n}\\ \vdots & \vdots & & \vdots \\ {a}_{n1}& {a}_{n2}& \cdots & {a}_{nn}\end{array}\right)$$be the coefficient matrix of the system of differential equations

$$\begin{array}{rcl}{{x}^{\prime}}_{1}& =& {a}_{11}{x}_{1}+{a}_{12}{x}_{2}+\cdots +{a}_{1n}{x}_{n}\\ {{x}^{\prime}}_{2}& =& {a}_{21}{x}_{1}+{a}_{22}{x}_{2}+\cdots +{a}_{2n}{x}_{n}\\ \vdots & & \\ {{x}^{\prime}}_{n}& =& {a}_{n1}{x}_{1}+{a}_{n2}{x}_{2}+\cdots +{a}_{nn}{x}_{n}.\end{array}$$Suppose that

`A`is diagonalizable and that the distinct eigenvalues of`A`are ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$. Prove that a differentiable function $x:\text{}R\to {\text{R}}^{n}$ is a solution to the system if and only if`x`is of the form$$x(t)={e}^{{\lambda}_{1}t}{z}_{1}+{e}^{{\lambda}_{2}t}{z}_{2}+\cdots +{e}^{{\lambda}_{k}t}{z}_{k},\text{}$$where ${z}_{i}\in {\text{E}}_{{\lambda}_{i}}$ for $i=1,\text{}2,\text{}\dots ,\text{}k$. Use this result to prove that the set of solutions to the system is an

`n`-dimensional real vector space.Let $C\in {\text{M}}_{m\times n}(R)$, and let

`Y`be an $n\times p$ matrix of differentiable functions. Prove $(CY{)}^{\prime}=C{Y}^{\prime}$, where ${({Y}^{\prime})}_{ij}={{Y}^{\prime}}_{ij}$ for all`i`,`j`.

Exercises 18 through 20 are concerned with *simultaneous diagonalization.*

# Definitions.

*Two linear operators* T *and* U *on a finite-dimensional vector space* V *are called simultaneously diagonalizable if there exists an ordered basis* $\beta $ *for* V *such that both* ${[\text{T}]}_{\beta}$ *and* ${[\text{U}]}_{\beta}$ *are diagonal matrices. Similarly, * $A,\text{}B\in {\text{M}}_{n\times n}(F)$ *are called simultaneously diagonalizable if there exists an invertible matrix* $Q\in {\text{M}}_{n\times n}(F)$ *such that both* ${Q}^{-1}AQ$ *and* ${Q}^{-1}BQ$ *are diagonal matrices*.

(a) Prove that if T and U are simultaneously diagonalizable linear operators on a finite-dimensional vector space V, then the matrices ${[\text{T}]}_{\beta}$ and ${[\text{U}]}_{\beta}$ are simultaneously diagonalizable for any ordered basis $\beta $.

(b) Prove that if

`A`and`B`are simultaneously diagonalizable matrices, then ${\text{L}}_{A}$ and ${\text{L}}_{B}$ are simultaneously diagonalizable linear operators.

(a) Prove that if T and U are simultaneously diagonalizable operators, then T and U commute (i.e., $\text{TU}=\text{UT}$).

(b) Show that if

`A`and`B`are simultaneously diagonalizable matrices, then`A`and`B`commute.

The converses of (a) and (b) are established in Exercise 25 of Section 5.4.

Let T be a diagonalizable linear operator on a finite-dimensional vector space, and let

`m`be any positive integer. Prove that T and ${\text{T}}^{m}$ are simultaneously diagonalizable.

Exercises 21 through 24 are concerned with direct sums.

Let ${\text{W}}_{1},\text{}{\text{W}}_{2},\text{}\dots ,\text{}{\text{W}}_{k}$ be subspaces of a finite-dimensional vector space V such that

$$\sum _{i=1}^{k}{\text{W}}_{i}=\text{V}.$$Prove that V is the direct sum of ${\text{W}}_{1},\text{}{\text{W}}_{2},\text{}\dots ,\text{}{\text{W}}_{k}$ if and only if

$$\text{dim}(\text{V})={\displaystyle \sum _{i=1}^{k}\text{dim}({\text{W}}_{i})}.$$Let V be a finite-dimensional vector space with a basis $\beta $, and let ${\beta}_{1},\text{}{\beta}_{2},\text{}\dots ,\text{}{\beta}_{k}$ be a partition of $\beta $ (i.e., ${\beta}_{1},\text{}{\beta}_{2},\text{}\dots ,\text{}{\beta}_{k}$ are subsets of $\beta $ such that $\beta ={\beta}_{1}\cup {\beta}_{2}\cup \cdots \cup {\beta}_{k}$ and ${\beta}_{i}\cap {\beta}_{j}=\varnothing $ if $i\ne j$). Prove that $\text{V}=\text{span}({\beta}_{1})\oplus \text{span}({\beta}_{2})\oplus \cdots \oplus \text{span}({\beta}_{k})$.

Let T be a linear operator on a finite-dimensional vector space V, and suppose that the distinct eigenvalues of T are ${\lambda}_{1},\text{}{\lambda}_{2},\text{}\dots ,\text{}{\lambda}_{k}$. Prove that

$$\text{span}(\{x\in \text{V}:\text{}x\text{isaneigenvectorofT}\})={\text{E}}_{{\lambda}_{1}}\oplus {\text{E}}_{{\lambda}_{2}}\oplus \cdots \oplus {\text{E}}_{{\lambda}_{k}}.$$Let ${\text{W}}_{1},\text{}{\text{W}}_{2},\text{}{\text{K}}_{1},\text{}{\text{K}}_{2},\text{}\dots ,\text{}{\text{K}}_{p},\text{}{\text{M}}_{1},\text{}{\text{M}}_{2},\text{}\dots ,\text{}{\text{M}}_{q}$ be subspaces of a vector space V such that ${\text{W}}_{1}={\text{K}}_{1}\oplus {\text{K}}_{2}\oplus \cdots \oplus {\text{K}}_{p}$ and ${\text{W}}_{2}={\text{M}}_{1}\oplus {\text{M}}_{2}\oplus \cdots \oplus {\text{M}}_{q}$. Prove that if ${\text{W}}_{1}\cap {\text{W}}_{2}=\left\{0\right\}$, then

$${\text{W}}_{1}+{\text{W}}_{2}={\text{W}}_{1}\oplus {\text{W}}_{2}={\text{K}}_{1}\oplus {\text{K}}_{2}\oplus \cdots \oplus {\text{K}}_{p}\oplus {\text{M}}_{1}\oplus {\text{M}}_{2}\oplus \cdots \oplus {\text{M}}_{q}.$$