§ 24. Diagonalizability

Algebraic Multiplicity and Dimension of Eigenspace

Theorem 24.1 : Distinct eigenvalues give linearly independent eigenvectors.

Let \(T\) be a linear operator on a vector space \(V\), and let \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) be distinct eigenvalues of \(T\). If \(v_{1}, v_{2}, \ldots, v_{k}\) are eigenvectors of \(T\) such that \(\lambda_{i}\) corresponds to \(v_{i}\) \((1 \leq i \leq k)\), then \(\left\{v_{1}, v_{2}, \ldots, v_{k}\right\}\) is linearly independent.

Proof

The proof is by mathematical induction on \(k\). Suppose that \(k=1\). Then \(v_{1} \neq 0\) since \(v_{1}\) is an eigenvector, and hence \(\left\{v_{1}\right\}\) is linearly independent. Now assume that the theorem holds for \(k-1\) distinct eigenvalues, where \(k-1 \geq 1\), and that we have \(k\) eigenvectors \(v_{1}, v_{2}, \ldots, v_{k}\) corresponding to the distinct eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\). We wish to show that \(\left\{v_{1}, v_{2}, \ldots, v_{k}\right\}\) is linearly independent. Suppose that \(a_{1}, a_{2}, \ldots, a_{k}\) are scalars such that

\[ a_{1} v_{1}+a_{2} v_{2}+\cdots+a_{k} v_{k}=0 \]

Applying \(T-\lambda_{k} I\) to both sides of (1), we obtain

\[ a_{1}\left(\lambda_{1}-\lambda_{k}\right) v_{1}+a_{2}\left(\lambda_{2}-\lambda_{k}\right) v_{2}+\cdots+a_{k-1}\left(\lambda_{k-1}-\lambda_{k}\right) v_{k-1}=0 \]

By the induction hypothesis \(\left\{v_{1}, v_{2}, \ldots, v_{k-1}\right\}\) is linearly independent, and hence

\[ a_{1}\left(\lambda_{1}-\lambda_{k}\right)=a_{2}\left(\lambda_{2}-\lambda_{k}\right)=\cdots=a_{k-1}\left(\lambda_{k-1}-\lambda_{k}\right)=0 . \]

Since \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) are distinct, it follows that \(\lambda_{i}-\lambda_{k} \neq 0\) for \(1 \leq i \leq k-1\). So \(a_{1}=a_{2}=\cdots=a_{k-1}=0\), and (1) therefore reduces to \(a_{k} v_{k}=0\). But \(v_{k} \neq 0\) and therefore \(a_{k}=0\). Consequently \(a_{1}=a_{2}=\cdots=a_{k}=0\), and it follows that \(\left\{v_{1}, v_{2}, \ldots, v_{k}\right\}\) is linearly independent.

Corollary 24.2 : \(n\) distinct eigenvalues imply diagonalizability.

Let \(T\) be a linear operator on an \(n\)-dimensional vector space \(V\). If \(T\) has \(n\) distinct eigenvalues, then \(T\) is diagonalizable.

Proof

Suppose that \(T\) has \(n\) distinct eigenvalues \(\lambda_{1}, \ldots, \lambda_{n}\). For each \(i\) choose an eigenvector \(v_{i}\) corresponding to \(\lambda_{i}\). By Theorem 24.1, \(\left\{v_{1}, \ldots, v_{n}\right\}\) is linearly independent, and since \(\operatorname{dim}(V)=n\), this set is a basis for \(V\). Thus, by Theorem 23.3, \(T\) is diagonalizable.

The converse of Theorem 24.1 is false. That is, it is not true that if \(T\) is diagonalizable, then it has \(n\) distinct eigenvalues.

Definition 24.3 : Polynomial Splitting over a Field

A polynomial \(f(t)\) in \(\mathrm{P}(F)\) splits over \(F\) if there are scalars \(c, a_{1}, \ldots, a_{n}\) (not necessarily distinct) in \(F\) such that

\[ f(t)=c\left(t-a_{1}\right)\left(t-a_{2}\right) \cdots\left(t-a_{n}\right) \]

Theorem 24.4 : Characteristic polynomial of a diagonalizable operator splits.

The characteristic polynomial of any diagonalizable linear operator splits.

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 \([T]_{\beta}=D\) is a diagonal matrix. Suppose that

\[ D=\left(\begin{array}{cccc} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_{2} & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{array}\right) \]

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

\[ \begin{aligned} f(t) & =\operatorname{det}(D-t I)=\operatorname{det}\left(\begin{array}{cccc} \lambda_{1}-t & 0 & \cdots & 0 \\ 0 & \lambda_{2}-t & \cdots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & \cdots & \lambda_{n}-t \end{array}\right) \\ & =\left(\lambda_{1}-t\right)\left(\lambda_{2}-t\right) \cdots\left(\lambda_{n}-t\right)=(-1)^{n}\left(t-\lambda_{1}\right)\left(t-\lambda_{2}\right) \cdots\left(t-\lambda_{n}\right) . \end{aligned} \]

The converse of Theorem 24.4 is false; that is, the characteristic polynomial of \(T\) may split, but \(T\) need not be diagonalizable.

Definition 24.5 : Algebraic Multiplicity of an Eigenvalue

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

Definition 24.6 : Eigenspace

Let \(T\) be a linear operator on a vector space \(V\), and let \(\lambda\) be an eigenvalue of \(T\). Define \(E_{\lambda}=\{x \in V: T(x)=\lambda x\}=\mathrm{N}\left(T-\lambda I_{V}\right)\). The set \(E_{\lambda}\) is called the eigenspace of \(T\) corresponding to the eigenvalue \(\lambda\).

Analogously, we define the eigenspace of a square matrix \(A\) to be the eigenspace of \(L_{A}\).

Theorem 24.7 : Eigenspace dimension is at most its multiplicity.

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 \(1 \leq \operatorname{dim}\left(E_{\lambda}\right) \leq m\).

Proof

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

\[ A=\left(\begin{array}{cc} \lambda I_{p} & B \\ O & C \end{array}\right) \]

By Exercise 21.21, the characteristic polynomial of \(T\) is

\[ \begin{aligned} f(t) & =\operatorname{det}\left(A-t I_{n}\right)=\operatorname{det}\left(\begin{array}{cc} (\lambda-t) I_{p} & B \\ O & C-t I_{n-p} \end{array}\right) \\ & =\operatorname{det}\left((\lambda-t) I_{p}\right) \operatorname{det}\left(C-t I_{n-p}\right) \\ & =(\lambda-t)^{p} g(t) \end{aligned} \]

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 \(\operatorname{dim}\left(E_{\lambda}\right)=p\), and so \(\operatorname{dim}\left(E_{\lambda}\right) \leq m\).

Lemma 24.8

Let \(T\) be a linear operator, and let \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) be distinct eigenvalues of \(T\). For each \(i=1,2, \ldots, k\), let \(v_{i} \in E_{\lambda_{i}}\), the eigenspace corresponding to \(\lambda_{i}\). If

\[ v_{1}+v_{2}+\cdots+v_{k}=0 \]

then \(v_{i}=0\) for all \(i\).

Proof

Suppose otherwise. By renumbering if necessary, suppose that, for \(1 \leq m \leq k\), we have \(v_{i} \neq 0\) for \(1 \leq i \leq m\), and \(v_{i}=0\) for \(i>m\). Then, for each \(i \leq m\), \(v_{i}\) is an eigenvector of \(T\) corresponding to \(\lambda_{i}\) and

\[ v_{1}+v_{2}+\cdots+v_{m}=0 \]

But this contradicts Theorem 24.1, which states that these \(v_{i}\) s are linearly independent. We conclude, therefore, that \(v_{i}=0\) for all \(i\).

Theorem 24.9 : Union of linearly independent eigenspace subsets is independent.

Let \(T\) be a linear operator on a vector space \(V\), and let \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) be distinct eigenvalues of \(T\). For each \(i=1,2, \ldots, k\), let \(S_{i}\) be a finite linearly independent subset of the eigenspace \(E_{\lambda_{i}}\). Then \(S=S_{1} \cup S_{2} \cup \cdots \cup S_{k}\) is a linearly independent subset of \(V\).

Proof

Suppose that for each \(i\)

\[ S_{i}=\left\{v_{i 1}, v_{i 2}, \ldots, v_{i n_{i}}\right\} \]

Then \(S=\left\{v_{i j}: 1 \leq j \leq n_{i}, 1 \leq i \leq k\right\}\). Consider any scalars \(\left\{a_{i j}\right\}\) such that

\[ \sum_{i=1}^{k} \sum_{j=1}^{n_{i}} a_{i j} v_{i j}=0 \]

For each \(i\), let

\[ w_{i}=\sum_{j=1}^{n_{i}} a_{i j} v_{i j} . \]

Then \(w_{i} \in E_{\lambda_{i}}\) for each \(i\), and \(w_{1}+\cdots+w_{k}=0\). Therefore, by Lemma 24.8, \(w_{i}=0\) for all \(i\). But each \(S_{i}\) is linearly independent, and hence \(a_{i j}=0\) for all \(j\). We conclude that \(S\) is linearly independent.

Theorem 24.10 : Diagonalizability is characterized by eigenspace dimensions.

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\) such that the characteristic polynomial of \(T\) splits. Let \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) be the distinct eigenvalues of \(T\). Then

• (a) \(T\) is diagonalizable if and only if the multiplicity of \(\lambda_{i}\) is equal to \(\operatorname{dim}\left(E_{\lambda_{i}}\right)\) for all \(i\).
• (b) If \(T\) is diagonalizable and \(\beta_{i}\) is an ordered basis for \(E_{\lambda_{i}}\) for each \(i\), then \(\beta=\beta_{1} \cup \beta_{2} \cup \cdots \cup \beta_{k}\) is an ordered basis for \(V\) consisting of eigenvectors of \(T\).

Proof

For each \(i\), let \(m_{i}\) denote the multiplicity of \(\lambda_{i}\), \(d_{i}=\operatorname{dim}\left(E_{\lambda_{i}}\right)\), and \(n=\operatorname{dim}(V)\).

First, suppose that \(T\) is diagonalizable. Let \(\beta\) be a basis for \(V\) consisting of eigenvectors of \(T\). For each \(i\), let \(\beta_{i}=\beta \cap E_{\lambda_{i}}\), 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 \(n_{i} \leq d_{i}\) for each \(i\) because \(\beta_{i}\) is a linearly independent subset of a subspace of dimension \(d_{i}\), and \(d_{i} \leq m_{i}\) by Theorem 24.7. 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

\[ n=\sum_{i=1}^{k} n_{i} \leq \sum_{i=1}^{k} d_{i} \leq \sum_{i=1}^{k} m_{i}=n \]

It follows that

\[ \sum_{i=1}^{k}\left(m_{i}-d_{i}\right)=0 \]

Since \(\left(m_{i}-d_{i}\right) \geq 0\) 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 \(E_{\lambda_{i}}\), and let \(\beta=\beta_{1} \cup \beta_{2} \cup \cdots \cup \beta_{k}\). By Theorem 24.9, \(\beta\) is linearly independent. Furthermore, since \(d_{i}=m_{i}\) for all \(i\), \(\beta\) contains

\[ \sum_{i=1}^{k} d_{i}=\sum_{i=1}^{k} m_{i}=n \]

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

Test for Diagonalization

Concept 24.11 : Test for diagonalization

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.

1. The characteristic polynomial of \(T\) splits.
2. For each eigenvalue \(\lambda\) of \(T\), the multiplicity of \(\lambda\) equals \(n-\operatorname{rank}(T-\lambda I)\).

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 \(L_{A}\).

If \(T\) is a diagonalizable operator and \(\beta_{1}, \beta_{2}, \ldots, \beta_{k}\) 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 \([T]_{\beta}\) is a diagonal matrix.

When testing \(T\) for diagonalizability, it is usually easiest to choose a convenient basis \(\alpha\) for \(V\) and work with \(B=[T]_{\alpha}\). If the characteristic polynomial of \(B\) splits, then use condition 2 above to check if the multiplicity of each repeated eigenvalue of \(B\) equals \(n-\operatorname{rank}(B-\lambda I)\). (By Theorem 24.7, 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 \(F^{n}\) 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 Corollary 12.9 to find an invertible \(n \times n\) matrix \(Q\) and a diagonal \(n \times n\) matrix \(D\) such that \(Q^{-1} A Q=D\). 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 \(Q\).

Example 24.12 : Testing a matrix for diagonalizability

We test the matrix

\[ A=\left(\begin{array}{lll} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{array}\right) \in \mathrm{M}_{3 \times 3}(\mathbb{R}) \]

for diagonalizability.

The characteristic polynomial of \(A\) is \(\operatorname{det}(A-t I)=-(t-4)(t-3)^{2}\), 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

\[ A-\lambda_{2} I=\left(\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) \]

has rank 2, we see that \(3-\operatorname{rank}\left(A-\lambda_{2} I\right)=1\), which is not the multiplicity of \(\lambda_{2}\). Thus condition 2 fails for \(\lambda_{2}\), and \(A\) is therefore not diagonalizable.

Example 24.13 : Testing a linear operator for diagonalizability

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

\[ T(f(x))=f(1)+f^{\prime}(0) x+\left(f^{\prime}(0)+f^{\prime \prime}(0)\right) x^{2} \]

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

\[ B=\left(\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 2 \end{array}\right) \]

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,

\[ 3-\operatorname{rank}\left(B-\lambda_{1} I\right)=3-\operatorname{rank}\left(\begin{array}{lll} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{array}\right)=3-1=2 \]

which is equal to the multiplicity of \(\lambda_{1}\). Therefore \(T\) is diagonalizable. We now find an ordered basis \(\gamma\) for \(\mathbb{R}^{3}\) of eigenvectors of \(B\). We consider each eigenvalue separately.

The eigenspace corresponding to \(\lambda_{1}=1\) is

\[ E_{\lambda_{1}}=\left\{\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right) \in \mathbb{R}^{3}:\left(\begin{array}{lll} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=0\right\} \]

which is the solution space for the system

\[ x_{2}+x_{3}=0 \]

and has

\[ \gamma_{1}=\left\{\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right),\left(\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right)\right\} \]

as a basis. The eigenspace corresponding to \(\lambda_{2}=2\) is

\[ E_{\lambda_{2}}=\left\{\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right) \in \mathbb{R}^{3}:\left(\begin{array}{rrr} -1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & 1 & 0 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=0\right\} \]

which is the solution space for the system

\[ \begin{aligned} -x_{1}+x_{2}+x_{3} & =0 \\ x_{2} & =0, \end{aligned} \]

and has

\[ \gamma_{2}=\left\{\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)\right\} \]

as a basis. Let

\[ \gamma=\gamma_{1} \cup \gamma_{2}=\left\{\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right),\left(\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right),\left(\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right)\right\} . \]

Then \(\gamma\) is an ordered basis for \(\mathbb{R}^{3}\) 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

\[ \beta=\left\{1,-x+x^{2}, 1+x^{2}\right\} \]

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

\[ [T]_{\beta}=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array}\right) \]

Example 24.14 : Computing the power of a matrix using diagonalization

Let

\[ A=\left(\begin{array}{rr} 0 & -2 \\ 1 & 3 \end{array}\right) \]

We show that \(A\) is diagonalizable and find a \(2 \times 2\) matrix \(Q\) such that \(Q^{-1} A Q\) 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 24.1 to the operator \(L_{A}\), we see that \(A\) is diagonalizable. Moreover,

\[ \gamma_{1}=\left\{\binom{-2}{1}\right\} \quad \text { and } \quad \gamma_{2}=\left\{\binom{-1}{1}\right\} \]

are bases for the eigenspaces \(E_{\lambda_{1}}\) and \(E_{\lambda_{2}}\), respectively. Therefore

\[ \gamma=\gamma_{1} \cup \gamma_{2}=\left\{\binom{-2}{1},\binom{-1}{1}\right\} \]

is an ordered basis for \(\mathbb{R}^{2}\) consisting of eigenvectors of \(A\). Let

\[ Q=\left(\begin{array}{rr} -2 & -1 \\ 1 & 1 \end{array}\right) \]

the matrix whose columns are the vectors in \(\gamma\). Then, by the corollary to Theorem 12.6,

\[ D=Q^{-1} A Q=\left[L_{A}\right]_{\beta}=\left(\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right) \]

To find \(A^{n}\) for any positive integer \(n\), observe that \(A=Q D Q^{-1}\). Therefore

\[ \begin{aligned} A^{n} & =\left(Q D Q^{-1}\right)^{n} \\ & =\left(Q D Q^{-1}\right)\left(Q D Q^{-1}\right) \cdots\left(Q D Q^{-1}\right) \\ & =Q D^{n} Q^{-1} \\ & =Q\left(\begin{array}{rr} 1^{n} & 0 \\ 0 & 2^{n} \end{array}\right) Q^{-1} \\ & =\left(\begin{array}{rr} -2 & -1 \\ 1 & 1 \end{array}\right)\left(\begin{array}{rr} 1 & 0 \\ 0 & 2^{n} \end{array}\right)\left(\begin{array}{rr} -1 & -1 \\ 1 & 2 \end{array}\right)=\left(\begin{array}{rr} 2-2^{n} & 2-2^{n+1} \\ -1+2^{n} & -1+2^{n+1} \end{array}\right) . \end{aligned} \]

Direct Sums

Definition 24.15 : Sum of Subspaces

Let \(W_{1}, W_{2}, \ldots, W_{k}\) be subspaces of a vector space \(V\). We define the sum of these subspaces to be the set

\[ \left\{v_{1}+v_{2}+\cdots+v_{k}: v_{i} \in W_{i} \text { for } 1 \leq i \leq k\right\} \]

which we denote by \(W_{1}+W_{2}+\cdots+W_{k}\) or \(\sum_{i=1}^{k} W_{i}\).

Definition 24.16 : Direct Sum of Subspaces

Let \(W_{1}, W_{2}, \ldots, W_{k}\) be subspaces of a vector space \(V\). We call \(V\) the direct sum of the subspaces \(W_{1}, W_{2}, \ldots, W_{k}\) and write \(V=W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}\), if

\[ V=\sum_{i=1}^{k} W_{i} \]

and

\[ W_{j} \cap \sum_{i \neq j} W_{i}=\{0\} \quad \text { for each } j(1 \leq j \leq k) \]

Theorem 24.17 : Equivalent conditions to a direct sum

Let \(W_{1}, W_{2}, \ldots, W_{k}\) be subspaces of a finite-dimensional vector space \(V\). The following conditions are equivalent.

• (a) \(V=W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}\).
• (b) \(V=\sum_{i=1}^{k} W_{i}\) and, for any vectors \(v_{1}, v_{2}, \ldots, v_{k}\) such that \(v_{i} \in W_{i}\) \((1 \leq i \leq k)\), if \(v_{1}+v_{2}+\cdots+v_{k}=0\), then \(v_{i}=0\) for all \(i\).
• (c) Each vector \(v \in V\) can be uniquely written as \(v=v_{1}+v_{2}+\cdots+v_{k}\), where \(v_{i} \in W_{i}\).
• (d) If \(\gamma_{i}\) is an ordered basis for \(W_{i}\) \((1 \leq i \leq k)\), then \(\gamma_{1} \cup \gamma_{2} \cup \cdots \cup \gamma_{k}\) is an ordered basis for \(V\).
• (e) For each \(i=1,2, \ldots, k\), there exists an ordered basis \(\gamma_{i}\) for \(W_{i}\) 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

\[ V=\sum_{i=1}^{k} W_{i} \]

Now suppose that \(v_{1}, v_{2}, \ldots, v_{k}\) are vectors such that \(v_{i} \in W_{i}\) for all \(i\) and \(v_{1}+v_{2}+\cdots+v_{k}=0\). Then for any \(j\)

\[ -v_{j}=\sum_{i \neq j} v_{i} \in \sum_{i \neq j} W_{i} \]

But \(-v_{j} \in W_{j}\) and hence

\[ -v_{j} \in W_{j} \cap \sum_{i \neq j} W_{i}=\{0\} \]

So \(v_{j}=0\), proving (b). Now assume (b). We prove (c). Let \(v \in V\). By (b), there exist vectors \(v_{1}, v_{2}, \ldots, v_{k}\) such that \(v_{i} \in W_{i}\) 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 \(w_{i} \in W_{i}\) for all \(i\). Then

\[ \left(v_{1}-w_{1}\right)+\left(v_{2}-w_{2}\right)+\cdots+\left(v_{k}-w_{k}\right)=0 \]

But \(v_{i}-w_{i} \in W_{i}\) 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 \(W_{i}\). Since

\[ V=\sum_{i=1}^{k} W_{i} \]

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 \(v_{i j} \in \gamma_{i}\) \((j=1,2, \ldots, m_{i})\) and \((i=1,2, \ldots, k)\) and scalars \(a_{i j}\) such that

\[ \sum_{i, j} a_{i j} v_{i j}=0 \]

For each \(i\), set

\[ w_{i}=\sum_{j=1}^{m_{i}} a_{i j} v_{i j} . \]

Then for each \(i\), \(w_{i} \in \operatorname{span}\left(\gamma_{i}\right)=W_{i}\) and

\[ w_{1}+w_{2}+\cdots+w_{k}=\sum_{i, j} a_{i j} v_{i j}=0 \]

Since \(0 \in W_{i}\) 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

\[ 0=w_{i}=\sum_{j=1}^{m_{i}} a_{i j} v_{i j} \]

for each \(i\). But each \(\gamma_{i}\) is linearly independent, and hence \(a_{i j}=0\) 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 \(W_{i}\) such that \(\gamma_{1} \cup \gamma_{2} \cup \cdots \cup \gamma_{k}\) is an ordered basis for \(V\). Then

\[ V=\operatorname{span}\left(\gamma_{1} \cup \gamma_{2} \cup \cdots \cup \gamma_{k}\right) \]

\[ =\operatorname{span}\left(\gamma_{1}\right)+\operatorname{span}\left(\gamma_{2}\right)+\cdots+\operatorname{span}\left(\gamma_{k}\right)=\sum_{i=1}^{k} W_{i} \]

by repeated applications of Exercise 4.14. Fix \(j(1 \leq j \leq k)\), and suppose that, for some nonzero vector \(v \in V\),

\[ v \in W_{j} \cap \sum_{i \neq j} W_{i} \]

Then

\[ v \in W_{j}=\operatorname{span}\left(\gamma_{j}\right) \quad \text { and } \quad v \in \sum_{i \neq j} W_{i}=\operatorname{span}\left(\bigcup_{i \neq j} \gamma_{i}\right) . \]

Hence \(v\) is a nontrivial linear combination of both \(\gamma_{j}\) and \(\left(\bigcup_{i \neq j} \gamma_{i}\right)\), 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 6.2, and so we conclude that

\[ W_{j} \cap \sum_{i \neq j} W_{i}=\{0\} \]

proving (a).

Theorem 24.18 : A linear operator is diagonalizable if and only if the space is a direct sum of eigenspaces.

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 \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) 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 \(E_{\lambda_{i}}\). By Theorem 24.10, \(\gamma_{1} \cup \gamma_{2} \cup \cdots \cup \gamma_{k}\) is a basis for \(V\), and hence \(V\) is a direct sum of the \(E_{\lambda_{i}}\) 's by Theorem 24.17.

Conversely, suppose that \(V\) is a direct sum of the eigenspaces of \(T\). For each \(i\), choose an ordered basis \(\gamma_{i}\) of \(E_{\lambda_{i}}\). By Theorem 24.17, 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.

Application : Systems of Differential Equations

Example 24.19 : Solving a system of differential equations by diagonalization

Consider the system of differential equations

\[ \begin{aligned} x_{1}^{\prime} & =3 x_{1}+x_{2}+x_{3} \\ x_{2}^{\prime} & =2 x_{1}+4 x_{2}+2 x_{3} \\ x_{3}^{\prime} & =-x_{1}-x_{2}+x_{3}, \end{aligned} \]

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 \(x: \mathbb{R} \rightarrow \mathbb{R}^{3}\) be the function defined by

\[ x(t)=\left(\begin{array}{l} x_{1}(t) \\ x_{2}(t) \\ x_{3}(t) \end{array}\right) \]

The derivative of \(x\), denoted \(x^{\prime}\), is defined by

\[ x^{\prime}(t)=\left(\begin{array}{c} x_{1}^{\prime}(t) \\ x_{2}^{\prime}(t) \\ x_{3}^{\prime}(t) \end{array}\right) \]

Let

\[ A=\left(\begin{array}{rrr} 3 & 1 & 1 \\ 2 & 4 & 2 \\ -1 & -1 & 1 \end{array}\right) \]

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

It can be verified that for

\[ Q=\left(\begin{array}{rrr} -1 & 0 & -1 \\ 0 & -1 & -2 \\ 1 & 1 & 1 \end{array}\right) \quad \text { and } \quad D=\left(\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right) \]

we have \(Q^{-1} A Q=D\). Substitute \(A=Q D Q^{-1}\) into \(x^{\prime}=A x\) to obtain \(x^{\prime}=Q D Q^{-1} x\) or, equivalently, \(Q^{-1} x^{\prime}=D Q^{-1} x\). The function \(y: \mathbb{R} \rightarrow \mathbb{R}^{3}\) defined by \(y(t)=Q^{-1} x(t)\) can be shown to be differentiable (see Exercise 24.16), and \(y^{\prime}=Q^{-1} x^{\prime}\). Hence the original system can be written as \(y^{\prime}=D y\).

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

\[ y(t)=\left(\begin{array}{l} y_{1}(t) \\ y_{2}(t) \\ y_{3}(t) \end{array}\right) \]

we can rewrite \(y^{\prime}=D y\) as

\[ \left(\begin{array}{l} y_{1}^{\prime}(t) \\ y_{2}^{\prime}(t) \\ y_{3}^{\prime}(t) \end{array}\right)=\left(\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right)\left(\begin{array}{l} y_{1}(t) \\ y_{2}(t) \\ y_{3}(t) \end{array}\right)=\left(\begin{array}{l} 2 y_{1}(t) \\ 2 y_{2}(t) \\ 4 y_{3}(t) \end{array}\right) \]

The three equations

\[ \begin{aligned} & y_{1}^{\prime}=2 y_{1} \\ & y_{2}^{\prime}=2 y_{2} \\ & y_{3}^{\prime}=4 y_{3} \end{aligned} \]

are independent of each other, and thus can be solved individually. It is easily seen that the general solution to these equations is \(y_{1}(t)=c_{1} e^{2 t}, y_{2}(t)=c_{2} e^{2 t}\), and \(y_{3}(t)=c_{3} e^{4 t}\), where \(c_{1}, c_{2}\), and \(c_{3}\) are arbitrary constants. Finally,

\[ \begin{aligned} \left(\begin{array}{l} x_{1}(t) \\ x_{2}(t) \\ x_{3}(t) \end{array}\right) & =x(t)=Q y(t)=\left(\begin{array}{rrr} -1 & 0 & -1 \\ 0 & -1 & -2 \\ 1 & 1 & 1 \end{array}\right)\left(\begin{array}{l} c_{1} e^{2 t} \\ c_{2} e^{2 t} \\ c_{3} e^{4 t} \end{array}\right) \\ & =\binom{-c_{1} e^{2 t}-c_{2} e^{2 t}-2 c_{3} e^{4 t} e^{4 t}}{c_{1} e^{2 t}+c_{2} e^{2 t}+c_{3} e^{4 t}} \end{aligned} \]

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

\[ x(t)=e^{2 t}\left[c_{1}\left(\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right)+c_{2}\left(\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right)\right]+e^{4 t}\left[c_{3}\left(\begin{array}{r} -1 \\ -2 \\ 1 \end{array}\right)\right] \]

The expressions in brackets are arbitrary vectors in \(E_{\lambda_{1}}\) and \(E_{\lambda_{2}}\), respectively, where \(\lambda_{1}=2\) and \(\lambda_{2}=4\). Thus the general solution of the original system is \(x(t)=e^{2 t} z_{1}+e^{4 t} z_{2}\), where \(z_{1} \in E_{\lambda_{1}}\) and \(z_{2} \in E_{\lambda_{2}}\).

Theorem 24.20 : Generalization of solving a system of differential equations by diagonalization

Let

\[ A=\left(\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{array}\right) \]

be the coefficient matrix of the system of differential equations

\[ \begin{aligned} x_{1}^{\prime} & =a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n} \\ x_{2}^{\prime} & =a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n} \\ \vdots & \\ x_{n}^{\prime} & =a_{n 1} x_{1}+a_{n 2} x_{2}+\cdots+a_{n n} x_{n} . \end{aligned} \]

Suppose that \(A\) is diagonalizable and that the distinct eigenvalues of \(A\) are \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\). Then a differentiable function \(x: \mathbb{R} \rightarrow \mathbb{R}^{n}\) is a solution to the system if and only if \(x\) is of the form

\[ x(t)=e^{\lambda_{1} t} z_{1}+e^{\lambda_{2} t} z_{2}+\cdots+e^{\lambda_{k} t} z_{k} \]

where \(z_{i} \in E_{\lambda_{i}}\) for \(i=1,2, \ldots, k\). Moreover, the set of solutions to the system is an \(n\)-dimensional real vector space.

Proof

Let \(V=\mathbb{R}^{n}\) and view \(A\) as the matrix of the linear operator \(T=L_{A}:V\rightarrow V\). The system can be written compactly as

\[ x^{\prime}(t)=A x(t). \]

• (Only if)
  Assume \(x:\mathbb{R}\to\mathbb{R}^{n}\) is a differentiable solution of \(x^{\prime}=Ax\). Since \(A\) is diagonalizable and has distinct eigenvalues \(\lambda_{1},\ldots,\lambda_{k}\), we have a direct-sum decomposition

  \[ \mathbb{R}^{n}=E_{\lambda_{1}}\oplus E_{\lambda_{2}}\oplus\cdots\oplus E_{\lambda_{k}} . \]

  Therefore, for each \(t\), there exist unique vectors \(x_{i}(t)\in E_{\lambda_{i}}\) such that

  \[ x(t)=x_{1}(t)+x_{2}(t)+\cdots+x_{k}(t). \]

  Let \(P_{i}:\mathbb{R}^{n}\to E_{\lambda_{i}}\) denote the projection onto \(E_{\lambda_{i}}\) along the direct sum of the other eigenspaces, so that \(x_{i}(t)=P_{i}(x(t))\). Because \(P_{i}\) is linear, we have

  \[ x_{i}^{\prime}(t)=(P_{i}x)^{\prime}(t)=P_{i}(x^{\prime}(t))=P_{i}(A x(t)). \]

  Also, each \(P_{i}\) commutes with \(A\). To see this, let \(v=v_{1}+\cdots+v_{k}\) with \(v_{j}\in E_{\lambda_{j}}\). Then

  \[ A(P_{i}v)=A(v_{i})=\lambda_{i}v_{i} \]

  and

  \[ P_{i}(Av)=P_{i}\!\left(\sum_{j=1}^{k}A(v_{j})\right)=P_{i}\!\left(\sum_{j=1}^{k}\lambda_{j}v_{j}\right)=\lambda_{i}v_{i}. \]

  Hence \(P_{i}A=AP_{i}\), and therefore

  \[ x_{i}^{\prime}(t)=P_{i}(A x(t))=A(P_{i}x(t))=A x_{i}(t). \]

  But \(x_{i}(t)\in E_{\lambda_{i}}\), so \(A x_{i}(t)=\lambda_{i}x_{i}(t)\). Thus

  \[ x_{i}^{\prime}(t)=\lambda_{i}x_{i}(t). \]

  Define \(z_{i}(t)=e^{-\lambda_{i}t}x_{i}(t)\). Then

  \[ z_{i}^{\prime}(t)= -\lambda_{i}e^{-\lambda_{i}t}x_{i}(t)+e^{-\lambda_{i}t}x_{i}^{\prime}(t) =e^{-\lambda_{i}t}\bigl(x_{i}^{\prime}(t)-\lambda_{i}x_{i}(t)\bigr)=0. \]

  Hence each \(z_{i}(t)\) is constant, say \(z_{i}(t)\equiv z_{i}\), and \(z_{i}\in E_{\lambda_{i}}\) because \(x_{i}(t)\in E_{\lambda_{i}}\) for all \(t\). Therefore \(x_{i}(t)=e^{\lambda_{i}t}z_{i}\) and

  \[ x(t)=\sum_{i=1}^{k}x_{i}(t)=\sum_{i=1}^{k}e^{\lambda_{i}t}z_{i}. \]

• (If)
  Conversely, suppose

  \[ x(t)=\sum_{i=1}^{k}e^{\lambda_{i}t}z_{i} \]

  where each \(z_{i}\in E_{\lambda_{i}}\). Then \(A z_{i}=\lambda_{i}z_{i}\), and differentiating term-by-term gives

  \[ x^{\prime}(t)=\sum_{i=1}^{k}\lambda_{i}e^{\lambda_{i}t}z_{i} =\sum_{i=1}^{k}e^{\lambda_{i}t}(\lambda_{i}z_{i}) =\sum_{i=1}^{k}e^{\lambda_{i}t}(A z_{i}) =A\sum_{i=1}^{k}e^{\lambda_{i}t}z_{i} =A x(t). \]

  Hence \(x\) is a solution.

• (Vector space and dimension)
  Let \(\mathcal{S}\) denote the set of all solutions \(x:\mathbb{R}\to\mathbb{R}^{n}\) of \(x^{\prime}=Ax\). If \(x,y\in\mathcal{S}\) and \(c\in\mathbb{R}\), then \((x+y)^{\prime}=x^{\prime}+y^{\prime}=Ax+Ay=A(x+y)\) and \((c x)^{\prime}=c x^{\prime}=cAx=A(cx)\), so \(\mathcal{S}\) is a real vector space.

  Define a map \(\Psi:\mathcal{S}\to\mathbb{R}^{n}\) by \(\Psi(x)=x(0)\). This map is linear. By the representation proved above, for any solution \(x(t)=\sum_{i=1}^{k}e^{\lambda_{i}t}z_{i}\) we have

  \[ \Psi(x)=x(0)=\sum_{i=1}^{k}z_{i}. \]

  Because \(\mathbb{R}^{n}=E_{\lambda_{1}}\oplus\cdots\oplus E_{\lambda_{k}}\), every vector \(v\in\mathbb{R}^{n}\) can be written uniquely as \(v=\sum_{i=1}^{k}z_{i}\) with \(z_{i}\in E_{\lambda_{i}}\). For such \(v\), define

  \[ x_{v}(t)=\sum_{i=1}^{k}e^{\lambda_{i}t}z_{i}. \]

  Then \(x_{v}\in\mathcal{S}\) and \(\Psi(x_{v})=v\), so \(\Psi\) is onto. Also, if \(\Psi(x)=0\), then \(x(0)=0\), so \(\sum_{i=1}^{k}z_{i}=0\), which implies each \(z_{i}=0\) by uniqueness of the direct-sum decomposition, hence \(x(t)\equiv 0\); thus \(\Psi\) is one-to-one. Therefore \(\Psi\) is an isomorphism, and

  \[ \operatorname{dim}(\mathcal{S})=\operatorname{dim}(\mathbb{R}^{n})=n. \]

Exercise

Exercise 24.9

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 \([T]_{\beta}\) 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 26.32.

Exercise 24.10

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

Exercise 24.11

Let \(A\) be an \(n \times n\) matrix that is similar to an upper triangular matrix and has the distinct eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\) with corresponding multiplicities \(m_{1}, m_{2}, \ldots, m_{k}\). Prove the following statements.

• (a) \(\operatorname{tr}(A)=\sum_{i=1}^{k} m_{i} \lambda_{i}\)
• (b) \(\operatorname{det}(A)=\left(\lambda_{1}\right)^{m_{1}}\left(\lambda_{2}\right)^{m_{2}} \cdots\left(\lambda_{k}\right)^{m_{k}}\).

Exercise 24.12

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 \(T^{-1}\) (Exercise 23.8). Prove that the eigenspace of \(T\) corresponding to \(\lambda\) is the same as the eigenspace of \(T^{-1}\) corresponding to \(\lambda^{-1}\).
• (b) Prove that if \(T\) is diagonalizable, then \(T^{-1}\) is diagonalizable.

Exercise 24.13

Let \(A \in \mathrm{M}_{n \times n}(F)\). Recall from Exercise 23.14 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 \(E_{\lambda}\) and \(E_{\lambda}^{\prime}\) 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 \(\lambda\), \(\operatorname{dim}\left(E_{\lambda}\right)=\operatorname{dim}\left(E_{\lambda}^{\prime}\right)\).
• (c) Prove that if \(A\) is diagonalizable, then \(A^{t}\) is also diagonalizable.

Exercise 24.16

Let \(C \in \mathrm{M}_{m \times n}(\mathbb{R})\), and let \(Y\) be an \(n \times p\) matrix of differentiable functions. Prove \((C Y)^{\prime}=C Y^{\prime}\), where \(\left(Y^{\prime}\right)_{i j}=Y_{i j}^{\prime}\) for all \(i, j\).

Definition 24.21 : Simultaneous Diagonalizability

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 \([T]_{\beta}\) and \([U]_{\beta}\) are diagonal matrices.

Similarly, \(A, B \in \mathrm{M}_{n \times n}(F)\) are called simultaneously diagonalizable if there exists an invertible matrix \(Q \in \mathrm{M}_{n \times n}(F)\) such that both \(Q^{-1} A Q\) and \(Q^{-1} B Q\) are diagonal matrices.

Exercise 24.17

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

Exercise 24.18

• (a) Prove that if \(T\) and \(U\) are simultaneously diagonalizable operators, then \(T\) and \(U\) commute (i.e., \(T U=U T\)).
• (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 26.25.

Exercise 24.19

Let \(T\) be a diagonalizable linear operator on a finite-dimensional vector space, and let \(m\) be any positive integer. Prove that \(T\) and \(T^{m}\) are simultaneously diagonalizable.

Exercise 24.22

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and suppose that the distinct eigenvalues of \(T\) are \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\). Prove that

\[ \operatorname{span}(\{x \in V: x \text { is an eigenvector of } T\})=E_{\lambda_{1}} \oplus E_{\lambda_{2}} \oplus \cdots \oplus E_{\lambda_{k}} . \]