Diagonalization by Similarity Transformations

Diagonalize $\mathbf{A} = \begin{pmatrix} -8 & -6 \\ 12 & 10 \end{pmatrix}$ with a similarity transformation, or else explain why $\mathbf{A}$ can't be diagonalized.

Show that similar matrices need not have the same eigenvectors by giving an example of two matrices that are similar but have different eigenspaces.

$\lambda =2$ is an eigenvalue for $\mathbf{A} = \begin{pmatrix} 3 & 2 & 1 \\ 0 & 2 & 0 \\ -2 & -3 & 0 \end{pmatrix}$. Find $\operatorname {alg\, mult}_{\mathbf{A}}(\lambda )$ as well as $\operatorname {geo\, mult}_{\mathbf{A}}(\lambda )$. Can you conclude anything about the diagonalizability of $\mathbf{A}$ from these results?

If $\mathbf{B} = \mathbf{P}^{-1}\mathbf{A}\mathbf{P}$, explain why $\mathbf{B}^k = \mathbf{P}^{-1}\mathbf{A}^k\mathbf{P}$.

Compute $\lim_{n\to \infty } \mathbf{A}^n$ for $\mathbf{A} = \begin{pmatrix} 7/5 & 1/5 \\ -1 & 1/2 \end{pmatrix}$.

Let $\left\{ \overrightarrow {x}_1, \overrightarrow {x}_2, \ldots , \overrightarrow {x}_t\right\}$ be a set of linearly independent eigenvectors for $\mathbf{A}_{n\times n}$ associated with respective eigenvalues $\left\{ \lambda_1, \lambda_2, \ldots , \lambda_t\right\}$, and let $\mathbf{X}$ be any $n \times (n-t)$ matrix such that $\mathbf{P}_{n\times n} = (\overrightarrow {x}_1 \mid \cdots \mid \overrightarrow {x}_t \mid \mathbf{X})$ is nonsingular. Prove that if $\mathbf{P}^{-1} = \begin{pmatrix} \overrightarrow {y}_1^{*} \\ \vdots \\ \overrightarrow {y}_t^{*} \\ \mathbf{Y}^{*} \end{pmatrix}$, where the $\overrightarrow {y}_i^{*}$'s are rows and $\mathbf{Y}^{*}$ is $(n-t) \times n$, then $\left\{ \overrightarrow {y}_1^{*}, \overrightarrow {y}_2^{*}, \ldots , \overrightarrow {y}_t^{*}\right\}$ is a set of linearly independent left-hand eigenvectors associated with $\left\{ \lambda_1, \lambda_2, \ldots , \lambda_t\right\}$, respectively (i.e., $\overrightarrow {y}_i^{*}\mathbf{A} = \lambda_i \overrightarrow {y}_i^{*}$).

Let $\mathbf{A}$ be a diagonalizable matrix, and let $\rho (\star )$ denote the spectral radius. Prove that $\lim_{k\to \infty } \mathbf{A}^k = \mathbf{0}$ if and only if $\rho (\mathbf{A}) < 1$.

Note: This result also holds for nondiagonalizable matrices, as demonstrated elsewhere in the text.

Apply the technique used to prove Schur's triangularization theorem to construct an orthogonal matrix $\mathbf{P}$ such that $\mathbf{P}^T\mathbf{A}\mathbf{P}$ is upper triangular for $\mathbf{A} = \begin{pmatrix} 13 & -9 \\ 16 & -11 \end{pmatrix}$.

Verify the Cayley--Hamilton theorem for $\mathbf{A} = \begin{pmatrix} 1 & -4 & -4 \\ 8 & -11 & -8 \\ -8 & 8 & 5 \end{pmatrix}$.

Hint: This is the matrix from Example 7.2.1 on p. 507.

Since each row sum in the following symmetric matrix $\mathbf{A}$ is $4$, it's clear that $\overrightarrow {x} = (1,1,1,1)^T$ is both a right-hand and left-hand eigenvector associated with $\lambda =4 \in \sigma (\mathbf{A})$. Use the deflation technique (rescale $\overrightarrow {x}$ to unit length $\overrightarrow {y}$, build an orthogonal matrix $\mathbf{P} = [\overrightarrow {y} \mid \mathbf{X}]$ having $\overrightarrow {y}$ as its first column, and then the remaining eigenvalues of $\mathbf{A}$ are exactly the eigenvalues of the smaller matrix $\mathbf{X}^T\mathbf{A}\mathbf{X}$) to determine the remaining eigenvalues of

Explain why $\mathbf{A}\mathbf{G}_i = \mathbf{G}_i\mathbf{A} = \lambda_i \mathbf{G}_i$ for the spectral projector $\mathbf{G}_i$ associated with the eigenvalue $\lambda_i$ of a diagonalizable matrix $\mathbf{A}$.

Prove that $\mathbf{A} = \overrightarrow {c}_{n\times 1}\overrightarrow {d}_{1\times n}^T$ is diagonalizable if and only if $\overrightarrow {d}^T\overrightarrow {c} \ne 0$.

Prove that $\mathbf{A} = \begin{pmatrix} \mathbf{W} & \mathbf{0} \\ \mathbf{0} & \mathbf{Z} \end{pmatrix}$ is diagonalizable if and only if $\mathbf{W}_{s\times s}$ and $\mathbf{Z}_{t\times t}$ are each diagonalizable.

Prove that if $\mathbf{A}\mathbf{B} = \mathbf{B}\mathbf{A}$, then $\mathbf{A}$ and $\mathbf{B}$ can be simultaneously triangularized by a unitary similarity transformation---i.e., $\mathbf{U}^{*}\mathbf{A}\mathbf{U} = \mathbf{T}_1$ and $\mathbf{U}^{*}\mathbf{B}\mathbf{U} = \mathbf{T}_2$ for some unitary matrix $\mathbf{U}$.

Hint: Recall Exercise 7.1.20 along with the development of Schur's triangularization theorem.

For diagonalizable matrices, prove that $\mathbf{A}\mathbf{B} = \mathbf{B}\mathbf{A}$ if and only if $\mathbf{A}$ and $\mathbf{B}$ can be simultaneously diagonalized---i.e., $\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}_1$ and $\mathbf{P}^{-1}\mathbf{B}\mathbf{P} = \mathbf{D}_2$ for some $\mathbf{P}$.

Hint: If $\mathbf{A}$ and $\mathbf{B}$ commute, then so do $\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \begin{pmatrix} \lambda_1\mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{D} \end{pmatrix}$ and $\mathbf{P}^{-1}\mathbf{B}\mathbf{P} = \begin{pmatrix} \mathbf{W} & \mathbf{X} \\ \mathbf{Y} & \mathbf{Z} \end{pmatrix}$.

Explain why the following ``proof'' of the Cayley--Hamilton theorem is not valid: $p(\lambda ) = \operatorname {det}\left(\mathbf{A}-\lambda \mathbf{I}\right) \implies p(\mathbf{A}) = \operatorname {det}\left(\mathbf{A}-\mathbf{A}\mathbf{I}\right) = \operatorname {det}\left(\mathbf{0}\right) = 0$.

Show that the eigenvalues of the finite difference matrix

are $\lambda_j = 4\sin^2\dfrac {j\pi }{2(n+1)}$, $1 \le j \le n$.

Example 7.2.5 establishes that the eigenvalues of this finite difference matrix are $\lambda_j = 2 - 2\cos \dfrac {j\pi }{n+1}$, $1 \le j \le n$.

Let $\mathbf{N} = \begin{pmatrix} 0 & 1 & & \\ & \ddots & \ddots & \\ & & \ddots & 1 \\ & & & 0 \end{pmatrix}_{n\times n}$.

A Toeplitz matrix having the form

is called a circulant matrix. If $p(x) = c_0+c_1 x+\cdots +c_{n-1}x^{n-1}$, and if $\left\{ 1,\xi ,\xi^2,\ldots ,\xi^{n-1}\right\}$ are the $n$th roots of unity, then the results of Exercise 5.8.12 (p. 379) insure that

in which $\mathbf{F}_n$ is the Fourier matrix of order $n$. Verify these facts for the circulant below by computing its eigenvalues and eigenvectors directly:

Suppose that $(\lambda , \overrightarrow {x})$ and $(\mu , \overrightarrow {y}^{*})$ are right-hand and left-hand eigenpairs for $\mathbf{A} \in \mathbb {R}^{n\times n}$---i.e., $\mathbf{A}\overrightarrow {x}=\lambda \overrightarrow {x}$ and $\overrightarrow {y}^{*}\mathbf{A}=\mu \overrightarrow {y}^{*}$. Explain why $\overrightarrow {y}^{*}\overrightarrow {x}=0$ whenever $\lambda \ne \mu$.

Consider $\mathbf{A} \in \mathbb {R}^{n\times n}$.

Consider $\mathbf{A} \in \mathbb {R}^{n\times n}$ with $\lambda \in \sigma (\mathbf{A})$.

For $\mathbf{A} \in \mathbb {R}^{n\times n}$ with $\sigma (\mathbf{A}) = \left\{ \lambda_1,\lambda_2,\ldots ,\lambda_k\right\}$, show $\mathbf{A}$ is diagonalizable if and only if $\mathbb {R}^{n} = N(\mathbf{A}-\lambda_1\mathbf{I}) \oplus N(\mathbf{A}-\lambda_2\mathbf{I}) \oplus \cdots \oplus N(\mathbf{A}-\lambda_k\mathbf{I})$.

Hint: Recall Exercise 5.9.14.

Schur's triangularization theorem insures that every square matrix $\mathbf{A}$ is unitarily similar to an upper-triangular matrix---say, $\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\mathbf{T}$. But even when $\mathbf{A}$ is real, $\mathbf{U}$ and $\mathbf{T}$ may have to be complex if $\mathbf{A}$ has some complex eigenvalues. However, the matrices (and the arithmetic) can be constrained to be real by settling for a block-triangular result with $2\times 2$ or scalar entries on the diagonal. Prove that for each $\mathbf{A} \in \mathbb {R}^{n\times n}$ there exists an orthogonal matrix $\mathbf{P} \in \mathbb {R}^{n\times n}$ and real matrices $\mathbf{B}_{ij}$ such that

If $\mathbf{B}_{jj} = [\lambda_j]$ is $1\times 1$, then $\lambda_j \in \sigma (\mathbf{A})$, and if $\mathbf{B}_{jj}$ is $2\times 2$, then $\sigma (\mathbf{B}_{jj}) = \left\{ \lambda_j, \overline{\lambda_j}\right\} \subseteq \sigma (\mathbf{A})$.

When $\mathbf{A} \in \mathbb {R}^{n\times n}$ is diagonalizable by a similarity transformation $\mathbf{S}$, then $\mathbf{S}$ may have to be complex if $\mathbf{A}$ has some complex eigenvalues. Analogous to Exercise 7.2.25, we can stay in the realm of real numbers by settling for a block-diagonal result with $1\times 1$ or $2\times 2$ entries on the diagonal. Prove that if $\mathbf{A} \in \mathbb {R}^{n\times n}$ is diagonalizable with real eigenvalues $\left\{ \rho_1,\ldots ,\rho_r\right\}$ and complex eigenvalues $\left\{ \lambda_1,\overline{\lambda_1},\lambda_2,\overline{\lambda_2},\ldots ,\lambda_t,\overline{\lambda_t}\right\}$ with $2t+r=n$, then there exists a nonsingular $\mathbf{P} \in \mathbb {R}^{n\times n}$ and $\mathbf{B}_j$'s $\in \mathbb {R}^{2\times 2}$ such that

and where $\mathbf{B}_j$ has eigenvalues $\lambda_j$ and $\overline{\lambda_j}$.

For $\mathbf{A} \in \mathcal{C}^{n\times n}$, prove that $\overrightarrow {x}^{*}\mathbf{A}\overrightarrow {x}=0$ for all $\overrightarrow {x} \in \mathcal{C}^{n\times 1} \implies \mathbf{A}=\mathbf{0}$. Show that $\overrightarrow {x}^T\mathbf{A}\overrightarrow {x}=0$ for all $\overrightarrow {x} \in \mathbb {R}^{n\times 1} \centernot \implies \mathbf{A}=\mathbf{0}$, even if $\mathbf{A}$ is real.