91Ó°ÊÓ

If \(A=\left(\begin{array}{cc}A_{0} & 0 \\ 0 & B_{0}\end{array}\right)\) is a block diagonal matrix with diagonal block matrices \(A_{0}\) and \(B_{0},\) prove that $$ e^{A t}=\left(\begin{array}{cc} e^{A_{0} t} & 0 \\ 0 & e^{B_{0} t} \end{array}\right) $$

Find the Jordan canonical form \(J\) for the matrix \(A\). You need not determine an invertible matrix \(S\) such that \(S^{-1} A S=J\). \(A=\left[\begin{array}{lllll}0 & 0 & 0 & 1 & 4 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]\).

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rrrr}0 & 1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\\0 & 0 & 0 & -1 \\\ 0 & 0 & 1 & 0\end{array}\right]$$.

Prove that the eigenvalues of an upper (or lower) triangular matrix are just the diagonal elements of the matrix.

Use some form of technology to determine the eigenvalues and a basis for each eigenspace of the given matrix. Hence, determine the dimension of each eigenspace and state whether the matrix is defective or nondefective. $$\diamond A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$

Solve the initial-value problem \(\mathbf{x}^{\prime}=A \mathbf{x},\) where \(A=\left[\begin{array}{rr}-2 & -1 \\ 1 & -4\end{array}\right], \quad \mathbf{x}(0)=\left[\begin{array}{r}0 \\ -1\end{array}\right]\).

(a) Let \(J\) be a Jordan block. Prove that the Jordan canonical form of the matrix \(J^{T}\) is \(J\) (b) Let \(A\) be an \(n \times n\) matrix. Prove that \(A\) and \(A^{T}\) have the same Jordan canonical form.

Use some form of technology to determine the eigenvalues and eigenvectors of \(A\) in the following manner: (1) Form the matrix \(A-\lambda I.\) (2) Solve the characteristic equation \(\operatorname{det}(A-\lambda I)=0\) to determine the eigenvalues of \(A.\) (3) For each eigenvalue \(\lambda_{i}\) found in \((2),\) solve the system \(\left(A-\lambda_{i} I\right) \mathbf{v}=\mathbf{0}\) to determine the eigenvectors of \(A.\) $$\diamond A=\left[\begin{array}{ll}3 & 1 \\\2 & 4\end{array}\right]$$