Chapter 7 - Eigenvalues and Eigenvectors - 7.6 Jordan Canonical Forms - Problems - Page 486: 13

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Given: $p(\lambda)=(3-\lambda)^2(-2-\lambda)^3\lambda^2$ Find the eigenvalues: $\lambda_1=3$ with multiplicity of 2 $\lambda_2=-2$ with multiplicity of 3 $\lambda_3=0$ with multiplicity of 2 The possibilities for $\lambda_1=3$ with multiplicity of 2 are: One block of $2 \times 2$ matrix; Two blocks of $1\times 1$ The possibility for $\lambda_1=-2$ with multiplicity of 3 are: One block of $3 \times 3$; One blocks of $2 \times 2$ & one block of $1 \times 1$; Three blocks of $1 \times 1$. The possibilities for $\lambda_1=0$ with multiplicity of 2 are: One block of $2 \times 2$ matrix; Two blocks of $1\times 1$ The set S of matrices include: $J_1=\begin{bmatrix} 3 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 3 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -2 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -2 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $J_2=\begin{bmatrix} 3 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 3 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -2 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -2 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $J_3=\begin{bmatrix} 3 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 3 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -2 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $J_4=\begin{bmatrix} 3 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 3 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -2 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $J_1=\begin{bmatrix} 3 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 3 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -2 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & -2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $J_6=\begin{bmatrix} 3 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 3 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & -2 & 0& 0 & 0 & 0\\ 0 & 0 & 0 & -2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ There are $3 \times 2=6$ possible Jordan canonical forms.