Example · Undergraduate
Terms
| Term | This example | Why |
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| Jordan normal form |
\(P = \begin{bmatrix} 1 & 1 \\ -1 & 0 \end{bmatrix}\), \(J = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}\)
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The matrix \(A = \begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix}\) has the eigenvalue $2$ twice, but its eigenvectors are only the multiples of $(1, -1)$, so it is not diagonalizable, and $J$ is a single $2 \times 2$ Jordan block. The columns of $P$ are the eigenvector $v_1 = (1, -1)$ and a vector $v_2 = (1, 0)$ with $(A - 2I) v_2 = v_1$. To check, \(A P = P J = \begin{bmatrix} 2 & 3 \\ -2 & -1 \end{bmatrix}\).
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| eigenvalue |
$2$ (twice)
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The eigenvalue of \(A = \begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix}\) is $2$, because \(A \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).
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