| LU decomposition |
None, without exchanging rows
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In any product $L U$, the top-left entry $0$ would force $L$ or $U$ to be singular, but $A$ is invertible. Exchanging the rows gives $P A = I = I \cdot I$.
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| QR decomposition |
$Q = A$, $R = I$
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The matrix is already orthogonal.
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| singular value decomposition |
$U = A$, $\Sigma = I$, $V = I$
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| Jordan–Chevalley decomposition |
$D = A$, $N = 0$
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| PLU decomposition |
$P = A$, $L = I$, $U = I$
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The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) has no LU decomposition, but it has the PLU decomposition $A = A \cdot I \cdot I$, because $A$ is itself the permutation matrix that exchanges the two rows.
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| polar decomposition |
$Q = A$, $S = I$
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The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) is orthogonal, so its polar decomposition is $A = A \cdot I$.
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| Cholesky decomposition |
None, since $A$ has the negative eigenvalue $-1$
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The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) is symmetric but has no Cholesky decomposition, because its eigenvalue $-1$ is negative, so it is not positive definite.
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| eigendecomposition |
\(Q = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\), $\Lambda = \operatorname{diag}(1, -1)$
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The matrix \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) is symmetric, so it has an eigendecomposition with an orthogonal $Q$: it fixes $(1, 1)$ and flips $(1, -1)$.
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| Schur decomposition |
$T = \Lambda$, as in the eigendecomposition
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| Jordan normal form |
$J = \Lambda$, as in the eigendecomposition
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| Toeplitz decomposition |
$H = A$, $K = 0$
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The Toeplitz decomposition of \(A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\) has $H = A$ and $K = 0$, because $A$ is symmetric.
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