| LU decomposition |
\(L = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\), \(U = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}\)
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Subtracting $2$ times row $1$ from row $2$ gives $U$, and $L$ records the multiplier $2$.
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| Jordan–Chevalley decomposition |
$D = A$, $N = 0$
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The matrix \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) has distinct eigenvalues, so it is diagonalizable. Thus its Jordan–Chevalley decomposition has $D = A$ and $N = 0$.
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| PLU decomposition |
$P = I$, with $L$ and $U$ as in the LU decomposition
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The matrix \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) has an LU decomposition, so $P = I$ works. With partial pivoting, the rows are exchanged first instead, because $4$ is larger than $2$.
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| singular value decomposition |
$\Sigma = \operatorname{diag}\bigl(\sqrt{15 + \sqrt{221}}, \sqrt{15 - \sqrt{221}}\bigr) \approx \operatorname{diag}(5.46, 0.37)$
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The singular values of \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) are the square roots of the eigenvalues $15 \pm \sqrt{221}$ of \(A^{\top} A = \begin{bmatrix} 20 & 14 \\ 14 & 10 \end{bmatrix}\). Their product is $\lvert \det A \rvert = 2$.
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| polar decomposition |
\(Q = \frac{1}{\sqrt{34}} \begin{bmatrix} 5 & -3 \\ 3 & 5 \end{bmatrix}\), \(S = \frac{1}{\sqrt{34}} \begin{bmatrix} 22 & 14 \\ 14 & 12 \end{bmatrix}\)
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The polar factor $S$ of \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) is the positive semidefinite square root of \(A^{\top} A = \begin{bmatrix} 20 & 14 \\ 14 & 10 \end{bmatrix}\), and $Q = A S^{-1}$ is a rotation.
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| Cholesky decomposition |
None, since $A$ is not symmetric
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| eigendecomposition |
\(P = \begin{bmatrix} 1 & 1 \\ \frac{1 + \sqrt{17}}{2} & \frac{1 - \sqrt{17}}{2} \end{bmatrix}\), $D = \operatorname{diag}\bigl(\frac{5 + \sqrt{17}}{2}, \frac{5 - \sqrt{17}}{2}\bigr)$
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The matrix \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) has the distinct eigenvalues $\lambda = \frac{5 \pm \sqrt{17}}{2}$, with eigenvectors $(1, \lambda - 2)$, so it has an eigendecomposition. The eigenvectors are not perpendicular, because $A$ is not symmetric.
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| Schur decomposition |
\[T = \begin{bmatrix} \frac{5 + \sqrt{17}}{2} & -3 \\ 0 & \frac{5 - \sqrt{17}}{2} \end{bmatrix}\]
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In the Schur decomposition of \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\), the first column of $Q$ is a unit eigenvector for $\frac{5 + \sqrt{17}}{2}$, and the second column is perpendicular to it. The entry $-3$ above the diagonal is nonzero because $A$ is not normal.
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| Jordan normal form |
$J = D$, as in the eigendecomposition
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| Toeplitz decomposition |
\(H = \begin{bmatrix} 2 & 5/2 \\ 5/2 & 3 \end{bmatrix}\), \(K = \begin{bmatrix} 0 & -3/2 \\ 3/2 & 0 \end{bmatrix}\)
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The Toeplitz decomposition of \(A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}\) splits $A$ into its symmetric part $H = \frac{1}{2}(A + A^{\top})$ and its skew-symmetric part $K = \frac{1}{2}(A - A^{\top})$, which average and split the off-diagonal entries $1$ and $4$.
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