| eigenvalue |
$1$ (twice)
|
|
| singular value |
$\varphi$ and $1 / \varphi$, where $\varphi = \frac{1 + \sqrt{5}}{2}$
|
|
| LU decomposition |
$L = I$, $U = A$
|
|
| QR decomposition |
$Q = I$, $R = A$
|
|
| Jordan–Chevalley decomposition |
\(D = I\), \(N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)
|
The matrix \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) is a Jordan block, which is not diagonalizable, so its Jordan–Chevalley decomposition has a nonzero nilpotent part $N = A - I$. $I$ commutes with every matrix, so $D = I$ and $N$ commute.
|
| PLU decomposition |
$P = I$, $L = I$, $U = A$
|
|
| singular value decomposition |
$\Sigma = \operatorname{diag}(\varphi, 1 / \varphi)$, where $\varphi = \frac{1 + \sqrt{5}}{2}$
|
The singular values of \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) are the golden ratio $\varphi$ and $1 / \varphi$, the square roots of the eigenvalues $\frac{3 \pm \sqrt{5}}{2}$ of \(A^{\top} A = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\). Both eigenvalues of $A$ are $1$, so the singular values differ from the eigenvalues. # TODO: Check the singular values.
|
| polar decomposition |
\(Q = \frac{1}{\sqrt{5}} \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix}\), \(S = \frac{1}{\sqrt{5}} \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}\)
|
The polar factor $S$ of \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) is the positive semidefinite square root of \(A^{\top} A = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\), and $Q = A S^{-1}$ is a rotation. The eigenvalues of $S$ are the singular values $\varphi$ and $1 / \varphi$ of $A$.
|
| Cholesky decomposition |
None, since $A$ is not symmetric
|
|
| eigendecomposition |
None, since $A$ is not diagonalizable
|
The matrix \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) has no eigendecomposition, because its only eigenvalue is $1$ and its eigenvectors are the multiples of $(1, 0)$, so it does not have two independent eigenvectors.
|
| Schur decomposition |
$Q = I$, $T = A$
|
|
| Jordan normal form |
$P = I$, $J = A$
|
The matrix \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) is already a Jordan block, so it is its own Jordan normal form.
|
| Toeplitz decomposition |
\(H = \begin{bmatrix} 1 & 1/2 \\ 1/2 & 1 \end{bmatrix}\), \(K = \begin{bmatrix} 0 & 1/2 \\ -1/2 & 0 \end{bmatrix}\)
|
The Toeplitz decomposition of \(A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) has the symmetric part \(H = \begin{bmatrix} 1 & 1/2 \\ 1/2 & 1 \end{bmatrix}\) and the skew-symmetric part \(K = \begin{bmatrix} 0 & 1/2 \\ -1/2 & 0 \end{bmatrix}\). $H$ and $K$ do not commute, because $A$ is not normal.
|