The matrix $\begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}$

Terms

Term	This example	Why
LU decomposition	$L = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix}$, $U = \begin{bmatrix} 4 & 2 \\ 0 & 2 \end{bmatrix}$
PLU decomposition	$P = I$, with $L$ and $U$ as in the LU decomposition
Cholesky decomposition	\[L = \begin{bmatrix} 2 & 0 \\ 1 & \sqrt{2} \end{bmatrix}\]	The matrix $A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}$ is symmetric with positive eigenvalues $\frac{7 \pm \sqrt{17}}{2}$, so it has a Cholesky decomposition. Its factor $L$ is the LU factor $U$ with each row divided by the square root of its diagonal entry, transposed.
eigendecomposition	$\Lambda = \operatorname{diag}\bigl(\frac{7 + \sqrt{17}}{2}, \frac{7 - \sqrt{17}}{2}\bigr)$, with $Q$ orthogonal
singular value decomposition	$\Sigma = \Lambda$, $U = V = Q$, as in the eigendecomposition	The matrix $A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}$ is symmetric positive definite, so its eigendecomposition $A = Q \Lambda Q^{\top}$ is also a singular value decomposition.
polar decomposition	$Q = I$, $S = A$	The matrix $A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}$ is already symmetric positive definite, so its polar decomposition is $A = I \cdot A$.
Schur decomposition	$T = \Lambda$, as in the eigendecomposition
Jordan normal form	$J = \Lambda$, as in the eigendecomposition
Jordan–Chevalley decomposition	$D = A$, $N = 0$
Toeplitz decomposition	$H = A$, $K = 0$
symmetric matrix	✓
positive definite	✓	The matrix $A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}$ is positive definite, because it is symmetric and $x^{\top} A x = 4 x_1^2 + 4 x_1 x_2 + 3 x_2^2 = (2 x_1 + x_2)^2 + 2 x_2^2,$ which is positive unless $x_2 = 0$ and $2 x_1 + x_2 = 0$, that is, unless $x = 0$.
positive semidefinite	✓

Distinctions using this example

Positive definite vs. positive semidefinite: The matrix $\begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}$ is positive definite, because its leading principal minors $4$ and $4 \cdot 3 - 2 \cdot 2 = 8$ are positive. So it is also positive semidefinite.
Types of matrix decompositions: The matrix $A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}$ is symmetric positive definite, so it has every decomposition in the table, including a Cholesky decomposition. Its eigendecomposition, Schur decomposition, Jordan normal form, and singular value decomposition all coincide, and its polar and Toeplitz decompositions are trivial.

All examples · All distinctions · All definitions

Term	This example	Why
LU decomposition	\(L = \begin{bmatrix} 1 & 0 \\ 1/2 & 1 \end{bmatrix}\), \(U = \begin{bmatrix} 4 & 2 \\ 0 & 2 \end{bmatrix}\)
PLU decomposition	$P = I$, with $L$ and $U$ as in the LU decomposition
Cholesky decomposition	\[L = \begin{bmatrix} 2 & 0 \\ 1 & \sqrt{2} \end{bmatrix}\]	The matrix \(A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\) is symmetric with positive eigenvalues $\frac{7 \pm \sqrt{17}}{2}$, so it has a Cholesky decomposition. Its factor $L$ is the LU factor $U$ with each row divided by the square root of its diagonal entry, transposed.
eigendecomposition	$\Lambda = \operatorname{diag}\bigl(\frac{7 + \sqrt{17}}{2}, \frac{7 - \sqrt{17}}{2}\bigr)$, with $Q$ orthogonal
singular value decomposition	$\Sigma = \Lambda$, $U = V = Q$, as in the eigendecomposition	The matrix \(A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\) is symmetric positive definite, so its eigendecomposition $A = Q \Lambda Q^{\top}$ is also a singular value decomposition.
polar decomposition	$Q = I$, $S = A$	The matrix \(A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\) is already symmetric positive definite, so its polar decomposition is $A = I \cdot A$.
Schur decomposition	$T = \Lambda$, as in the eigendecomposition
Jordan normal form	$J = \Lambda$, as in the eigendecomposition
Jordan–Chevalley decomposition	$D = A$, $N = 0$
Toeplitz decomposition	$H = A$, $K = 0$
symmetric matrix	✓
positive definite	✓	The matrix \(A = \begin{bmatrix} 4 & 2 \\ 2 & 3 \end{bmatrix}\) is positive definite, because it is symmetric and \(x^{\top} A x = 4 x_1^2 + 4 x_1 x_2 + 3 x_2^2 = (2 x_1 + x_2)^2 + 2 x_2^2,\) which is positive unless $x_2 = 0$ and $2 x_1 + x_2 = 0$, that is, unless $x = 0$.
positive semidefinite	✓