The matrix $\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$

Terms

Term	This example	Why
domain	$\reals^3$	The matrix $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ has $3$ columns, so the domain of $x \mapsto A x$ is $\reals^3$.
codomain	$\reals^2$	The matrix $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ has $2$ rows, so the codomain of $x \mapsto A x$ is $\reals^2$.
column space	$\reals^2$	The column space of $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ is all of $\reals^2$, because the first two columns are the standard basis vectors $(1, 0)$ and $(0, 1)$.
kernel	\[\ker A = \operatorname{span} \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}\]	The kernel of $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ is a line in $\reals^3$, because $A x = 0$ means $x_1 = -2 x_3$ and $x_2 = -3 x_3$, so every solution is a multiple of $(2, 3, -1)$.
eigenvalue	None	The matrix $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ has no eigenvalues, because only square matrices have eigenvalues.
singular value	$\sqrt{14}$ and $1$	The singular values of $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ are $\sqrt{14}$ and $1$, the square roots of the eigenvalues of $A A^{\top} = \begin{bmatrix} 5 & 6 \\ 6 & 10 \end{bmatrix}$. That matrix has trace $15$ and determinant $50 - 36 = 14$, so its eigenvalues are the roots of $\lambda^2 - 15 \lambda + 14 = (\lambda - 14)(\lambda - 1)$. The $3 \times 3$ matrix $A^{\top} A$ has the same nonzero eigenvalues, plus $0$.
full rank	✓

Term	This example	Why
domain	$\reals^3$	The matrix \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) has $3$ columns, so the domain of $x \mapsto A x$ is $\reals^3$.
codomain	$\reals^2$	The matrix \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) has $2$ rows, so the codomain of $x \mapsto A x$ is $\reals^2$.
column space	$\reals^2$	The column space of \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) is all of $\reals^2$, because the first two columns are the standard basis vectors $(1, 0)$ and $(0, 1)$.
kernel	\[\ker A = \operatorname{span} \begin{bmatrix} 2 \\ 3 \\ -1 \end{bmatrix}\]	The kernel of \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) is a line in $\reals^3$, because $A x = 0$ means $x_1 = -2 x_3$ and $x_2 = -3 x_3$, so every solution is a multiple of $(2, 3, -1)$.
eigenvalue	None	The matrix \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) has no eigenvalues, because only square matrices have eigenvalues.
singular value	$\sqrt{14}$ and $1$	The singular values of \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) are $\sqrt{14}$ and $1$, the square roots of the eigenvalues of \(A A^{\top} = \begin{bmatrix} 5 & 6 \\ 6 & 10 \end{bmatrix}\). That matrix has trace $15$ and determinant $50 - 36 = 14$, so its eigenvalues are the roots of $\lambda^2 - 15 \lambda + 14 = (\lambda - 14)(\lambda - 1)$. The $3 \times 3$ matrix $A^{\top} A$ has the same nonzero eigenvalues, plus $0$.
full rank	✓