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

Terms

Term	This example	Why
eigenvalue	$i$ and $-i$	The eigenvalues of $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ are $i$ and $-i$, because $A \begin{bmatrix} 1 \\ -i \end{bmatrix} = \begin{bmatrix} i \\ 1 \end{bmatrix} = i \begin{bmatrix} 1 \\ -i \end{bmatrix}$ and $A \begin{bmatrix} 1 \\ i \end{bmatrix} = \begin{bmatrix} -i \\ 1 \end{bmatrix} = -i \begin{bmatrix} 1 \\ i \end{bmatrix}$. $A$ rotates every nonzero real vector by $90°$, so it has no real eigenvalues.
singular value	$1$ and $1$	The singular values of $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ are $1$ and $1$, because $A^{\top} A = I$, whose eigenvalues are $1$ and $1$. A rotation does not change the length of any vector.
transpose	\[A^{\top} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\]	For the rotation by $90°$, the transpose and the inverse are the same matrix, because the rotation is orthogonal.
determinant	$\det A = 0 \cdot 0 - (-1) \cdot 1 = 1$
trace	$\operatorname{tr} A = 0 + 0 = 0$
invertible matrix	\[A^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\]
array	✓
matrix	✓
orthogonal matrix	✓	The rotation $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ is orthogonal, because its inverse, the rotation by $-90°$, is its transpose.
skew symmetric matrix	✓	The rotation $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ is skew-symmetric, because $A^{\top} = -A$.
linear map	✗
symmetric matrix	✗

Array vs. matrix vs. linear map: The matrix represents rotation by $90°$ in the standard basis, but it is a grid of numbers, not a function.
Eigenvalue vs. singular value

Term	This example	Why
eigenvalue	$i$ and $-i$	The eigenvalues of \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) are $i$ and $-i$, because \(A \begin{bmatrix} 1 \\ -i \end{bmatrix} = \begin{bmatrix} i \\ 1 \end{bmatrix} = i \begin{bmatrix} 1 \\ -i \end{bmatrix}\) and \(A \begin{bmatrix} 1 \\ i \end{bmatrix} = \begin{bmatrix} -i \\ 1 \end{bmatrix} = -i \begin{bmatrix} 1 \\ i \end{bmatrix}\). $A$ rotates every nonzero real vector by $90°$, so it has no real eigenvalues.
singular value	$1$ and $1$	The singular values of \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) are $1$ and $1$, because $A^{\top} A = I$, whose eigenvalues are $1$ and $1$. A rotation does not change the length of any vector.
transpose	\[A^{\top} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\]	For the rotation by $90°$, the transpose and the inverse are the same matrix, because the rotation is orthogonal.
determinant	$\det A = 0 \cdot 0 - (-1) \cdot 1 = 1$
trace	$\operatorname{tr} A = 0 + 0 = 0$
invertible matrix	\[A^{-1} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\]
array	✓
matrix	✓
orthogonal matrix	✓	The rotation \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) is orthogonal, because its inverse, the rotation by $-90°$, is its transpose.
skew symmetric matrix	✓	The rotation \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) is skew-symmetric, because $A^{\top} = -A$.
linear map	✗
symmetric matrix	✗