If A is a square matrix, then $(A^T - A)$ is-

For a square matrix $A$, $A^T$ denotes the transpose of matrix $A$, obtained by swapping rows with columns.

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, then $A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

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To determine the type of matrix $(A^T - A)$ is, check if it is skew-symmetric.

A matrix $B$ is skew-symmetric if $B^T = -B$

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Let $B = A^T - A$

Find the transpose of $B$:

$B^T = (A^T - A)^T$

Since $(X - Y)^T = X^T - Y^T$:

$B^T = (A^T)^T - A^T$

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Using the property $(A^T)^T = A$:

$B^T = A - A^T$

$B^T = -(A^T - A)$

$B^T = -B$

This proves that $(A^T - A)$ is skew-symmetric.

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$(A^T - A)$ cannot be a null matrix since that would only occur if $A^T = A$ for all square matrices, which is not true.

$(A^T - A)$ cannot be an identity matrix as there is no reason this difference would always produce the identity pattern.

$(A^T - A)$ cannot be symmetric since we proved $B^T = -B$, not $B^T = B$.

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Therefore, $(A^T - A)$ is a skew-symmetric matrix.

Note: $A^T + A$ is always symmetric, while $A^T - A$ is always skew-symmetric.