If $A$ and $B$ are symmetric matrices of the same order, then $A B-B A$ is a :

If $A$ and $B$ are symmetric matrices of the same order, then $AB-BA$ is a skew-symmetric matrix.

Recall that a symmetric matrix $X$ has the property that $X = X^T$, where $X^T$ is the transpose of $X$.

A skew-symmetric matrix $Y$ has the property that $Y = -Y^T$.

Let's find the transpose of $AB - BA$:

$(AB - BA)^T = (AB)^T - (BA)^T$

For any matrices $P$ and $Q$, we know that $(PQ)^T = Q^T P^T$. So:

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

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

Since $A$ and $B$ are symmetric matrices, $A = A^T$ and $B = B^T$.

Substituting into our transpose expression:

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

Since $(AB - BA)^T = -(AB - BA)$, by definition $AB - BA$ is skew-symmetric.

Note: This means all diagonal elements will be zero, and corresponding off-diagonal elements will be negatives of each other.