Proofs on Symmetric and Skew-Symmetric Matrices

This lesson works through three short proofs about symmetric and skew-symmetric matrices. Each one rests on one key property of the transpose: the transpose of a product reverses the order of the factors, so $(XY)^{\mathsf T} = Y^{\mathsf T} X^{\mathsf T}$.

Recall: the transpose of a product

For any matrices whose product is defined,

$(XY)^{\mathsf T} = Y^{\mathsf T} X^{\mathsf T}, \qquad (XYZ)^{\mathsf T} = Z^{\mathsf T} Y^{\mathsf T} X^{\mathsf T}.$

Also recall that $A$ is symmetric when $A^{\mathsf T} = A$, and skew-symmetric when $A^{\mathsf T} = -A$.

Proof 1: $B^{\mathsf T} A B$ is symmetric when $A$ is symmetric

Given that $A$ is symmetric, so $A^{\mathsf T} = A$. Take the transpose of $B^{\mathsf T} A B$ and reverse the order of the factors:

$\left(B^{\mathsf T} A B\right)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T} \left(B^{\mathsf T}\right)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T} B.$

Since $A^{\mathsf T} = A$, this becomes

$\left(B^{\mathsf T} A B\right)^{\mathsf T} = B^{\mathsf T} A B.$

The matrix equals its own transpose, so $B^{\mathsf T} A B$ is symmetric.

Proof 2: $B^{\mathsf T} A B$ is skew-symmetric when $A$ is skew-symmetric

Now $A$ is skew-symmetric, so $A^{\mathsf T} = -A$. The same expansion gives

$\left(B^{\mathsf T} A B\right)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T} B.$

Replacing $A^{\mathsf T}$ with $-A$,

$\left(B^{\mathsf T} A B\right)^{\mathsf T} = B^{\mathsf T} (-A) B = -\,B^{\mathsf T} A B.$

The transpose equals the negative of the matrix, so $B^{\mathsf T} A B$ is skew-symmetric.

Proof 3: $AB - BA$ is skew-symmetric when $A$ and $B$ are symmetric

Given that both $A$ and $B$ are symmetric, so $A^{\mathsf T} = A$ and $B^{\mathsf T} = B$. Transpose the difference term by term:

$(AB - BA)^{\mathsf T} = (AB)^{\mathsf T} - (BA)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T} - A^{\mathsf T} B^{\mathsf T}.$

Using $A^{\mathsf T} = A$ and $B^{\mathsf T} = B$,

$(AB - BA)^{\mathsf T} = BA - AB = -(AB - BA).$

The transpose is the negative of the original, so $AB - BA$ is skew-symmetric.

Key takeaways

• The transpose of a product reverses the order of the factors, which is the engine behind every proof here.
• $B^{\mathsf T} A B$ inherits the symmetry of $A$: it is symmetric when $A$ is symmetric and skew-symmetric when $A$ is skew-symmetric.
• For symmetric $A$ and $B$, the commutator-style difference $AB - BA$ is always skew-symmetric.