Suppose A is a symmetric \(n \times n\) matrix and B is any \(n \times m\) matrix. Show that \({B^T}AB\), \({B^T}B\), and \(B{B^T}\) are symmetric matrices.

Since A is a symmetric matrix, so \({A^T} = A\).

Let \(C = {B^T}AB\).

Apply transpose on both sides of the equation \(C = {B^T}AB\).

\(\begin{aligned}{}{C^T} &= {\left( {{B^T}AB} \right)^T}\\ &= {B^T}{A^T}{\left( {{B^T}} \right)^T}\\ &= {B^T}AB\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{A^T} = A} \right)\\ &= C\end{aligned}\)

Thus, the matrix \({B^T}AB\) is symmetric.

Let \(D = {B^T}B\).

Apply transpose on both sides of the equation \(D = {B^T}B\).

\(\begin{aligned}{}{D^T} &= {\left( {{B^T}B} \right)^T}\\ &= {B^T}{\left( {{B^T}} \right)^T}\\ &= {B^T}B\\ &= D\end{aligned}\)

The matrix \({B^T}B\) is symmetric.

Let \(P = B{B^T}\).

Apply transpose on both sides of the equation \(P = B{B^T}\).

\(\begin{aligned}{}{P^T} &= {\left( {B{B^T}} \right)^T}\\ &= {\left( {{B^T}} \right)^T}{B^T}\\ &= B{B^T}\end{aligned}\)

The matrix \(B{B^T}\) is symmetric.