Question:
If $A$ is symmetric matrix, then $B^{\top} A B$ is
Solution:
It is given that, $A$ is symmetric matrix.
$\therefore A^{T}=A$ ...(1)
Now,
$\left(B^{T} A B\right)^{T}$
$\left(B^{T} A B\right)^{T}$
$=B^{T} A^{T}\left(B^{T}\right)^{T} \quad\left[\right.$ For any matrices $\left.X, Y, Z,(X Y Z)^{\top}=Z^{\top} Y^{\top} X^{\top}\right]$
$=B^{T} A B$ [Using (1)]
Since $\left(B^{T} A B\right)^{T}=B^{T} A B$, so the matrix $B^{\top} A B$ is symmetric.
If $A$ is symmetric matrix, then $B^{\top} A B$ is symmetric