Question

Show that the matrix B' AB is symmetric or skew-symmetric according to A which is symmetric or skew -symmetric.

Answer 1

We suppose that A is a symmetric matrix, then A'=A
Consider (B' AB)'=(B'(AB))'=(AB)'(B')' $[\because (AB{)}^{\prime }=B^{\prime }{A}^{\prime }]$
$=B^{\prime }{A}^{\prime }\left(B\right)[\because (B^{\prime }{)}^{\prime }=B]$
$=B^{\prime }\left(A^{\prime }B\right)=B^{\prime }\left(AB\right)[\because A^{\prime }=A]$
$\therefore (B^{\prime }AB{)}^{\prime }=B^{\prime }A$
Which show that B' AB is a symmetric matrix.
Now, we suppose that A is a skew-symmetric matrix.
Then, A'=-A
Consider (B'AB)'=(B'(AB))'=(AB)'(B') $[\because (AB{)}^{\prime }=B^{\prime }{A}^{\prime }and(A^{\prime }{)}^{\prime }=A]$
=(B'A')B=B'(-A)B=-B'AB $[\because A^{\prime }=-A]$
$\therefore (B^{\prime }AB{)}^{\prime }AB$ which shows that B' AB is a skew -symmetric matrix.