Question

Show that the matrix $B^T$$AB$ is symmetric or skewsymmetric according when $A$ is symmetric or skew symmetric.

Answer 1

we need to prove,

$B^{\prime }AB$ is symmetric if $A$ is symmetric

and

$B^{\prime }AB$ is skew is symmetric if $A$ is skew symmetric:

Let $A$ be a symmetric matrix,then

$A^{\prime }=A-------\left(1\right)$

Taking $(B^{\prime }AB{)}^{\prime }$ Let $AB=P$

$\Rightarrow (B,P{)}^{\prime }$

$\Rightarrow p{.}^{\prime }\left(B^{\prime }\right)\left[\right(AB{)}^{\prime }=B{.}^{\prime }A]$

$\Rightarrow p^1.\left(B{)}^1\right[\left(A{B}^1\right)=B^1.A]$

$\Rightarrow p^1.B\left[\right(B^1)=B]$

putting $p=AB$

$=\left(AB{)}^1\right(B)$

$=B^{\prime }{A}^{\prime }\left(B\right)[:(AB)=B^1.A^{\prime }]$

$=B^{1}AB\left[USING1\right]$

$\therefore (B^{1}AB{)}^1=B^{1}AB.$

Thus, $B^{\prime }AB$ is a symmetric matrix.

proving $B^{1}AB$ is skew symmetric if $A$ is skew symmetric

Let $A$ be a skew symmetric, then

$A^1=-A-----\left(2\right)$

$Taking\left(B^{1}AB\right)$ Putting $P=AB$

Let $AB=P$ $\left(AB{)}^1\right(B)$

$=(BP{)}^1$ $=B^{\prime }{A}^{\prime }.\left(B\right)$

$=p\left(B^1{)}^1\right[(AB{)}^1=B^1.A^1]$ $=B^{\prime }{A}^{\prime }.\left(B\right)[\because (AB{)}^1=B^1.A^1]$

$=P^{1}B\left[\right(B{)}^1=B]$ $=B^1(-A)B\left[USING\right(2\left)\right]$

$=-B^{1}AB$

$\therefore (B^{1}AB{)}^1=-B^1.AB$

Thus,$B^{1}AB$ is skew symmetric matrix .

Hence, matrix $B^{1}AB$is symmetric or skew symmetric

according as A is symmetric or skew symmetric.