Question

Prove that the matrix $B'AB$ is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Hint: Use the condition of a matrix A to be symmetric than it must satisfy $A' = A$ and if a matrix A is skew symmetric than it must satisfy $A' = - A$, where $A'$ refers to the transpose of matrix A. First consider A as symmetric and evaluate $B'AB$ by taking its transpose, then consider A as skew symmetric and evaluate $B'AB$ by taking its transpose.

Now we have to comment upon $B'AB$ being symmetric or skew symmetric depending upon A is symmetric or skew symmetric.

$(1)$ Now letâ€™s first consider A as symmetric
So if A is symmetric than $A' = A$â€¦â€¦â€¦â€¦ (1)

Now we have $B'AB$â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)

Letâ€™s transpose equation (2) so we get
${\left( {B'AB} \right)^\prime }$

Now this can be written as
$\Rightarrow {\left( {B'\left[ {AB} \right]} \right)^\prime }$â€¦â€¦â€¦â€¦â€¦â€¦.. (3)

Using the property of transpose that is ${\left( {PQ} \right)^\prime } = Q'P'$ in equation (3) we get
$\Rightarrow {\left( {AB} \right)^\prime }{\left( {{B^\prime }} \right)^\prime }$â€¦â€¦â€¦â€¦â€¦â€¦.. (4)

Now using the property of transpose that ${\left( {P'} \right)^\prime } = P$ and another property mentioned above that is ${\left( {PQ} \right)^\prime } = Q'P'$ in equation (4) we get
$\Rightarrow B'A'B$

But A was considered as a symmetric matrix hence $A' = A$, using this we can say
$\Rightarrow B'A'B = B'AB$

Thus we can say that after transposing $B'AB$ we get $B'AB$ that is
${\left( {B'AB} \right)^\prime } = BAB'$â€™

Now clearly using equation (1) which explains the concept of symmetric matrix we can say that if A is symmetric then $B'AB$ is also symmetric.
$(2)$ Now letâ€™s first consider A as skew symmetric
So if A is skew symmetric than $A' = - A$â€¦â€¦â€¦â€¦ (5)

Letâ€™s transpose equation (2) so we get
${\left( {B'AB} \right)^\prime }$

Now this can be written as
$\Rightarrow {\left( {B'\left[ {AB} \right]} \right)^\prime }$â€¦â€¦â€¦â€¦â€¦â€¦.. (6)

Using the property of transpose that is ${\left( {PQ} \right)^\prime } = Q'P'$ in equation (6) we get
$\Rightarrow {\left( {AB} \right)^\prime }{\left( {{B^\prime }} \right)^\prime }$â€¦â€¦â€¦â€¦â€¦â€¦.. (7)

Now using the property of transpose that ${\left( {P'} \right)^\prime } = P$and another property mentioned above that is ${\left( {PQ} \right)^\prime } = Q'P'$ in equation (7) we get
$\Rightarrow B'A'B$

But A was considered as a skew symmetric matrix hence $A' = - A$, using this we can say
$\Rightarrow B'A'B = - B'AB$

Thus we can say that after transposing $B'AB$ we get $- B'AB$ that is
${\left( {B'AB} \right)^\prime } = - BAB'$â€™

Now clearly using equation (5) which explains the concept of skew symmetric matrix we can say that if A is skew symmetric then $B'AB$ is also skew symmetric.

Note: Whenever we face such types of problems the key point is to have a good grasp over the properties of transpose of a matrix, some of them are stated above in solution. This will help in getting the right track to reach the answer.