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Question: Prove that the matrix \(B'AB\) is symmetric or skew symmetric according as A is symmetric or skew sy...

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

Explanation

Solution

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

Complete step-by-step answer:

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

(1)(1) Now let’s first consider A as symmetric
So if A is symmetric than A′=AA' = A………… (1)

Now we have B′ABB'AB………………….. (2)

Let’s transpose equation (2) so we get
(B′AB)′{\left( {B'AB} \right)^\prime }

Now this can be written as
⇒(B′[AB])′\Rightarrow {\left( {B'\left[ {AB} \right]} \right)^\prime }……………….. (3)

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

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

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

Thus we can say that after transposing B′ABB'AB we get B′ABB'AB that is
(B′AB)′=BAB′{\left( {B'AB} \right)^\prime } = BAB'’

Now clearly using equation (1) which explains the concept of symmetric matrix we can say that if A is symmetric then B′ABB'AB is also symmetric.
(2)(2) Now let’s first consider A as skew symmetric
So if A is skew symmetric than A′=−AA' = - A………… (5)

Let’s transpose equation (2) so we get
(B′AB)′{\left( {B'AB} \right)^\prime }

Now this can be written as
⇒(B′[AB])′\Rightarrow {\left( {B'\left[ {AB} \right]} \right)^\prime }……………….. (6)

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

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

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

Thus we can say that after transposing B′ABB'AB we get −B′AB - B'AB that is
(B′AB)′=−BAB′{\left( {B'AB} \right)^\prime } = - BAB'’

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