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

Last updated date: 20th Mar 2023

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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.

Complete step-by-step answer:

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

$(1)$ Now let’s first consider A as symmetric

So if A is symmetric than $A' = A$………… (1)

Now we have $B'AB$………………….. (2)

Let’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 }$……………….. (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 }$……………….. (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

${\left( {B'AB} \right)^\prime } = BAB'$’

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 let’s first consider A as skew symmetric

So if A is skew symmetric than $A' = - A$………… (5)

Let’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 }$……………….. (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 }$……………….. (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

${\left( {B'AB} \right)^\prime } = - BAB'$’

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.

Complete step-by-step answer:

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

$(1)$ Now let’s first consider A as symmetric

So if A is symmetric than $A' = A$………… (1)

Now we have $B'AB$………………….. (2)

Let’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 }$……………….. (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 }$……………….. (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

${\left( {B'AB} \right)^\prime } = BAB'$’

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 let’s first consider A as skew symmetric

So if A is skew symmetric than $A' = - A$………… (5)

Let’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 }$……………….. (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 }$……………….. (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

${\left( {B'AB} \right)^\prime } = - BAB'$’

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.

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