Case 1: When A is a symmetric matrix i.e.

A = A’……. (1)

where A’ is the transpose of A

To prove: B’AB is also a symmetric matrix.

Calculating the transpose of B’AB

(B’AB)’= B’A’(B’)’ (By property of transpose i.e. (AB)’ = B’A’)

= B’A’B (By property of transpose i.e. (A’)’ = A)

= B’AB (from (1))

It satisfies the condition of symmetric matrix as matrix B’AB is equal to its transpose.

Hence B’AB is a symmetric matrix when A is symmetric.

Case 2: When A is a skew symmetric matrix i.e.

A = -A’……. (2)

where A’ is the transpose of A.

To prove: B’AB is also a skew symmetric matrix.

Calculating the transpose of B’AB

(B’AB)’= B’A’(B’)’ (By property of transpose i.e. (AB)’ = B’A’)

= B’A’B (By property of transpose i.e. (A’)’ = A)

= B’(-A)B (from (2))

= - (B’AB)

It satisfies the condition of skew symmetric matrix as matrix (B’AB) is equal to its transpose.

Hence (B’AB) is a skew symmetric matrix when A is skew symmetric.

∴ Both results are proved.