Q.If $A$, $B$ are square matrices of same order and $B$ is a skew-symmetric matrix, show that $A'BA$ is skew symmetric.

Concept: Symmetric and Skew-Symmetric Matrices — a matrix $P$ is skew-symmetric if $P' = -P$.

Step 1: We need to check whether $(A'BA)' = -A'BA$.

Step 2: Using the reversal property of transpose, $(A'BA)' = A' B' (A')' = A' B' A$. …

Idea

We test whether $A'BA$ satisfies the definition of skew-symmetry, $X' = -X$. Two transpose rules do all the work:

• $(XY)' = Y'X'$ — the order reverses under transposition.
• $(X')' = X$ — transposing twice returns the original.

The only special fact needed about $B$ is that it is skew-symmetric, i.e. $B' = -B$. Nothing is assumed about $A$.

Proof

Take the transpose of $A'BA$ and reverse the order of the three factors:

$(A'BA)' = A' \, B' \, (A')' = A' \, B' \, A.$ …

Method: Proving an expression is skew-symmetric via its transpose

To prove a matrix expression $M$ is skew-symmetric, show $M'=-M$. Using the transpose reversal law and the given symmetry/skew-symmetry of the inner matrix turns this into a short algebraic argument.

Steps

Step 1: State the target.

$M$ is skew-symmetric exactly when

Here $M=A'BA$, so aim to show $(A'BA)'=-A'BA$.

Step 2: Transpose the product with the reversal law.

Apply $(XYZ)'=Z'Y'X'$:

$(A'BA)'=A'\,B'\,(A')'=A'B'A.$ …

Common Mistakes

Mistake 1: Reversing the triple product incorrectly.

Why it's wrong: $(A'BA)'=A'B'(A')'$, i.e. the factors reverse to give $A'B'A$; writing it as $ABA'$ or $A B' A'$ is wrong. Correct approach: apply $(XYZ)'=Z'Y'X'$ carefully, and use $(A')'=A$.

Mistake 2: Using $B'=B$ instead of $B'=-B$.

Why it's wrong: $B$ is skew-symmetric, so its transpose is $-B$; treating it as symmetric loses the crucial minus sign and would "prove" the wrong conclusion. Correct approach: substitute $B'=-B$. …