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

Concept: Symmetric and Skew-Symmetric Matrices — we check the transpose of $B'AB$ and compare it to the original.

Step 1: Take the transpose of $B'AB$:

since $(B')' = B$ and $(AB)' = B'A'$.

Step 2: If $A$ is symmetric, $A' = A$. Then:

so $B'AB$ equals its own transpose — it is symmetric.

Step 3: If $A$ is skew symmetric, $A' = -A$. Then:

so $B'AB$ is the negative of its transpose — it is skew symmetric.

The core idea here is simple: we want to check whether $B'AB$ is symmetric or skew symmetric, depending on the nature of $A$. The only tool we need is the behaviour of the transpose operation — specifically, how it interacts with matrix multiplication. There is no need to expand entries or work element-by-element; a clean algebraic proof is far more elegant and exam-friendly.

Let’s walk through it.

1. Start with the expression we need to examine.

   We are given $B'AB$, where $B$ is any matrix (presumably of compatible dimensions) and $A$ is either symmetric or skew symmetric. We want to determine the nature of $B'AB$ — that is, whether it equals its own transpose or the negative of its transpose.

2. Take the transpose of $B'AB$.

   Recall the reversal rule for transposes of products:

Applying this to $B'AB$ (where $X = B'$, $Y = A$, $Z = B$), we get:

And since $(B')' = B$, this simplifies to:

3. Now use the given property of $A$.
   • Case 1: $A$ is symmetric. By definition, $A' = A$. Substituting:

 This is exactly the original matrix. So $B'AB$ is symmetric.

• Case 2: $A$ is skew symmetric. By definition, $A' = -A$. Substituting:

 This is the negative of the original matrix. So $B'AB$ is skew symmetric.

Thus, the nature of $A$ (symmetric or skew symmetric) is inherited by $B'AB$ under the transformation $B'(\cdot)B$.

Method: Determining the type of $B'AB$ from the type of $A$

To classify $B'AB$, transpose it with the reversal law and substitute the property of $A$; the symmetric or skew nature of $A$ is inherited by $B'AB$.

Steps

Step 1: Transpose using the reversal law

$(B'AB)'=B'A'(B')'=B'A'B$, since $(B')'=B$ and the order reverses.

Step 2: Substitute the nature of $A$

If $A'=A$: $(B'AB)'=B'AB$, so it is symmetric. If $A'=-A$: $(B'AB)'=-B'AB$, so it is skew symmetric.

Step 3: State the conclusion

$B'AB$ is symmetric when $A$ is symmetric and skew symmetric when $A$ is skew symmetric, for any $B$ of compatible order.

Common Mistakes

Mistake 1: Reversing the triple product incorrectly

Why it's wrong: $(B'AB)'=B'A'(B')'=B'A'B$; the rule is $(XYZ)'=Z'Y'X'$, and forgetting to reverse — or mis-placing $B$ versus $B'$ — gives a wrong expression. Correct approach: apply $(XYZ)'=Z'Y'X'$ carefully and use $(B')'=B$.

Mistake 2: Assuming $B$ must be square or invertible

Why it's wrong: the result holds for any $B$ of compatible order — no inverse is used, only transposes. Correct approach: rely purely on the transpose rules, not on properties of $B$.