Q.$AA'$ is always a symmetric matrix for any matrix $A$.

The key idea is that the transpose of $AA'$ equals itself, which is the definition of a symmetric matrix.

Step 1: Take the transpose of $AA'$. Using the reversal property of transposes, $(AA')' = (A')' A'$.

Step 2: The double transpose of $A$ is $A$ itself, so $(A')' = A$. Thus $(AA')' = A A'$. …

Why This Works — The Core Idea

The question asks whether $AA'$ is always symmetric for any matrix $A$. This is a classic property that follows directly from how transposition interacts with matrix multiplication. The key insight: a matrix $M$ is symmetric if $M' = M$. So we just need to check whether $(AA')'$ equals $AA'$ itself — and the transpose of a product reverses the order.

This reversal is the engine behind the proof. Let's apply it step by step.

Step-by-Step Reasoning

1. Start with the definition of symmetry.

   A matrix $M$ is symmetric if $M' = M$. So we need to check whether $(AA')' = AA'$.

2. Take the transpose of $AA'$.

   Using the product rule for transposes:

3. Simplify the double transpose. The transpose of a transpose brings you back to the original matrix: $(A')' = A$. So:

4. Compare the result. We have $(AA')' = AA'$, which is exactly the condition for symmetry. Therefore $AA'$ is symmetric for any matrix $A$ — no restrictions on size or entries.

A Concrete Example (Optional)

Take a $2 \times 3$ matrix: …

Method: Checking Symmetry of a Constructed Matrix via the Transpose

Use this to prove a matrix built from products/transposes is symmetric. A matrix $M$ is symmetric exactly when $M'=M$, so compute $M'$ using the transpose rules and compare.

Steps

Step 1: State the symmetry test.

$M$ is symmetric $\iff M'=M$. So take the transpose of the given expression and simplify.

Step 2: Apply the reversal law and double-transpose rule.

Use $(XY)'=Y'X'$ and $(X')'=X$. For $M=AA'$:

Step 3: Conclude and note the general pattern. …

Common Mistakes

Mistake 1: Forgetting to reverse the order in the transpose of a product.

Why it's wrong: $(AA')'=(A')'A'$, not $A'(A')'$ in the wrong order; getting the reversal wrong derails the proof. Correct approach: always apply $(XY)'=Y'X'$.

Mistake 2: Believing $AA'$ is symmetric only for square $A$. …