Q.Show that $A'A$ and $AA'$ are both symmetric matrices for any matrix $A$.

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

Step 1: Take $A'A$. Its transpose is $(A'A)'$. By the reversal property of transposes, $(A'A)' = A' (A')'$.

Step 2: Since $(A')' = A$, we get $(A'A)' = A'A$. This is exactly the condition for symmetry. …

The key idea here is simple but powerful: when you multiply a matrix by its own transpose, the result is always symmetric. This isn't a coincidence — it's baked into the algebra of transposes.

Let's understand why. A symmetric matrix is one that equals its own transpose: $M = M'$. So to show $A'A$ is symmetric, we just need to prove $(A'A)' = A'A$. Similarly for $AA'$.

The only tool we need is the reversal rule: $(AB)' = B'A'$. When you transpose a product, you reverse the order and transpose each factor. That's the entire engine of this proof.

1. Start with $A'A$.

   Take its transpose: $(A'A)'$.

   By the reversal rule, this becomes $A' (A')'$.

   Now, the transpose of a transpose brings you back: $(A')' = A$.

   So $(A'A)' = A'A$.

   That's exactly the condition for symmetry. Hence $A'A$ is symmetric.

2. Now do the same for $AA'$.

   Transpose it: $(AA')'$.

   Reverse and transpose: $(A')' A' = A A'$.

   So $(AA')' = AA'$, which means $AA'$ is symmetric too. …

Method: Proving a matrix expression is symmetric via its transpose

To prove any expression $M$ is symmetric, you show $M'=M$; to prove it is skew-symmetric you show $M'=-M$. This turns a "prove symmetric" question into a short transpose computation.

Steps

Step 1: State the target condition.

A square matrix $M$ is symmetric exactly when

So to show $A'A$ is symmetric, aim to prove $(A'A)'=A'A$.

Step 2: Transpose the whole expression and apply the reversal law.

Use $(XY)'=Y'X'$ on the product. For $A'A$ this gives $(A'A)'=A'(A')'$.

Step 3: Collapse double transposes with $(A')'=A$. …

Common Mistakes

Mistake 1: Concluding $A'A=AA'$.

Why it's wrong: the proof shows each product is individually symmetric, not that they are equal to each other — in general $A'A$ and $AA'$ are different matrices (they can even have different orders when $A$ is rectangular). Correct approach: prove symmetry of each separately; do not claim they are the same.

Mistake 2: Applying the reversal law without reversing, e.g. writing $(A'A)'=(A')'A'=AA'$. …