Question

If $M$ is an arbitrary real $n \times n$ matrix, then which of the following matrices will have non-negative eigenvalues?

• A. $M^2$
• B. $MM^T$
• C. $M^TM$
• D. $(M^T)^2$

Hint:

Whenever you see $MM^T$ or $M^TM$, think of them as Gram matrices — they are always positive semi-definite and hence have non-negative eigenvalues.

Answer 1

The Correct Option is B, C

Step 1: Recall property of symmetric matrices.
For any real matrix $M$, both $MM^T$ and $M^TM$ are symmetric matrices. Symmetric matrices always have real eigenvalues.

Step 2: Positive semi-definiteness.
For any vector $x \in \mathbb{R}^n$: \[ x^T(MM^T)x = (M^Tx)^T(M^Tx) = \|M^Tx\|^2 \geq 0 \] This shows that all eigenvalues of $MM^T$ are non-negative.
Similarly, \[ x^T(M^TM)x = (Mx)^T(Mx) = \|Mx\|^2 \geq 0 \] Hence all eigenvalues of $M^TM$ are also non-negative.

Step 3: Why not $M^2$ or $(M^T)^2$?
In general, $M^2$ and $(M^T)^2$ need not be symmetric and can have negative or even complex eigenvalues. So they do not guarantee non-negative eigenvalues.
\[ \boxed{\text{Therefore, the correct options are (B) and (C).}} \]