Let $A$,$B$,$C$ be three different real $3 \times 3$ matricies with the following properties:
$A$ has the complex eigenvalue $\lambda=3-5i$
$B$ has eigenvalues $\lambda=0$, $\lambda=5$, $\lambda=-5$
$C=M M^T$ for some real $3 \times 2$ matrix $M$.
Which of the matrices are necessarily diagonalizable? In the case of complex eigenvectors, diagonalization is over $\mathbb{C}$.
$(A)$ Only $B$
$(B)$ Only $A$ and $B$
$(C)$ Only $B$ and $C$
$(D)$ All three of them
$(E)$ None of them
Okay so first of all, I am almost certain that $(B)$ is diagonalizable. There are $3$ distinct eigenvalues for a $3 \times 3$ matrix so it can definitely be diagonalized.
I know for $(A)$ that because we have a real matrix, that the complex numbers must come in conjugate pairs so there must be a 2nd complex eigen value and one more real value so that $(A)$ can diagonalized too.
I'm not sure about $(C)$ though. I know that $(C)$ is a $3\times 3$ matrix but I am not sure if that justifies in there being 3 distinct eigenvalues so thus I could conclude that all three are diagonalizable (Thus $(D)$ being the answer).
Is my reasoning relatively on the right track?
Best Answer
Our OP Future Math person has correctly argued that $A$ and $B$ are diagonalizable, $A$ over $\Bbb C$ and $B$ over $\Bbb R$; in each case for the reason that the matrix has $3$ distinct eigenvalues, hence $3$ linearly independent eigenvectors.
So what about the case
$C = MM^T? \tag 1$
here we don't know too much about the eigenvalues, but we may observe that
$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; \tag 2$
that is, $C$ is a real symmetric matrix; as such, it too may be diagonalized, whether or not the eigenvalues are distinct; that real symmetric matrices are diagonalizable is a well-known result.