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?

  • $\begingroup$ Well, $\;A\;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real... $\endgroup$ – DonAntonio Dec 22 '18 at 1:30
  • $\begingroup$ There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers. $\endgroup$ – Future Math person Dec 22 '18 at 1:42

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.

  • 1
    $\begingroup$ Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks! $\endgroup$ – Future Math person Dec 22 '18 at 1:43
  • $\begingroup$ @FutureMathperson: you are most welcome! And thanks for the "acceptance"! $\endgroup$ – Robert Lewis Dec 22 '18 at 1:50
  • 1
    $\begingroup$ It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:\mathbb{R}^3 \to \mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel. $\endgroup$ – Dionel Jaime Dec 22 '18 at 4:54
  • $\begingroup$ @DionelJaime: right you are! Cheers! $\endgroup$ – Robert Lewis Dec 22 '18 at 4:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.