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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?

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  • $\begingroup$ 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
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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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