# Diagonalizable matricies and eigenvalues

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?

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

$$C = MM^T? \tag 1$$
$$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; \tag 2$$
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.
• 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. – Dionel Jaime Dec 22 '18 at 4:54