# A real skew symmetric matrix of order $3$ is diagonalizable over $\Bbb{C}$

Let $$M$$ be a $$3$$ $$\times$$ $$3$$ skew symmetric matrix with real entries.

Then I need to show that $$M$$ is diagonalizable over $$\Bbb{C}$$.

This has been my attempt.

The characteristic polynomial will be of degree $$3$$ and will have real coefficients. So if there are complex roots of this polynomial, they will be in pairs.

Since a skew symmetric matrix can only have eigen values either $$0$$ or purely imaginary, we can conclude that $$0$$ will definitely be an eigen value of $$M$$ since the complex ones are in pairs.

So there are two possibilities:-

1. Eigen values of $$M$$ are $$z_1, z_2$$ and $$0$$ where $$z_1$$ and $$z_2$$ are complex numbers and conjugates of each other. In this case since eigen values are distinct, we can conclude that $$M$$ is diagonalizable over $$\Bbb{C}$$.
2. Eigen values are $$0, 0, 0$$. From here, I can't conclude that $$M$$ is diagonalizable over $$\Bbb{C}$$.

I want help in second case.

• If the eigenvalues are three zeroes, what is the dimension of the kernel? – Leo Jun 28 '20 at 10:40
• It can't be 0..it could be 1,2 or 3 – Gitika Jun 28 '20 at 11:22

First of all, the result is immediate if we apply the spectral theorem.

That notwithstanding: with your work, we have reduced our consideration to the case that $$A$$ is skew symmetric and has only zero as an eigenvalue. We can see that $$A$$ must be the zero matrix in this case as follows:

If $$A$$ is non-zero with a zero-eigenvalue, then it must hold that $$A^3 = 0$$, which means that we must have $$\operatorname{rank}(A^2) < \operatorname{rank}(A)$$. However, we note that $$A^TA$$ has the same rank as $$A$$. So, if $$A$$ is ske-symmetric, then $$\operatorname{rank}(A^2) = \operatorname{rank}(A(-A^T)) = \operatorname{rank}(-A^TA) = \operatorname{rank}(A^TA) = \operatorname{rank}(A).$$

• If A is non-zero with a zero eigen-value then why is it that A$^3$ = 0? – Gitika Jun 28 '20 at 12:55
• @Gitika Cayley-Hamilton theorem – Ben Grossmann Jun 28 '20 at 13:24
• Okay..but if A$^3$ = 0 then how did you get that inequality between ranks of A and A$^2$? – Gitika Jun 28 '20 at 15:01
• If the rank of $A$ is equal to that of $A^2$, then the restriction of $A$ to the image (column space) of $A$ is invertible which means that $A^3$ and $A^2$ have the same rank. – Ben Grossmann Jun 28 '20 at 15:23
• Another approach: the image of $A$ is an invariant subspace, so it must contain an eigenvector, which means that the image kernel of $A$ intersect, which means that the rank of $A^2$ is smaller than that of $A$ – Ben Grossmann Jun 28 '20 at 15:25

In the spirit of the proof of the spectral theorem you can argue as follows, in any finite dimension. If $$0$$ is the only eigenvalue, then (as long as the dimension is positive) you can find a real eigenvector, which by definition spans a subspace stable under the action of $$A$$. For any matrix, the orthogonal complement of an $$A$$-staple subspace is $$A^T$$-stable, but in the skew symmetric case that means $$-A$$-stable, or just $$A$$-stable. So you can restrict to that orthogonal complement, and the restriction of that action of $$A$$ to it still has only $$0$$ as eigenvalue. So you can continue by induction on the dimension, finding new eigenvectors until you hit dimension $$0$$ and you have established a basis of eigenvectors for $$0$$. Which of course means that you had $$A=0$$ to begin with.