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:-

*

*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 $\Bbb{C}$.

*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.
 A: 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
$$
\operatorname{rank}(A^2) = \operatorname{rank}(A(-A^T)) = \operatorname{rank}(-A^TA) = \operatorname{rank}(A^TA) = \operatorname{rank}(A).
$$
A: 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.
