Some questions about a $3 \times 3 $ real skew-symmetric matrix

Question

Let $A = -A^T$ be a $3 \times 3$ real skew-symmetric matrix.

1. Prove that $A$ has an eigenvalue $\lambda = 0$ and two other pure imaginary eigenvalues, and that the eigenvectors of $A$ are orthogonal in $\mathbb{C}^3$.

2. Find $e^A$, with: $$ A = \left[ \begin{matrix} 0 & 1 & 0 \\ -1 &0 & 2 \\ 0 & -2 & 0 \end {matrix} \right ]$$

I've found a similar question here: Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$

In one of the answers it is said: "In general, a real skew-symmetric 3×3 matrix K looks like: $$\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end {matrix} \right ]$$ "

I don't understand why the general form of the matrix has that diagonal of zeros. I would have thought about something like: $$\left[ \begin{matrix} d & a & b \\ -a & e & c \\ -b & -c & f \end {matrix} \right ]$$

Wich of course has much worse determinant & characteristic polynomial...

Answer 1

Note that$$-\begin{pmatrix}d & a & b \\ -a & e & c \\ -b & -c & f\end{pmatrix}^T=\begin{pmatrix}-d & a & b \\ -a & -e & c \\ -b & -c & -f\end{pmatrix},$$which will be equal to$$\begin{pmatrix}d & a & b \\ -a & e & c \\ -b & -c & f\end{pmatrix}$$if and only if $d=e=f=0$. Can you take it from here?

Answer 2

If you negate every value on the diagonal, you must have the same value. That is only true of zero. Therefore you get 0 on the diagonal.

The eigenvalues of a real matrix must be either real or in complex conjugate pairs. If you have three eigenvalues, one of them must be a real one. Since you can prove, that skew syymetric matrices have all imaginary eigenvalues, you can directly see, that there is only one number, that is both on the real and the imaginary axis. That is 0.

Answer 3

$iA$ is a Hermitian matrix.

Hermitian matrices are always diagonalizable and always have real eigenvalues, and have orthogonal eigen-vectors.

The eigenvalues of $A$ are either $0$ or pure imaginary.

The characteristic equation of $A$ will be a degree 3 polynomial with real coefficients. It must have one real root. And it must be $0.$