General form and basis for $B_2$

Question

I am trying to find the general form of the element $B_2$ and work out the basis for $B_2$ also. Would somebody be able to start me off please? I don't want the answer I just wanna know the method of doing it. $$B_2=\{X\in M_5(\mathbb{C})|X^TS+SX=0\}$$ $$S\in M_5(\mathbb{C})$$ $$S= \begin{bmatrix} 1&0&0&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ \end{bmatrix}$$

Answer 1

You can find a basis by computing $E_{ij}^TS+SE_{ij}$ for each $i,j$ (where $E_{ij}$ is the standard unit matrix with $1$ in the $(i,j)$-entry and $0$'s elsewhere). Once you've computed these expressions, you can find certain combinations that yield $0$.

For example, consider $E_{12} = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$. We get $E_{12}^TS+SE_{12} = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ (so $E_{12}$ itself isn't in $B_2$).

Now try other standard unit vectors and you'll see that $E_{41}$ yields the same thing as $E_{12}$. Therefore, $A=E_{12}-E_{41}=\begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ yields $A^TS+SA=0$ and thus $A=E_{12}-E_{41} \in B_2$.