# A subset of $SL(2,\mathbb{C})$

Let $\mathcal{H}$ be the real vector space of $2 \times 2$ complex Hermitian matrices. Set

$$\mathcal{K} := \{A \in SL(2,\mathbb{C}) : \forall H \in \mathcal{H}, \space A^{-1}H = HA^* \}.$$

Here $A^*$ denotes the conjugate transpose of $A$. I'm trying to determine which matrices are exactly the elements of $\mathcal{K}$. It seems to me that $\space \mathcal{K} = \{I_2,-I_2\}$, but all my attempts to prove this have failed.

Any ideas would be appreciated.

Fix $A\in\mathcal K$. Taking $H=I_2$ you get that $A^*=A^{-1}$. Thus $A^*H=HA^*$ for all Hermitian $H$, which implies $AH=HA$ for all $H$ (take the conjugate transpose of each side). Because $M_2(\mathbb C)$ is the complex linear span of $\mathcal H$, this implies that $A$ commutes with every $2$-by-$2$ matrix, hence $A$ is a scalar multiple of the identity. Because $\det(A)=1$, the scalar is $\pm1$.
• I see, Jonas. I didn't know that $M_2(\mathbb{C})$ was the complex linear span of $\mathcal{H}$. – ragrigg Oct 24 '13 at 4:43
• @ragrigg: Yep, if $X$ is in $M_n(\mathbb C)$, then $A=\frac12(X+X^*)$ and $B=\frac1{2i}(X-X^*)$ are Hermitian, and $X=A+iB$. (The $1$-by-$1$ case gives the real and imaginary parts of a complex number.) – Jonas Meyer Oct 24 '13 at 4:45
• mmmm... I did know that. Never came to my mind to apply it here. I concluded that any $A \in \mathcal{K}$ commutes with all $H \in \mathcal{H}$, but got stuck there. – ragrigg Oct 24 '13 at 4:51
• @ragrigg: Working with $\mathcal H$ directly, you could show that commuting with $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ implies that a matrix is diagonal, and that a diagonal matrix that commutes with $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ is scalar. – Jonas Meyer Oct 24 '13 at 4:54
• But $\left[ \begin{array}{} 0 & 1 \\ 0 & 1 \end{array} \right]$ is not in $\mathcal{H}$, right?... GOT IT. You edited the previous comment at the same time I was asking this. – ragrigg Oct 24 '13 at 5:07