Let \(W\) be the set of all \(2 \times 2\) complex Hermitian matrices, that is, the set of \(2 \times 2\) complex matrices \(A\) such that \(A_{ij} = \overline{A_{ji}}\) (the bar denoting complex conjugation). As we pointed out in Example 6 of Chapter 2, \(W\) is a vector space over the field of real numbers, under the usual operations. Verify that \[ (x, y, z, t) \rightarrow \begin{bmatrix} t + x & y + iz \\ y - iz & t - x \end{bmatrix} \] is an isomorphism of \(\RR^4\) onto \(W\).
Solution
Define \[ \Phi(x,y,z,t)= \begin{bmatrix} t+x&y+iz\\ y-iz&t-x \end{bmatrix}. \] Each entry depends real-linearly on \((x,y,z,t)\), so \(\Phi\) is real-linear, and its image is Hermitian.
If \(\Phi(x,y,z,t)=0\), then \(t+x=t-x=0\) and \(y+iz=0\). Since \(x,y,z,t\) are real, these equations give \(x=y=z=t=0\). Thus \(\Phi\) is injective.
Conversely, every Hermitian matrix has the form \[ A=\begin{bmatrix}a&c+id\\c-id&b\end{bmatrix} \qquad(a,b,c,d\in\mathbb R). \] Taking \[ t=\frac{a+b}{2},\qquad x=\frac{a-b}{2},\qquad y=c,\qquad z=d \] gives \(\Phi(x,y,z,t)=A\). Hence \(\Phi\) is surjective and therefore an isomorphism \(\mathbb R^4\cong W\).