2 Star-preserving maps
In this section, we define what it means for a linear map to be real (also known as star-preserving) on Hilbert spaces \(\mathcal{H}_1,\mathcal{H}_2\).
We say \(A\in \mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)\) is real (star-preserving) when \(A(a^*)={A(a)}^*\) for each \(a\in \mathcal{H}_1\).
We define the map \(\cdot ^{\operatorname {r}}\) as the self-invertible anti-linear automorphism \((\mathcal{H}_1\to \mathcal{H}_2)\cong (\mathcal{H}_1\to \mathcal{H}_2)\) given by
Let \(A\in \mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)\). Then \(A\) is real if and only if \(A^{\operatorname {r}}=A\).
Clearly \(A^{\operatorname {r}}(x)={A(x^*)}^*=A(x)\) if and only if \(A(x^*)={A(x)}^*\) for all \(x\in \mathcal{H}\), which means \(A\) is real.
When \(f\in \mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)\) and \(g\in \mathcal{B}(\mathcal{H}_3,\mathcal{H}_1)\), then we get \({(f\circ g)}^{\operatorname {r}}={f^{\operatorname {r}}}\circ {{g^{\operatorname {r}}}}\).
Straightforward computation.
Given Hilbert spaces \(\mathcal{H}_1,\mathcal{H}_2,\mathcal{H}_3,\mathcal{H}_4\), and linear maps \(x\colon \mathcal{H}_1\to \mathcal{H}_2\) and \(y\colon \mathcal{H}_3\to \mathcal{H}_4\), we clearly get \({(x\otimes y)}^{\operatorname {r}}={x}^{\operatorname {r}}\otimes {y}^{\operatorname {r}}\).
Straightforward computation.
Given \(A\in \mathcal{B}(\mathcal{H})\), we get \(\operatorname{Spectrum}({A}^{\operatorname {r}})=\overline{\operatorname{Spectrum}(A)}\).
In fact, \(x\in \ker (A-\lambda \operatorname {id})\) if and only if \(x^*\in \ker ({A}^{\operatorname {r}}-\bar{\lambda }\operatorname {id})\).
For any \(x\in \mathcal{H}\), we have \({A}^{\operatorname {r}}(x^*)={A(x)}^*\). So if \(x\) is an eigenvector of \(A\) with eigenvalue \(\lambda \), then clearly \({A}^{\operatorname {r}}(x^*)=\bar{\lambda }x^*\), so \(x^*\) is an eigenvector of \({A}^{\operatorname {r}}\) with eigenvalue \(\bar{\lambda }\). If, on the other hand, \(x^*\) is an eigenvector of \({A}^{\operatorname {r}}\) with eigenvalue \(\bar{\lambda }\), then \({A(x)}^*=\bar{\lambda }x^*\), and so \(A(x)=\lambda x\), which means \(x\) is an eigenvector of \(A\) with eigenvalue \(\lambda \).