Is the anti-unitary square root of the identity operator always diagonalizable?

Question

Let $\mathscr{H}$ be a complex Hilbert space (possibly infinite-dim) and $J:\mathscr{H} \rightarrow \mathscr{H}$ be an anti-unitary operator such that $J^2 = I$ is equal to the identity operator. Then is $J$ always diagonalizable by an orthonormal basis?

Answer 1

Let $J:\Bbb C^2\to\Bbb C^2$ be given by: $$J(x_1, x_2) = (-i\overline{x_1}, (1-i)\overline{x_1}-\overline{x_2})$$ then $$J^2 (x_1,x_2)= (x_1, (1-i)(ix_1)-((1+i)x_1-x_2)) = (x_1, (i+1)x_1-(1+i)x_1 +x_2)=(x_1,x_2)$$

My claim is that this map is not diagonalisable by an orthogonal basis (but it diagonalisable, for example by $(1,0)$ and $(1,1)$). To see that we need to find its eigenvectors. First $(0,x_1)$ is always an eigenvector for any $x_2\in \Bbb C$, but $(x_1,0)$ is not and as such there can be no ONB of eigenvectors that contains $(0,x_2)$. So if there were an ONB every element of it must have non-zero $x_1$ component.

I'm just going to write the long and uninspiring computation out. Suppose $(x_1,x_2)$ were an eigenvector, then:

$$J(x_1,x_2)=(-i\overline{x_1}, (1-i)\overline{x_1}-\overline{x_2})\overset!= (\lambda x_1, \lambda x_2),$$ whence $\lambda = -i\overline{x_1}\frac{1}{x_1}$ follows. Plug this into the second component to get: $$(1-i)\overline{x_1} -\overline{x_2}+i\overline{x_1}\frac{x_2}{x_1}=0,$$ which be reformulated to $$\overline{x_1}(1-i+\overline{x_2}/\overline{x_1} i-\frac{x_2}{x_1}) =0,$$ ie $$1-i = \frac{x_2}{x_1}- i\overline{(\frac{x_2}{x_1})},$$ multpilying by $\frac 1{\sqrt{i}}$ yields: $$-i\sqrt2=i^{-1/2}-i^{1/2}=i^{-1/2} \frac{x_2}{x_1}-i^{1/2}\overline{(\frac{x_2}{x_1})}=2i\mathrm{Im}(i^{-1/2}\frac{x_2}{x_1})$$ This just means $$-1 = -\mathrm{Re}(x_2/x_1)+\mathrm{Im}(x_2/x_1)$$ This means that $$x_1/x_2=(a+1)+i a$$ for some real $a$. Finally: $$x_1= (a+1+ia)x_2.$$

So the eigenvectors are of the form $(0,x)$ and $(\ (a+1+ia)x, x\ )$ with $x\in \Bbb C$ and $a\in\Bbb R$ arbitrary. Can you build an ONB from this? Well $$\langle ((a+1+ia) x, x) , ((b+1+ib)y,y)\rangle = \overline{x}y( (a+1-ia)(b+1+ib) +1)\overset!=0$$ Now if you just write that out and look at the imaginary component you find $a=b$. Do that substitution and the real component becomes $2(1+a+a^2)$, which admits no real roots. Hence no two eigenvectors are orthogonal.