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Given the Hilbert space $\mathbb{C}^2$.

Consider bounded opertors: $$N:\mathbb{C}^2\to\mathbb{C}^2:\quad\|N\|<\infty$$

Then there are some with: $$N\neq N^*\quad N^*N=NN^*$$

What examples are there?

Reference

This is a lemma for: Polar Decomposition

Community
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freishahiri
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  • Boundedness condition is automatically satisfied, of course. (Looks just better to read, IMHO.) – freishahiri Jul 17 '15 at 13:03
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    $\lambda I$ for $\lambda\notin\mathbb{R}$. And $U^{\star}\left[\begin{array}{cc}\lambda & 0 \ 0 & \mu\end{array}\right]U$ where $U^{\star}U=I$ and not both $\lambda$, $\mu$ are real. – Disintegrating By Parts Jul 17 '15 at 13:55
  • @TrialAndError: Ah right so it reduces to finding unitaries instead. Do you have some nice examples in mind? – freishahiri Jul 17 '15 at 14:00
  • @TrialAndError: The idea is really clever: Deriving all unitaries is so much easier than all normals. :D – freishahiri Jul 17 '15 at 19:50
  • Yes, especially in 2-d because negative reciprocal slope is all you need to find a complementary orthogonal vector. That is, $(a,b) \perp (-\overline{b},\overline{a})$ and the second vector is unique up to a unimodular scalar. So you start with a unit vector and the second is determined up to a unimodular constant. – Disintegrating By Parts Jul 17 '15 at 20:44

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In $M_2(\mathbb C)$, the normal operators are precisely the unitarily diagonalizable ones. So, any non-selfadjoint normal operator is, as TrialAndError mentioned, of the form $$ N=U^*\,\begin{bmatrix}\lambda&0\\0&\mu\end{bmatrix}\,U $$ with at least one of $\lambda,\mu$ not real (if both were real, $N=N^*$).

The general $2\times 2$ unitary is of the form $$ U=\begin{bmatrix}e^{ia}\cos t&e^{ib}\sin t\\ e^{ic}\sin t&-e^{i(b-a+c)}\cos t\end{bmatrix},\ \ \ \text{ with }t, a, b, c\in[0,2\pi), $$

Martin Argerami
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  • Can you explain in short how to work out this general form in a clever way, please? – freishahiri Jul 17 '15 at 19:28
  • You just write $U^U=I=UU^$ in terms of the entries. You look at the equations in terms of the absolute values of the entries and that's how you determine that the absolute values should be $\cos t$, $\sin t$, $\sin t$, $\cos t$. Then you write the equations again and look at the relations between the arguments. – Martin Argerami Jul 17 '15 at 19:31
  • Ah ok easy peasy thanks!!! :D (I was getting so lost there yesterday) – freishahiri Jul 17 '15 at 19:41
  • It seems enough to have $0\leq t\leq\frac{\pi}{2}$, or?? (Absolute value of sine and cosine mirror after that range.) – freishahiri Jul 17 '15 at 19:53
  • Yes, you are right. – Martin Argerami Jul 17 '15 at 20:07