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Let $E\overline\otimes F$ be the complex Hilbert tensor product.

Let $\newcommand{\mc}{\mathcal}T\in\mc L(E), S\in\mc L(F)$ be such that $T\otimes I = I\otimes S$. Why there exists $\alpha\in\mathbb{C}$ such that $T=S=\alpha I$ ?

If $E$ and $F$ are complex Banach spaces we have the following proof:

Proof in Banach spaces: Pick $f\in F, f^*\in F^*$ with $f^*(f)=1$. Then $$ T(e) = f^*(f) T(e) = f^*(S(f)) e \qquad (e\in E). $$ Thus $T = \alpha I$ where $\alpha = f^*(S(f))\in\mathbb C$. Then $\alpha I\otimes I = I\otimes S$ so $S=\alpha I$.

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The idea is similar to the one in Banach spaces you posted. Since we are in Hilbert spaces we have $\mathcal F\simeq\mathcal F^*$ with the Riesz isomorphism $\Psi z:=\langle z,\cdot\rangle$ which we can use. Also it's wise to consider vectors form the unit ball (as your $f^*(f)=1$ condition translates to $\langle f,f\rangle=\Vert f\Vert^2=1$.)

Let any $x\in\mathcal E$ and fix $y_0\in\mathcal F$ with $\Vert x\Vert_{\mathcal E}=\Vert y_0\Vert_{\mathcal F}=1$ be given. Then

$$ \langle x,Tx\rangle_{\mathcal E} =\langle x\otimes y_0,(T\otimes\operatorname{id}_{\mathcal F})(x\otimes y_0)\rangle_{\mathcal E\otimes \mathcal F}=\langle x\otimes y_0,(\operatorname{id}_{\mathcal E}\otimes S)(x\otimes y_0)\rangle_{\mathcal E\otimes \mathcal F}=\langle y_0,Sy_0\rangle_{\mathcal F} $$

because $(A\otimes B)(x\otimes y_0)=Ax\otimes By_0$ by the definition of the tensor product of bounded linear operators. So this means $\langle x,Tx\rangle_{\mathcal E}=\alpha$ on the unit ball of $\mathcal E$ where $\alpha:=\langle y_0,Sy_0\rangle$ is a constant independent of $x$. Now we get

$$ \langle z,(T-\alpha \operatorname{id}_{\mathcal E})z\rangle=\Vert z\Vert^2\Big\langle\frac{z}{\Vert z\Vert},T\frac{z}{\Vert z\Vert}\Big\rangle-\alpha\langle z,\operatorname{id}_{\mathcal E}z\rangle=\Vert z\Vert^2\alpha-\alpha\Vert z\Vert^2=0 $$

for all $z\in\mathcal E$ (where $z=0$ is the obvious case). As we are in a complex Hilbert space, by a well known result this means $T=\alpha \operatorname{id}_{\mathcal E}$. This yields

$$ \langle y,Sy\rangle=\langle x\otimes y,(\operatorname{id}\otimes S)(x\otimes y)\rangle=\langle x\otimes y,(T\otimes\operatorname{id})(x\otimes y)\rangle=\langle x,Tx\rangle=\langle y_0,Sy_0\rangle $$

for any $y\ (,x)$ from the unit ball of $\mathcal F\ (,\mathcal E)$ so with the same argument as above, we see that $S$ has to be a multiple of the identity as well which concludes the proof.

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  • $\begingroup$ Thank you for your answer. Afer you have shown that $T = \alpha I$. We get $\alpha I\otimes I = I\otimes S$ and so $I\otimes( \alpha I - S)=0$. We can conclude that $( \alpha I - S)=0$? $\endgroup$
    – Student
    Mar 4, 2018 at 20:26
  • $\begingroup$ As I said you do the same trick as for $T$, due to the last equation in my answer we know that $S$ is constant on the unit ball of $\mathcal F$ so we extend that to all of $\mathcal F$ via $\langle z,(S-\alpha\operatorname{id})z\rangle =0$ for any $z\in\mathcal F$ and thus get $S=\alpha\operatorname{id}$. $\endgroup$
    – Frederik vom Ende
    Mar 4, 2018 at 20:38
  • $\begingroup$ Yes I understand your proof but I think afer we show that $T=\alpha I$ we can deduce without repeating the first idea that $S=\alpha I$ $\endgroup$
    – Student
    Mar 4, 2018 at 20:44
  • $\begingroup$ Oh I see what you mean. Yeah that also works, as $\operatorname{id}\otimes (\alpha\operatorname{id}-S)=0$, we know - because of $\Vert A\otimes B\Vert=\Vert A\Vert\cdot\Vert B\Vert$ - that either $\Vert\operatorname{id}\Vert=0$ (obviously that can't be as $\operatorname{id}$ has norm 1) or $\Vert\alpha\operatorname{id}-S\Vert=0$. The latter implies that $\alpha\operatorname{id}-S=0$ because of the norm property. $\endgroup$
    – Frederik vom Ende
    Mar 4, 2018 at 20:48

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