# When is the space of linear and bounded operators between Hilbert spaces a HIlbert space?

Let $$H_{1}$$ and $$H_{2}$$ be two Hilbert spaces. My question is, when is the set of linear and bounded operators $$\mathcal{L}(H_{1}, H_{2})$$ with the usual norm a Hilbert space? I think I've proved that whenever $$H_{1}$$ or $$H_{2}$$ are $$1$$-dimensional the space of linear and bounded operators is a Hilbert space. But, what happens in arbitrary dimensions?

• Since you excluded $1$-dimensional spaces, you should try $H_1=H_2=\mathbb R^2$. Note that Hilbert spaces are characterized by the parallelogram law. May 16 at 16:11
In higher dimension, i.e. $$\dim H_1\ge 2$$ and $$\dim H_2\ge 2,$$ the space $$\mathcal{L}(H_1,H_2)$$ is not a Hilbert space. For example, let $$H_1=H_2=\mathbb{C}^2$$ and $$A=\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \qquad B=\begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$$ Then $$\|A+\lambda B\|=1$$ for $$0<\lambda \le 1.$$ Assume for a contradiction that the operator norm is associated with an inner product. Then $$\displaylines{1=\|A+\lambda B\|^2=\langle A+\lambda B,A+\lambda B\rangle\\ =\|A\|^2+2\lambda\, \Re \langle B,A\rangle +\lambda^2\|B\|^2}$$ The expression on the right hand side is constant for $$0<\lambda\le 1.$$ Hence $$\|B\|=0.$$