$S,T$ are bounded operators $S,T:H\rightarrow H$ when $H$ an Hilbert space over $\mathbb{C}$ then $S=T\Leftrightarrow\forall h\in H:( Sh,h )=( Th,h )$

Question

The general property is, that for $S,T:H\rightarrow K$ bounded linear transformation between two Hilbert spaces, $S=T \Leftrightarrow \forall h\in H , \forall k\in K : \langle Sh,k\rangle = \langle Th,k\rangle$ . This is pretty simple to demonstrate since the first direction ($\Rightarrow$) is obvious, and for the second direction (using the fact the equality holds for all $k,h$ and the definition of the inner product): $\langle Sh,k\rangle - \langle Th,k\rangle = 0 \Rightarrow \langle (S-T)h,k\rangle =0 \Rightarrow (S-T)h = 0_K \Rightarrow (S-T) \equiv 0 (operator) \Rightarrow S=T$

I was told that for bounded linear operators, i.e $T,S : H\rightarrow H$, $H$ must be an Hilbert space over $\mathbb{C}$ in order for this property to hold ( $S=T\Leftrightarrow\forall h\in H:\langle Sh,h \rangle=\langle Th,h \rangle$). But I don't understand why does one need to specify whether $H$ is over $\mathbb{C}$ or $\mathbb{R}$ it seems to me that the general case holds for operators (over $\mathbb{R}$ and $\mathbb{C}$), does my above explanation for the general case misses some crucial point?

Answer 1

Let $W=S-T$. Suppose $\langle Wx , x \rangle =0$ for all $x$. Then $$\langle W(x+y) , (x+y) \rangle =0$$ and $$\langle W(x-y) , (x-y) \rangle =0$$ for all $x$ and $y$. Subtract the second equation from the first to get $$\langle Wx , y \rangle +\langle Wy , x \rangle =0.$$ Now replace $x$ by $ix$. Using the fact that inner product is conjugate linear in the second variable you will see that $$\langle Wx , y \rangle -\langle Wy , x \rangle =0.$$ It follows that $$\langle Wx , y \rangle =\langle Wy , x \rangle =0.$$ Hence $W=0$.

This argument fails for the case of real scalars and a counterexmaple is provided by a rotation by $90^{0}$ in $\mathbb R^{2}$.