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?


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 $\mathbb R^{2}$.

  • $\begingroup$ Thanks. May you please explain why the case of $S,T : H \rightarrow K$ ($H\neq K)$ doesn't fail for real scalars? $\endgroup$ Aug 6 '19 at 7:01
  • $\begingroup$ Define $S,T: \mathbb R^{2} \to \mathbb R^{2}$ by $T(x,y)=(y,-x)$ and $S(x,y)=0$ for all $(x,y) \in \mathbb R^{2}$. Then you can see that $\langle S(x,y) , (x,y) \rangle =\langle T(x,y) , (x,y) \rangle$ for all $(x,y) \in \mathbb R^{2}$ but $S \neq T$. In my proof replacing $x$ by $ix$ was a crucial step and you cannot do this when the scalar field is $\mathbb R$. $\endgroup$ Aug 6 '19 at 7:20
  • $\begingroup$ Thanks, I understood it. I asked for the case of linear transformation (i.e different spaces domain and range) which I cited at the beginning of the question $H\neq K $ , and $\forall h\in H , \forall k \in K \langle (S-T)h , k \rangle = 0 \Rightarrow S=T $, does it hold also for $H,K$ which are spaces over $\mathbb{R}$? and if so, what is the difference between the cases which allows to build counter example as you demonstrated. $\endgroup$ Aug 6 '19 at 7:49
  • $\begingroup$ Of course $\langle (S-T)h , k \rangle $ for all $h \in H, k \in K$ implies $(S-T)h=0$ for all $h \in H$ which implies $S=T$. I don't quite understand what you are asking when $H \neq K$. $\endgroup$ Aug 6 '19 at 7:58

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