Prove that the linear transformations are the same.

Question

I have this lemma:

If X is a complex inner product space and $S,T \in B(X)$ are such that $(Sz,z)=(Tz,z)\forall z \in X$, then $S=T$.

$B(x)$ is the set of bounded linear operators from X to X.

$(,)$ is the inner-product function.

I want to prove this lemma. But I do not know how to prove it.

I tried something like this:

Assume for contradiction that the transformations are not the same.

Then there is an $z'$, such that $(Sz',z')=(Tz',z')$, but $Sz'\ne Tz'$ . Then $z'\ne0$, $Sz'-Tx'\ne0 $, but $z'$ and $Sz'-Tz'$ are orthogonal. But I don't see how to continue.

Any tips?

Answer 1

Let $Q=S-T$. By assumption, $\langle Qv,v \rangle =0$ for all $v=\alpha x + \beta y$. Now, $$ 0=\langle Q(\alpha x+y),\alpha x+y \rangle = |\alpha|^2 \langle Qx,x \rangle + \langle Qy,y \rangle + \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle \\ = \alpha \langle Qx,y \rangle + \bar{\alpha} \langle Qy,x \rangle. $$ Choosing first $\alpha =1$ and then $\alpha =i$, we get $$ \langle Qx,y \rangle + \langle Qy,x \rangle=0 $$ and $$ \langle Qx,y \rangle - \langle Qy,x \rangle =0. $$ Sum and conclude that $\langle Qx,y \rangle =0$ for all $x$ and all $y$. Hence $Q=0$.

You should observe that the statement is false in a real inner product space: even in $\mathbb{R}^2$ there are matrices $Q$ for which $Qx$ is always orthogonal to $x$, and yet $Q$ is non-trivial.

Answer 2

By considering $S-T$, it suffices to show that if $(Sz,z)=0$ for all $z$, then $S=0$. To show that $S=0$, it suffices to show that $(Sx,y)=0$ for all $x,y\in X$. Now think about what $(Sz,z)=0$ tells you when $z=ax+by$ is a linear combination of $x$ and $y$. By varying $a$ and $b$, can you show that $(Sx,y)$ must vanish?