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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?

user119615
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  • Let $U = S-T$. Then $(Uz,z) = 0$ for all $z$. Recall the polarisation identity for sesquilinear forms. – Daniel Fischer Jul 25 '15 at 16:18
  • @DanielFischer What do you mean, I haven't hear about polarisation identity, but it says on wikipedia that if the parallelogram law holds, then there is an innerproduct on the space, how does that help? – user119615 Jul 25 '15 at 16:22
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    Sigh. It's more general than what wikipedia lists. It's what Eric Wofsey also hints at in his answer. For a sesquilinear form $\sigma$, you can recover $\sigma(x,y)$ as a combination of values of the form $\sigma(w,w)$ where $w$ runs through some linear combinations of $x$ and $y$. (For real scalars and bilinear forms, you need symmetry, but not in the complex case.) – Daniel Fischer Jul 25 '15 at 16:30
  • See http://math.stackexchange.com/questions/57350/for-complex-matrices-if-langle-ax-x-rangle-langle-bx-x-rangle-for-all-x – Jonas Meyer Jul 26 '15 at 00:12

2 Answers2

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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.

Siminore
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  • Thank you, but in the first line, after 0=, in the second place, you mean $\alpha x+y$, not $Q(\alpha x+y)$? – user119615 Jul 25 '15 at 16:40
  • @user119615 Thank you, my fault! – Siminore Jul 25 '15 at 16:43
  • @Siminore Could you please explain why, for a given $\alpha, \beta$ pair, the corresponding $x$ and $y$ values are the same? I'm a little tripped up by your usage of $x$ and $y$. Said differently, aren't the corresponding $x$ and $y$ for $\alpha =1, \beta = 1$ different than the corresponding $x$ and $y$ for $\alpha = i, \beta = 1$? When you define $v=\alpha x + \beta y$, in order to make $v$ arbitrary, don't you necessarily impose a relationship between "$x$ and $\alpha$" and "$y$ and $\beta$? As such the $x, y$ in your second to last equation are different than the $x,y$ in the final one? – S.C. Oct 07 '20 at 10:54
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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?

Eric Wofsey
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