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Let $V$ be a a Euclidean or unitary vector space and $f,g$ be linear endomorphisms from $V$ such that $\langle f(x),x\rangle=\langle g(x),x\rangle,\,\forall x\in V$ holds. Does $f=g$ hold? Does it hold if $V$ is finite-dimensional?


I thought that this looked like a good application of the polarisation identities since the terms in there look similar to the above terms but I haven't been able to show the validity of the claim above with this or anything else so far (nor have I been able to disprove it).

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No. Take $f,g\colon\Bbb R^2\longrightarrow\Bbb R^2$ defined by $f(x,y)=(-y,x)$ and $g(x,y)=(y,-x)$. For each $v\in\Bbb R^2$, you have$$\bigl\langle f(v),v\bigr\rangle=\bigl\langle g(v),v\bigr\rangle=0.$$However, $f\ne g$. In fact, unless $v=(0,0)$, $f(v)\ne g(v)$.

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As José Carlos Santos pointed it out, it's not true in a real Hilbert-space. But on a complex Hilbert-space it's true, thanks to the following polarization identity: $$(Ax|y)=\frac{1}{4}\sum_{k=0}^3 i^k(A(x+i^ky)|x+i^ky)$$

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