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Proposition. If $V$ is a complex inner-product space and $T$ is a linear operator on $V$ such that $$\langle Tv, v\rangle = 0$$ for all $v \in V$, then $T = 0$. $$\begin{align*} \langle Tu, w\rangle &= \frac{\langle T(u+w), u+w\rangle - \langle T(u-w), u-w\rangle}{4} \\ &\qquad + \frac{\langle T(u+iw), u+iw\rangle - \langle T(u-iw), u-iw\rangle}{4}\,i \end{align*}$$

Can someone help me to understand why the left hand side equation is equal to right hand side equation?

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It is not very elegant but straightforward: $$\color{blue}{\frac{1}{4}[\langle T(v+w),v+w\rangle - \langle T(v-w),v-w\rangle ]} \\+ \color{green}{\frac{i}{4}[\langle T(v+iw),v+iw\rangle - \langle T(v-iw),v-iw\rangle ]}\\ = \color{blue}{\frac{1}{4}\left[\langle T(v),v\rangle+\langle T(v),w\rangle + \langle T(w),v\rangle + \langle T(w),w\rangle\right. } \\ \color{blue}{\left.-\langle T(v),v\rangle+\langle T(v),w\rangle +\langle T(w),v\rangle -\langle T(w),w\rangle\right]} \\ + \color{green}{\frac{i}{4}\left[\langle T(v),v\rangle -i\langle T(v),w\rangle + i\langle T(w),v\rangle + \langle T(w),w\rangle \right. } \\ \color{green}{\left.-\langle T(v),v\rangle -i\langle T(v),w\rangle +i\langle T(w),v\rangle -\langle T(w),w\rangle\right]} \\ = \color{blue}{\frac{1}{4}\left[2\langle T(v),w\rangle + 2\langle T(w),v\rangle\right]}+ \color{green}{\frac{i}{4}\left[-2i\langle T(v),w\rangle + 2i\langle T(w),v\rangle\right]} \\ = \color{blue}{\langle T(v),w\rangle} $$ Now suppose that $\langle T(u),u \rangle = 0 $ for every $u\in V$ and let $v\in V$, then by the above formula we get $\langle T(v),w \rangle = 0$ for every $w \in V$ since the terms are all of the form $\langle T(x),x \rangle$ with $x \in \{v+w, v-w, v+iw,v-iw\}$. In particular, this is true for $w = T(v)$ and thus by definition of the inner product we get $$0=\langle T(v), T(v) \rangle \implies T(v)=0.$$ Since this is true for every $v \in V$ it follows that $T = 0$.

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  • $\begingroup$ do u think if there is another way of proving? This way is just not intuitive to me. If it's an test question, I would never able to figure it out in this way $\endgroup$
    – ElleryL
    Aug 12 '14 at 21:16
  • $\begingroup$ @Hobbit6094 Unfortunately, I don't think so. It was taught to me in this way by a teacher that hates long computations and made all possible to avoid them. However this is an important formula, it is also a general case of the Polarization identity. $\endgroup$
    – Surb
    Aug 12 '14 at 21:20
  • $\begingroup$ @Hobbit6094 However, note that starting from the RHS of your equation and check that it is the same as the LHS is basically just using linearity of $T$ and the inner product. But it is true that going in the reverse direction is harder :). $\endgroup$
    – Surb
    Aug 12 '14 at 22:18

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