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This may be a slight duplicate but I am trying to show $T$ is the zero (linear) operator if $\langle T(x), x\rangle = 0$ $\forall x \in V$ where V is a complex inner product space.

I am using the Polarisation identity on $\langle T(x), y\rangle$ to show it must be $0$ but I always get a circular answer and am looking for help where I am going wrong please.

Given $x$, $y$ $\in V$

$ \langle T(x), y\rangle = {\rm Re}(\langle T(x), y\rangle ) + i\, {\rm Im}(\langle T(x), y\rangle )$

${\rm Re}( \langle T(x), y\rangle ) = \frac{1}{4}\cdot[\,||T(x) + y||^2 - ||T(x) - y||^2]$

Which, after tidying

$= \frac{1}{2}\cdot[\, \langle T(x), y\rangle + \langle y, T(x)\rangle ]$

Likewise,

${\rm Im}( \langle T(x), y\rangle ) = \frac{1}{4}\cdot[\,||T(x) + iy||^2 - ||T(x) - iy||^2]$

$= \frac{1}{2}\cdot[\, -i\,\langle T(x), y\rangle + i\,\langle y, T(x)\rangle ]$

$\therefore\, i\cdot\,{\rm Im}( \langle T(x), y\rangle ) = \frac{1}{2}\cdot[\, \,\langle T(x), y\rangle - \,\langle y, T(x)\rangle ]$

So adding these gives me back $\langle T(x), y\rangle$ when I wanted 0.

Frederik vom Ende
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number8
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    You are using the wrong polarization identity. See: https://math.stackexchange.com/questions/2033703/if-tx-mathbin-x-0-for-all-x-then-t-0 – Kavi Rama Murthy May 03 '24 at 08:47
  • Thanks, but for this after manipulating, I get

    $\frac{1}{4}\cdot [<T(x+y), x+y> + i\cdot<T(x+iy), x+iy>]$ So, $<T(x+y), x+y> = <T(x), y> + <T(y), x>$ And, $i\cdot<T(x+iy), x+iy> = <T(x), y> - <T(y), x>$

    which gives me a non-zero answer. Am I allowed to just use the fact because $<T(x), x> = 0$, $<T(x+y), x+y> = 0$ because $x+y \in V$, therefore T is the zero operator?

     – number8 May 03 '24 at 09:17
  • Your calculation is correct but it goes back in a circle to the original statement. You never get to a point, where you could use the assumption. – daw May 03 '24 at 09:43

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