Consequence of the polarization identity?
Solution 1
The point is that the polarization identity is true for any sesquilinear form on any complex vector space $V$. That is, if $g$ is any sesquilinear form on $V$ then we have
$$ g(x,y) = \frac{1}{4} \left( g(x + y, x + y)  g(x  y, x  y) + ig(x + iy, x + iy) ig(x  iy, x  iy) \right) $$
for all $x, y \in V$. Note that this identity implies that if $g(z,z) = 0$ for all $z \in V$ then $g \equiv 0$.
Now, let $(V, \left< \cdot, \cdot \right>)$ be a complex inner product space and $S \colon V \rightarrow V$ be a complex linear map. If we define $g(x,y) := \left< Sx, y \right>$ then $g$ is a sesquilinear form on $V$ and if $g(z,z) = \left< Sz, z \right> = 0$ for all $z \in V$ then $g \equiv 0$ which implies in particular that $g(x, Sx) = Sx^2 = 0$ for all $x \in V$ so $S = 0$.
Now apply this to $V = H$ and $S = T  T^{*}$.
Solution 2
You don't need the polarisation identity.
Suppose $\mathbb{H}$ is a complex Hilbert space and $\langle Ax,x\rangle = 0$ for all $x$. Then $A=0$.
We have $\langle A(x+y),x+y\rangle = 0 = \langle Ax,y\rangle + \langle Ay,x\rangle = 0$.
Replacing $y$ by $iy$ we get $\langle A(x+iy),x+iy\rangle = 0 = i(\langle Ax,y\rangle  \langle Ay,x\rangle) = 0$ and combining with the above we get $\langle Ax,y\rangle = 0$. Since this holds for all $x,y$ we see that $A=0$.
It is not true in a real Hilbert space, for example, $A =\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Then $\langle Ax,x\rangle = 0$ for all (real) $x$, but $A \neq 0$.
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M.G
Updated on August 01, 2022Comments

M.G over 1 year
Here is a proof which I do not fully understand.
Theorem : Let $H$ be a Hilbert space. A continuous linear map $T : H \rightarrow H$ is selfadjoint (hermitian) if and only if $$\big\langle T(x), x\big\rangle \in \mathbb{R},~~~~~(\forall x \in H)$$ Proof : ($\Rightarrow$) $$ \overline{\big\langle T(x), x\big\rangle} = \big\langle x, T(x)\big\rangle = \big\langle T^*(x),x \big\rangle = \big\langle T(x), x\big\rangle.$$ Therefore $$\big\langle T(x), x\big\rangle \in \mathbb{R}.$$ ($\Leftarrow$) Using the hypothesis we've got $$ \Big\langle (TT^*)(x), x\Big\rangle = \big\langle T(x), x\big\rangle  \big\langle x, T(x)\big\rangle = \big\langle T(x), x\big\rangle  \overline{\big\langle T(x), x\big\rangle} = 0$$ Therefore $$\Big\langle (TT^*)(x), x\Big\rangle = 0,~~~~~(\forall x \in H).$$ By polarization identity $$\Big\langle (TT^*)(x), y\Big\rangle = 0,~~~~~(\forall x,y \in H).$$ Since $(TT^*)(x)$ est orthogonal to every vecteur $y \in H$ we conclude that $(TT^*) = 0$. Therefore $T = T^*$.
I don't get the "by the polarization identity" bit. How is it that the polarization identity allows us let the right term of the inner product range over $H$ independently from the left term ?!
Edit : Seems related to Trouble with simple consequence of the polarization identity in a way I'm currently trying to figure out.

copper.hat almost 8 yearsDoes auto adjoint mean self adjoint?

M.G almost 8 yearsYes it does. I edited


M.G almost 8 yearsCrystal clear. Thank you !

Hrit Roy over 3 yearsIsn't it $\langle Ax,y\rangle + \langle Ay,x\rangle$?

copper.hat over 3 years@HritRoy: Thank for catching that, I made a mistake. Will delete shortly.

copper.hat over 3 years@HritRoy: Actually, it had a simple fix, see above.