# Consequence of the polarization identity?

1,288

## 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, 2022

• 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 self-adjoint (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 $$\Big\langle (T-T^*)(x), x\Big\rangle = \big\langle T(x), x\big\rangle - \big\langle x, T(x)\big\rangle = \big\langle T(x), x\big\rangle - \overline{\big\langle T(x), x\big\rangle} = 0$$ Therefore $$\Big\langle (T-T^*)(x), x\Big\rangle = 0,~~~~~(\forall x \in H).$$ By polarization identity $$\Big\langle (T-T^*)(x), y\Big\rangle = 0,~~~~~(\forall x,y \in H).$$ Since $(T-T^*)(x)$ est orthogonal to every vecteur $y \in H$ we conclude that $(T-T^*) = 0$. 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 years
• M.G almost 8 years
Yes it does. I edited
• M.G almost 8 years
Crystal clear. Thank you !
• Hrit Roy over 3 years
Isn'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.