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.

  • $\begingroup$ Does auto adjoint mean self adjoint? $\endgroup$
    – copper.hat
    Dec 23, 2015 at 19:36
  • $\begingroup$ Yes it does. I edited $\endgroup$
    – M.G
    Dec 23, 2015 at 19:37

2 Answers 2


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 $S = T - T^{*}$.

  • $\begingroup$ Crystal clear. Thank you ! $\endgroup$
    – M.G
    Dec 23, 2015 at 20:47

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 $A \neq 0$.

  • $\begingroup$ Isn't it $\langle Ax,y\rangle + \langle Ay,x\rangle$? $\endgroup$
    – Not Euler
    Jun 19, 2020 at 16:41
  • $\begingroup$ @HritRoy: Thank for catching that, I made a mistake. Will delete shortly. $\endgroup$
    – copper.hat
    Jun 19, 2020 at 17:08
  • $\begingroup$ @HritRoy: Actually, it had a simple fix, see above. $\endgroup$
    – copper.hat
    Jun 19, 2020 at 17:14
  • $\begingroup$ in $i(\langle Ax,y\rangle - \langle Ay,x\rangle) = 0$ , shouldn't we have a "+" instead? $\endgroup$ May 4 at 16:52
  • $\begingroup$ @SamaelManasseh No, the inner product is linear in one variable but conjugate linear in the other. $\endgroup$
    – copper.hat
    May 5 at 8:07

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