1

The question is in the title.

My attempt:

Since $\Vert F(x) \Vert = \Vert F^*(x) \Vert, \forall x \in \mathbb{C^n}$, then $\Vert F(x) \Vert^2 = \Vert F^*(x) \Vert^2, \forall x \in \mathbb{C^n}$. So, $\forall x \in \mathbb{C^n}$, we have: $$\Vert F(x) \Vert^2 = \langle F(x), F(x) \rangle = \langle F^*F(x), x \rangle$$ $$\Vert F^*(x) \Vert^2 = \langle F^*(x), F^*(x) \rangle = \langle FF^*(x), x \rangle$$ And we can conclude from the equality given that: $$\forall x \in \mathbb{C^n}: F^*F(x) = FF^*(x) \implies FF^* = F^*F$$

I think my argument is false, or at least insufficient. How can i improve it?

Yoxbox
  • 93
  • It is enough. On complex inner product spaces, linear operators are uniquely determined by their inner products. See https://math.stackexchange.com/questions/2103595/a-linear-operator-on-a-hilbert-space-is-uniquely-determined-by-the-values-lang – LSK21 Feb 15 '23 at 07:26
  • @LSK21 Thanks for the reference. I searched for a better explanation, as geetha290krm provided, since my maths prof would think my explanation is lacking. – Yoxbox Feb 15 '23 at 07:52

1 Answers1

2

You have proved that $\langle FF^*(x), x \rangle=\langle F^*F(x), x \rangle$ or $\langle (FF^*-F^*F)(x), x \rangle=0$ for all $x$. This implies that $\langle (FF^*-F^*F)(x), y \rangle=0$ for all $x$ and $y$ and you can take $y=(FF^*-F^*F)(x)$ to finish the proof. [Details below].

Supppose $\langle Tx, x \rangle=0$ for all $x$. Using the equations $\langle T(x+y), x+y \rangle=0$ and $\langle T(x-y), x-y \rangle=0$ conclude that $$\langle Tx, y \rangle=-\langle y, Tx \rangle.$$ Now use the equations $\langle T(x+iy), x+iy \rangle=0$ and $\langle T(x-iy), x-iy \rangle=0$ conclude that $$\langle Tx, y \rangle=\langle y, Tx \rangle.$$ Combining the two we get $\langle Tx, y \rangle=0$.

geetha290krm
  • 36,632