Proving that a isometry between Hilbert spaces is a isomorphism.

Question

In need show that if $\mathcal{H}, \mathcal{K}$ are Hilbert spaces and $T:\mathcal{H}\to \mathcal{K}$ is a isometric linear operator. i.e, $\|T(x)\|=\|x\|$ for all $x\in\mathcal{H}$ then $T$ is a isomorphism between Hilbert Spaces, i.e:

Note: The inner product is defined on a field $\mathbb{K}$.

1. $T$ is injective.
2. $T$ is surjective.
3. $\langle x,y\rangle=\langle Tx,Ty\rangle$ for all $x,y\in \mathcal{H}$.

I showed 1 and 3 but I can't to show part 2.

In 1, $T$ is injective from $T$ is isometry.

In 3, I showed using the polarization identity that $\langle x,y\rangle=Re(\langle Tx,Ty\rangle)+Im(\langle Tx,Ty\rangle)i=\langle Tx,Ty\rangle$. In this part is neccesary that I do separately the real case and complex case?

I Hope you can help me with part 2.

Answer 1

The statement may not be true without additional hypotheses and here is a counterexample:

$$ {\mathbb R}^n \ni (x_1,\cdots,x_n) \mapsto (x_1,\cdots,x_n,0,0,\cdots) \in l^2 $$

Answer 2

Hint:

It's an isomorphism between the Hilbert space $\ \mathcal{H}\ $ and the image of $\ T\ $ but not necessarily between $\ \mathcal{H}\ $ and $\ \mathcal{K}\ $, if $\ \mathcal{K}\ $ is merely some codomain of $\ T\ $. What you need to show is that the image of $\ T\ $ is a Hilbert space, which I don't think you'll find too difficult.