Can we deduce Surjectivity from a property of Unitary linear map?

Question

Many books define the unitary linear map $U: (\mathbb{H}_1,\left<\right>_1) \rightarrow (\mathbb{H}_2,\left<\right>_2)$ as a bijective linear map such that $\left< x,y \right>_1=\left<Ux,Uy \right>_2$.

And I have been trying to disprove that if $U: (\mathbb{H}_1,\left<\right>_1) \rightarrow (\mathbb{H}_2,\left<\right>_2)$ is a linear map such that $\left< x,y \right>_1=\left<Ux,Uy \right>_2$ then $U$ is surjective.

I am pretty sure that the statement above is false since definition specify the bijectivity of unitary map and otherwise they don't need to do that. I hope someone can help me to come up with the counter example.

Answer 1

Let $H_1 = \ell^2(\mathbb{N})$ and $H_2 = \ell^2(\mathbb{N})$, with standard basis elements $e_0,e_1,e_2,\cdots$. Then $U : H_1 \rightarrow H_2$ defined by $Ue_n = e_{2n}$ satisfies $\langle Ux,Uy\rangle_2 = \langle x,y\rangle_1$, but $U$ is not surjective.

Answer 2

Consider the space $\ell^2=L^2(\Bbb N)$ (where $\Bbb N$ carries the counting measure) and the right-shift operator that sends a sequence $(a_n)_{n \in \Bbb N}$ to the sequence $(b_n)_{n \in \Bbb N}$ given by $b_0=0$ and $b_{n}=a_{n-1}$ for $n\geq 1$.

Answer 3

Counterexamples are in order here. One is:

Take

$\Bbb H_2 = \Bbb H_1 \oplus \Bbb H_1, \tag 1$

and let the inner product on $\Bbb H_2$ be given by

$\langle (x_1, y_1), (x_2, y_2) \rangle_2 = \langle x_1, x_2 \rangle_1 + \langle y_1, y_2 \rangle_1, \tag 2$

where $(x_1, y_1), (x_2, y_2) \in \Bbb H_1 \oplus \Bbb H_1$; now take

$U:\Bbb H_1 \to \Bbb H_2, \; Ux = (x, 0); \tag 3$

then

$\langle Ux, Uy \rangle_2 = \langle (x, 0), (y, 0) \rangle_2 = \langle x, y \rangle_1 + \langle 0, 0 \rangle_1 = \langle x, y \rangle_1; \tag 4$

but it is easy to see that our map $U$ is not surjective, since for $0 \ne z \in \Bbb H_1$, $(0, z) \in \Bbb H_2$ is not in the image of $U$.