I've been interested in isometries of complex Hilbert spaces recently and wondered if it's possible to embed $\mathbb{C}^n$ isometrically into $\mathbb{C}^{n+1}$.
Rather than tackling the general problem first, I tried to construct an explicit isometry $U : \mathbb{C}^2 \rightarrow \mathbb{C}^3$, i.e. a $(3 \times 2)$-matrix $U$ satisfying
$$\langle Uv, Uw \rangle_{\mathbb{C}^3} = \langle v,w \rangle_{\mathbb{C}^2},$$
for all $v,w \in \mathbb{C}^2$, where $\langle \cdot, \cdot \rangle$ is the canonical (usual) inner product:
$$\langle v,w \rangle = \sum_j \overline{v_j}w_j.$$
However, it seems to me that such a matrix cannot exist -- which makes me question whether there is a general theorem stating that no isometry from $\mathbb{C}^n$ into $\mathbb{C}^{n+1}$ exists?
Can anybody correct me if I'm wrong or point me to somewhere in the litterature that discusses this in more detail?
Thank you in advance.
Of course you can construct such an embedding. You simply need to consider the following linear operator:
\begin{aligned} U\colon(z_1,\dots,z_n)\in\mathbb{C}^n\longrightarrow(z_1,\dots,z_n,0)\in\mathbb{C}^{n+1} \end{aligned} This will do the job, just check the property that you mention in your question(and that indeed it is a linear operator).