Bounded linear mappings in Hilbert space preserve orthogonality? My question is the title of this thread!
Assume we have a bounded, linear mapping $A:H\to H$ where $(H,\langle\cdot,\cdot\rangle)$
is a Hilbert space, and two non-zero elements that are orthogonal, $\langle\Bbb u,\Bbb v\rangle=0$. Is it then true that also $\langle A(\Bbb u),A(\Bbb v)\rangle =0$ ?
If so, how can I prove this?
 A: Preserving orthogonality is an extremely strong condition for an arbitrary mapping to have. Even most very nice mappings that fail this condition: for example, left-multiplication on column vectors by $\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ is a bijective, Hermetian, bounded, linear (and continuous) mapping in finite-dimensional real space, but notice:
$$\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \qquad\text{and}\qquad \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$$
while using the standard inner product shows $\left\langle\begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \end{bmatrix} \right\rangle = 2+2\neq 0$.
(However, it is also a very desirable property for a mapping to have, so it probably has a name: A unitary matrix will preserve all inner products; you can probably weaken this condition if you only care about the orthogonals.)
