# Orthogonality Preserving Linear Map.

• November 27th 2012, 09:28 AM
Orthogonality Preserving Linear Map.
I'm having trouble with the following problem:

Let $V$ be a real vector space equipped with an inner product $\langle\cdot,\cdot\rangle$ (and thus with a norm induced by that inner product).
Let $T:V\to V$ be a bijective linear transformation such that if $\langle u,v\rangle=0$, then $\langle T(u),T(v)\rangle=0$.

Given an orthonormal basis $\{b_1,...,b_n\}$ is $V$, it clearly is the case that $\{T(b_1),...,T(b_n)\}$ is an orthogonal basis of $V$ and hence $\left\{\frac{T(b_1)}{\|T(b_1)\|},...,\frac{T(b_n)} {\|T(b_n)\|}\right\}$ is an orthonormal basis.

I now have to show that $\|T(b_1)\|=\cdots=\|T(b_n)\|$, but everything I've tried so far has lead me to a dead end.

Any hint or tip from someone who knows how to prove this would be greatly appreciated (I'm not looking for someone to post a complete solution).
• November 27th 2012, 10:36 AM
Deveno
Re: Orthogonality Preserving Linear Map.
since our original basis is orthonormal:

$\|b_1\| = \|b_j\|$ for j = 2,...,n so:

$\|b_1\|^2 = \|b_j\|^2$

$\langle b_1,b_1 \rangle - \langle b_j,b_j \rangle = 0$

$\langle b_1,b_1 \rangle - \langle b_1,b_j \rangle + \langle b_j,b_1 \rangle - \langle b_j,b_j \rangle = 0$

$\langle b_1,b_1-b_j \rangle + \langle b_j,b_1-b_j \rangle = 0$

$\langle b_1+b_j,b_1-b_j \rangle = 0$

$\langle T(b_1+b_j),T(b_1-b_j) \rangle = 0$

i think it's clear where i'm going with this....