1. ## inner product space

Let $V$ be an inner product space, $T:V \rightarrow V$ be a linear operator such that $||T(x)||=||x||$ show that $T$ is injective.

2. Originally Posted by vincent
Let $V$ be an inner product space, $T:V \rightarrow V$ be a linear operator such that $||T(x)||=||x||$ show that $T$ is injective.
I hope I am not misreading the question... $x\ne y\implies \|x-y\|\ne 0$ and so $\|T(x-y)\|\ne 0$ and thus $T(x-y)\ne\bold{0}\implies T(x)\ne T(y)$.

This is a simpler case of an isometry.

3. aaaaaaaaaaaaaaaaaah, it's so easy, thanks a lot!

4. Originally Posted by vincent
aaaaaaaaaaaaaaaaaah, it's so easy, thanks a lot!
The thing that tripped me up is why we need it to be an inner product space? Surely (as you well know) every inner product space gives rise to a normed vector space, but we need the latter not the former. It's strange.

5. Don't we need the vector space to be an inner product space so that $||x||$ is defined?

6. Originally Posted by vincent
Don't we need the vector space to be an inner product space so that $||x||$ is defined?
No. While it is true that if $\left(V,\langle\cdot,\cdot\rangle\right)$ is an inner product space there is a canonical way to make it into a normed vector space $\left(V,\|\cdot\|\right)$ by defining $\|v\|=\sqrt{\langle v,v\rangle}$. The opposite is not necessarily true. There is no (at least not that I know of ) natural way of defining $\langle \cdot,\cdot\rangle$ given $\|\cdot\|$.

The obvious reason is that to go $\langle \cdot,\cdot\rangle\to\|\cdot\|$ notice we were reducing a bivalued function to a single valued function. There is an obvious way to do this by replacing both arguments of the bivalued function by one value. To go the other way is not so easy.

So, my point is that in general $\text{Inner product spaces}\subset\text{Normed vector spaces}$ and we only need the latter opposed to the former, so it's strange they gave us an "extra" condition.

For fun, see if you can find an easy inner product using the norm $\|(x_1,\cdots,x_n)\|=\max_{1\leqslant j\leqslant n}|x_j|$ with $(x_1,\cdots,x_n)\subseteq\mathbb{R}^n$

EDIT: An easy way to think about this is $\langle \cdot,\cdot\rangle\implies\text{ A notion of angles}$ and $\|\cdot\|\implies\text{ A notion of length}$. One clearly follows from the other but not vice versa.

7. or consider $T(x)=T(y)\implies T(x-y)=0$ so $\|x-y\|=0$ and this happens when $x=y,$ and $T$ is injective.