I am currently working on a problem to prove the following statement:

Suppose $T: \mathbb{R}^n \to \mathbb{R}^n$ is a linear transformation. Prove that $T$ is an isometry if and only if $T(v) \cdot T(w) = v \cdot w$

I've already written up a proof for the reverse statement (assuming $T$ is an isometry and showing it preserves the inner product) and now I need to prove the forward statement (assuming $T$ preserves the inner product and showing that it must be an isometry). I know this problem has, essentially, $2$ parts:

- Showing that the requirement of an isometry $\lvert T(v)-T(w) \rvert = \lvert v-w \rvert$ follows from the assumption that $T(v) \cdot T(w) = v \cdot w$, where $v,w \in \mathbb{R}^n$
- Proving that $T$ is a bijection. Since an isometry is a bijection that preserves distance.

I have a quick draft of a proof for part $1$ that I will write down below, that I believe is on the right track, but more than likely needs some work. For part $2$ I'm struggling to see how the condition that $T$ preserves inner product necessitates it be injective and surjective.

**Work for part $1$:**

Assume $T(v) \cdot T(w) = v \cdot w, \forall v,w \in \mathbb{R}^n$. Since vector spaces are closed under addition and additive inverses we can say that

$$T(v)-T(w) \in \mathbb{R}^n$$ $$v-w \in \mathbb{R}^n$$

Thus from our assumption we have

$$[T(v)-T(w)] \cdot [T(v)-T(w)] = [v-w] \cdot [v-w]$$

Taking the square root of both sides we have

$$ \lvert T(v)-T(w) \rvert = \sqrt{[T(v)-T(w)] \cdot [T(v)-T(w)]} = \sqrt{[v-w] \cdot [v-w]} = \lvert v-w \rvert$$

Which is exactly the requirement for a linear transformation to be an isometry.

**Proving $T$ is a bijection**

Now this part I seem to be struggling just getting started, so any guidance in that regard is much appreciated. I have collected some of the relevant facts and have been looking at them trying to see how it all comes together. I shall list them below:

- $T(av) = aT(v)$, for $a \in \mathbb{R}$ and $\forall v \in \mathbb{R}^n$
- $T(v+w) = T(v) + T(w)$ $\forall v,w \in \mathbb{R}^n$
- $T(v) \cdot T(w) = v \cdot w$

As previously mentioned I'm having trouble, specifically, seeing how property $3$ contributes to the requirement that $T$ is a bijection.

## Best Answer

$$\left\Vert Tx\right\Vert ^{2}=\left\langle Tx,Tx\right\rangle =\left\langle x,x\right\rangle =\left\Vert x\right\Vert ^{2}$$