# Proving a linear transformation that preserves the inner product is an isometry

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:

1. 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 $$v,w \in \mathbb{R}^n$$
2. 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:

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

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

• If $T(v)=0$, so that $T(v)\cdot T(v)=0$, then...
– Joe
May 10, 2022 at 22:22
• @Joe Then $v \cdot v = 0$ and hence $v = 0$? May 10, 2022 at 22:29
• I'm not sure what you're asking. We just showed that $T(v)=0 \implies v=0$, so $T$ is injective. Then, since $T$ maps from $\mathbb{R}^n \to \mathbb{R}^n$, it is also surjective.
– Joe
May 10, 2022 at 23:14
• Oh, I see. Well, for a linear transformation, $T(v_1)=T(v_2) \implies T(v_1 - v_2) = 0$. So if the null space is $\{0\}$, then the map is injective. Then we use the rank nullity theorem to deduce that it is surjective.
– Joe
May 11, 2022 at 16:13
• If you have $n$ vectors in $\mathbb{R}^n$, $v_1, \ldots, v_n$, then by linearity$$\sum_{i=1}^n c_i T(v_i) = ?$$You can use that (and the fact that $T$ is injective) to show that the image of $T$ has dimension $n$.
– Joe
May 12, 2022 at 23:24

## 1 Answer

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

• The supposition that $$T(v) \cdot T(v) = v \cdot v$$ could imply that $$T(v) = \pm v$$ though, couldn't it? Since $$T(v) \cdot T(v) = T(v_{1})^{2} + \cdots + T(v_{n})^{2}$$ and $$(v_{i})^{2} = (-v_{i})^{2}$$ May 11, 2022 at 14:01
• @NumericalDisintegration, $T(v)\cdot T(v) = v \cdot v$ could have many more solutions that just $T(v)=v$ or $T(v)=-v$. Simply consider $\mathbb{R}^2$ with $T$ equal to any rotation. However, if you're questioning whether or not $T$ is injective, you want to see if you can show that $T(v_1) = T(v_2) \implies v_1 = v_2$, not whether there is only one possible image of $v$ under $T$.
– Joe
May 11, 2022 at 16:51