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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.

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    $\begingroup$ If $T(v)=0$, so that $T(v)\cdot T(v)=0$, then... $\endgroup$
    – Joe
    May 10, 2022 at 22:22
  • $\begingroup$ @Joe Then $v \cdot v = 0$ and hence $v = 0$? $\endgroup$ May 10, 2022 at 22:29
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    $\begingroup$ 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. $\endgroup$
    – Joe
    May 10, 2022 at 23:14
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    $\begingroup$ 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. $\endgroup$
    – Joe
    May 11, 2022 at 16:13
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    $\begingroup$ 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 $n$. $\endgroup$
    – Joe
    May 12, 2022 at 23:24

1 Answer 1

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

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  • $\begingroup$ 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 $$ (v_{i})^{2} = (-v_{i})^{2}$$ $\endgroup$ May 11, 2022 at 14:01
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    $\begingroup$ @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 $T$. $\endgroup$
    – Joe
    May 11, 2022 at 16:51

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