# An Orthogonal Transformation from $\R^n$ to $\R^n$ is an Isomorphism

## Problem 592

Let $\R^n$ be an inner product space with inner product $\langle \mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{\trans}\mathbf{y}$ for $\mathbf{x}, \mathbf{y}\in \R^n$.

A linear transformation $T:\R^n \to \R^n$ is called orthogonal transformation if for all $\mathbf{x}, \mathbf{y}\in \R^n$, it satisfies
$\langle T(\mathbf{x}), T(\mathbf{y})\rangle=\langle\mathbf{x}, \mathbf{y} \rangle.$

Prove that if $T:\R^n\to \R^n$ is an orthogonal transformation, then $T$ is an isomorphism.

Contents

We give two proofs.
The second one uses a fact about the injectivity of linear transformations.

## Proof 1.

As $T$ is a linear transformation from $\R^n$ to itself, it suffices to show that $T$ is an injective linear transformation.

Suppose that $T(\mathbf{x})=T(\mathbf{y})$ for $\mathbf{x}, \mathbf{y}\in \R^n$.
We show that $\mathbf{x}=\mathbf{y}$.

We have
\begin{align*}
&\|\mathbf{x}-\mathbf{y}\|^2\\
&=(\mathbf{x}-\mathbf{y})^{\trans}(\mathbf{x}-\mathbf{y})\\
&=(\mathbf{x}^{\trans}-\mathbf{y}^{\trans})(\mathbf{x}-\mathbf{y})\\
&=\mathbf{x}^{\trans}\mathbf{x}-\mathbf{x}^{\trans}\mathbf{y}-\mathbf{y}^{\trans}\mathbf{x}+\mathbf{y}^{\trans}\mathbf{y}\\
&=\langle\mathbf{x}, \mathbf{x} \rangle-\langle\mathbf{x}, \mathbf{y} \rangle-\langle\mathbf{y}, \mathbf{x} \rangle+\langle\mathbf{y}, \mathbf{y} \rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{y}) \rangle-\langle T(\mathbf{y}), T(\mathbf{x}) \rangle+\langle T(\mathbf{y}), T(\mathbf{y}) \rangle\\
&\text{(since $T$ is an orthogonal transformation)}\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle+\langle T(\mathbf{x}), T(\mathbf{x}) \rangle\\
&\text{(since $T(\mathbf{x})=T(\mathbf{y})$)}\\
&=0.
\end{align*}
It follows that $\|\mathbf{x}-\mathbf{y}\|=0$ and hence $\mathbf{x}=\mathbf{y}$.
This proves that $T:\R^n\to \R^n$ is injective.

As $T$ is an injective linear transformation from the $n$-dimensional vector space $\R^n$ to itself, it is also surjective, and thus $T$ is an isomorphism.

## Proof 2.

Recall that the linear transformation $T$ is injective if and only if the null space $\calN(T)=\{\mathbf{0}\}$, that is, $T(\mathbf{x})=\mathbf{0}$ implies that $\mathbf{x}=\mathbf{0}$.
(See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for the proof of this fact.)

We use this fact to show that $T$ is injective.
Suppose that $T(\mathbf{x})=\mathbf{0}$.
Then we have
\begin{align*}
\|\mathbf{x}\|^2&=\langle \mathbf{x}, \mathbf{x}\rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x})\rangle &&\text{as $T$ is orthogonal}\\
&=\langle \mathbf{0}, \mathbf{0}\rangle=0 &&\text{as $T(\mathbf{x})=\mathbf{0}$}.
\end{align*}

It follows that the length $\|\mathbf{x}\|=0$, and hence $\mathbf{x}=\mathbf{0}$.
This proves that the null space $\calN(T)=\{\mathbf{0}\}$ and $T$ is injective.

As $T$ is an injective linear transformation from $\R^n$ to itself, it is also surjective, and hence $T$ is an isomorphism.

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