Isometry between $k$-dimensional subspace of $\mathbb{R}^n$ and $\mathbb{R}^k$ If $W\subset\mathbb{R}^n$ is a $k$-dimensional subspace of $\mathbb{R}^n$ with the usual Euclidean norm, then is there a result that says that there is an isometry between $W$ and $\mathbb{R}^k$? Could someone possibly provide either an easy proof or counterexample that would go along with this?
 A: There is always a linear isometry between two inner product spaces $V, W$ of the same finite dimension (over the same field $\Bbb{R}$ or $\Bbb{C}$). Use Gram-Schmidt to find orthonormal bases $B_1 = (e_1, \ldots, e_k)$ and $B_2 = (f_1, \ldots, f_k)$ for $V$ and $W$ respectively, and define the unique linear map $T$ that maps $e_i$ to $f_i$. Then, given any $v \in V$, there exist scalar $a_1, \ldots, a_k$ such that
$$v = a_1 e_1 + \ldots + a_k e_k.$$
Using Pythagoras's theorem,
\begin{align*}
\|v\|^2 &= \|a_1 e_1\|^2 + \ldots + \|a_k e_k\|^2 \\
&= |a_1|^2 \|e_1\|^2 + \ldots + |a_k|^2 \|e_k\|^2 \\
&= |a_1|^2 + \ldots + |a_k|^2.
\end{align*}
Using the linearity of $T$,
\begin{align*}
\|Tv\|^2 &= \|a_1 Te_1 + \ldots + a_k Te_k\|^2 \\
&= \|a_1 f_1 + \ldots + a_k f_k\|^2 \\
&= |a_1|^2 + \ldots + |a_k|^2,
\end{align*}
using Pythagoras's theorem again. In this way, we see $\|Tv\|^2 = \|v\|^2$, making $T$ an isometry. To see that $T$ preserves inner products, use the polarisation identities.
A: To flesh out Thomas Andrews' hint, since $W$ is $k$-dimensional it admits an orthonormal basis $B_W = \{\vec{u}_1, \cdots, \vec{u}_k\}$. Define the linear transformation $T: W \to \mathbb{R}^k$ such that $T(\vec{u}_i) = \hat{e}_i$, where $\hat{e}_i$ is the $i$th standard basis vector in $\mathbb{R}^k$.
To see why this is an isometry, you can note that $T$ is length-preserving, i.e. $\|T \vec{u}\| = \|\vec{u}\|$ for any $\vec{u} \in W$. If we let $\vec{u} = c_1 \vec{u}_1 + \cdots + c_k \vec{u}_k$ (recall the $\vec{u}_i$ form a basis), then we can see that
$$\|T \vec{u}\|= \|T(c_1 \vec{u}_1 + \cdots + c_k \vec{u}_k)\| = \|c_1 \hat{e}_1 + \cdots + c_k \hat{e}_k\| = \sqrt{\sum_{i = 1}^k c_i^2} = \|\vec{u}\|$$
This implies that $T$ preserves the dot-product as well (due to the polarization identity). $T$ is also clearly invertible, because its rank is $k$ and the dimension of the domain is also $k$. Hence, we are done. $\square$
