Show the inner product spaces are isometric

Question

Let $V$, $|| \bullet ||_V$ and let $W$, $|| \bullet ||_W$ be two $n$-dimensional normed linear spaces over $\mathbb R$. We know that $V,W$ are then isomorphic as vector spaces. We say that $V,W$ are isometric if there exists a linear map $T: V \to W$ such that $T$ is one to one and onto, and such that for all $v \in V$, $||T(v)||_w = ||v||_V$.

a. Assume additionally that $V,W$ are $n$-dimensional inner product spaces. Show that $V,W$ are then isometric.

We've done nothing in class regarding isometric stuff so I have no idea what this question is even asking of me.

Answer 1

Use Gram-Schmidt process to get an orthonormal basis $\{v_1, v_2, …, v_n\}$ for $V$ (this means $\langle v_i, v_j \rangle = 0$ for $i\neq j$ and $|| v_j||_{V} = 1$ for each $j$) and similarly orthonormal basis $\{w_1, w_2, …, w_n\}$ for $W$.

Now define the map $T: V\to W$ by sending $T(v_j)=w_j$ for each $1\leq j\leq n$ and extend by linearity. In other words, $T$ is given by $$ T(c_1 v_1 + \cdots + c_n v_n) = c_1 w_1 + c_2 w_2 + \cdots + c_n w_n $$ for any choice of scalars $c_1, c_2, …, c_n$. Then for each $v=c_1v_1+\cdots + c_n v_n$, we can use Pythagorean theorem to get $||v||^2_{V} = \sum_{j=1}^{n} |c_j|^2$. So by using Pythagorean theorem again: $$ ||T(v)||^2_{W} = || T(c_1v_1+\cdots + c_n v_n) ||^2_{W} = || c_1 w_1 + c_2 w_2 + \cdots + c_n w_n ||^2 = \sum_{j=1}^{n} |c_j|^2 = || v ||^2_{V} $$ which shows $||T(v)||_{W} = || v ||_{V}$. It is also easy to check that $T$ is one-to-one and onto.

The point of the problem is that two finite-dimensional inner product spaces of same dimension are not only isomorphic via invertible linear map, but via an invertible isometry (a map that also preserves inner product structure. This is what condition $||T(v)||^2_{W} = || v ||^2_{V}$ is saying).

Answer 2

Choose orthonormal bases $\{v_1, \dots , v_n\}$ and $\{w_1, \dots, w_n \}$ for $V$ and $W$, respectively. There is a unique linear transformation $T:V\to W$ such that $Tv_i=w_i$ for each $i$, and then $T$ is onto since each element of a basis of $W$ is in its range, hence one-to-one by the rank-nullity theorem. Further, if $v=a_1v_1+\cdots +a_nv_n$ is an arbitrary element of $V$, then $Tv=a_1w_1+\cdots +a_nw_n$ and by the Pythagorean theorem, $\|Tv\|^2=a_1^2+\cdots+a_n^2=\|v\|^2$.