Inner product spaces that are isometrically isomorphic I know this is a fundamental result in linear algebra, and although it is referenced in my textbook, it does not have a proof for it. I was wondering if someone could help me out:
Let $V$ and $W$ be inner product spaces. If $\dim(V)=\dim(W)$, then $V$ and $W$ are isometrically isomorphic.
 A: Let $\operatorname{dim}(V)=\operatorname{dim}(W)=n$. Let $\{v_i:i=1,n\}$ be an orthonormal basis of $V$, $\{w_i:i=1,n\}$ be an orthonormal basis of $W$. Following idea suggested by GEdgar define
$$
I:V\to W: v\mapsto \sum\limits_{i=1}^n\langle v,v_i\rangle w_i
$$
We claim that $I$ the desired isomorphism. Since $\{v_i:i=1,n\}$ is an orthonormal system, then $I(v_i)=w_i$. Since $\{v_i:i=1,n\}$ and $\{w_i:i=1,n\}$ are bases, then the last equality implies that $I$ is a linear isomorphism. It is remains to check it is isometric.
Let $v',v''\in V$, then
$$
\langle I(v'),I(v'')\rangle
=\left\langle\sum\limits_{i=1}^n \langle v',v_i\rangle w_i,  \langle v'',v_j\rangle w_j\right\rangle
=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\langle v',v_i\rangle \overline{\langle v'',v_j\rangle}\langle w_i,w_j\rangle\\
=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\langle v',v_i\rangle \overline{\langle v'',v_j\rangle} \delta_{ij}
=\sum\limits_{i=1}^n\langle v',v_i\rangle \overline{\langle v'',v_i\rangle}
=\left\langle v', \sum\limits_{i=1}^n \langle v'',v_i\rangle v_i\right\rangle
=\langle v',v''\rangle
$$ 
In particular
$$
\Vert I(v)\Vert=\left(\langle I(v),I(v)\rangle\right)^{1/2}=\left(\langle v,v\rangle\right)^{1/2}=\Vert v\Vert
$$
so $I$ is isometric. I want to emphsize that this proof may be easily generalized to Hilbert spaces of arbitrary Hilbert dimension.
