Isomorphism on inner product vector spaces

112 Views Asked by At

I have a question:

Let $V_1$ and $V_2$ be two finite-dimensional inner product vector spaces with the same dimension. Denote the inner product structures on them by $\langle ·, · \rangle_{V_1}$ and $\langle ·, · \rangle_{V_2}$ respectively.

Show that there is an isomorphism $\Phi:V_1 \rightarrow V_2$ such that for any $x,y \in V_1: \langle \Phi(x), \Phi(y) \rangle_{V_2} = \langle x,y \rangle_{V_1}$.

Any help would be appreciated

2

There are 2 best solutions below

0
On

Fix an orthonormal basis $(e_1,\dots, e_n)$ for $V_1$ and fix an orthonormal basis $(f_1, \dots, f_n)$ for $V_2$. These exist by the Gramm-Schmidt algorithm.

Define

$$\Phi: V_1 \to V_2: \sum_{k=1}^n a_k e_k \mapsto \sum_{k=1}^n a_k f_k$$

I leave it to you to check that this is the map you want.

0
On

Let $V_1, V_2$ finite-dimensional vector space. Thanks to the Gram–Schmidt process you can extract an orthonormal base $\mathcal{B}_1=\{v_1,...,v_n\}$ for $V_1$ and an orthonormal base $\mathcal{B}_2=\{w_1,...,w_n\}$ for $v_2$.

Now define $\Phi:V_1\longrightarrow V_2$ such that $\Phi(v_i)=w_i$ for $i=1,...,n$. $\Phi$ is an isomorphism and, by construction, $\langle \Phi(x),\Phi(y)\rangle_{V_2} = \langle x,y\rangle_{V_1}$ for all $x,y\in V_1$.