Showing that two generated spaces are isomorphic Let $m \in \mathbb{Z}^+$ and $\{ \alpha_1, \ldots , \alpha_m \}, \{ \beta_1, \ldots , \beta_m \}$ two sets of an inner product space $(V, \langle \cdot, \cdot \rangle).$ Suppose that $\langle \alpha_i, \alpha_j \rangle = \langle \beta_i, \beta_j \rangle$ for all $i,j \in \{ 1, \ldots, m \}.$
I need to prove that the subspaces $V=\text{gen} \{ \alpha_1, \ldots ,\alpha_m \}$ and $W=\text{gen} \{ \beta_1, \ldots ,\beta_m \}$ are isomorphic.
I've trying by using some properties of the orthogonal complement of a set, or trying to use Gram-Schmidt orthogonalization, but nothing really useful.
Can someone help me solving this problem? Thanks in advance. 
 A: Let $G = (g_{ij})_{m \times m}$ where $g_{ij} = \langle \alpha_i, \alpha_j \rangle = \langle \beta_i, \beta_j \rangle.$ 
G is a symmetric psd matrix so we can find a orthonormal basis $v_1,\dots,v_m$ of $\mathbb{R}^m$ such that
$$G = \sum_{i=1}^{r} \lambda_i v_i v_i^{\top}$$ where $\lambda_1 \geq \dots \lambda_r >0$ and $r \leq m$ is the rank of $G$.
Let $v_i = \begin{pmatrix} v_{1i} & \dots & v_{mi} \end{pmatrix}^{\top}.$
Define $$\tilde{\alpha}_i = \sum_{k=1}^{m}v_{ki}\alpha_k.$$
The span of $\tilde{\alpha}_i$'s is contained in the span $\alpha_i$'s.
Also since,
$$\sum_{i=1}^{m} v_{ti} \tilde{\alpha}_i = \sum_{i=1}^{m}\sum_{k=1}^{m} v_{ti}v_{ki}\alpha_k = \sum_{k=1}^{m} \alpha_k (\sum_{i=1}^m v_{ti}v_{ki}) = \alpha_t$$ for $1 \leq t \leq n$, so the span of $\alpha_i$'s and $\tilde{\alpha}_i$'s is the same.
It is easy to check that $\|\tilde{\alpha}_i\|^2 = v_i^{\top} G v_i = \lambda_i > 0$ for $1 \leq i \leq r$ and $\|\tilde{\alpha}_i\|^2 = 0$ for $i > r$ and $$\langle \tilde{\alpha}_i, \tilde{\alpha}_j \rangle = 0$$ for $1 \leq i,j \leq r$ and $i\neq j.$
So $\tilde{\alpha}_1,\dots,\tilde{\alpha}_r$ is an orthogonal basis of $V$.
Similarly defining $$\tilde{\beta}_i = \sum_{k=1}^{n}v_{ki}\beta_k,$$ for $1 \leq i \leq n$ we have $\tilde{\beta}_1,\dots,\tilde{\beta}_r$ is an orthogonal basis of $W$.
Let $T$ be the linear mapping from $V$ to $W$ such that $T(\tilde{\alpha}_i) = \tilde{\beta}_i$ for $1 \leq i \leq r$ then $T$ is an isomorphism.
