If $(\alpha_i,\alpha_j)=(\beta_i,\beta_j)$ for $0<i,j<n+1$, then ${\rm L}(\alpha_1, ..., \alpha_n)\cong {\rm L}(\beta_1, ..., \beta_n)$?

Question

Let $V$ be an inner product space over $\mathbb{R}$ with dimension $n$. Let $S$ be a subset of $V$ and $L(S)$ denote the span of $S$. If $(\alpha_i,\alpha_j)=(\beta_i,\beta_j)$ for every $i, j\in \{1, ...,n\}$, Then
$$L(\alpha_1, ..., \alpha_n)\cong L(\beta_1, ..., \beta_n)?$$

When $n=1$, the result is trivial.

Gram matrix may be useful to prove ${\rm dim} L(\alpha_1, ..., \alpha_n)={\rm dim} L(\beta_1, ..., \beta_n)$, but I do not know how to use it in detail. Can someone give me some hints to this problem?

Answer 1

One approach is as follows: note that the spaces in question are the images of the maps $A,B:\Bbb R^n \to V$ given by $$ A(x_1,\dots,x_n) = x_1 \alpha_1 + \cdots + x_n \alpha_n, \quad B(x_1,\dots,x_n) = x_1 \beta_1 + \cdots + x_n \beta_n. $$ Using the given information, conclude that $A^*A$ and $B^*B$ are the same map over $\Bbb R^n$ (where $A^*$ denotes the adjoint map of $A$).

Let $U$ denote the common column space of $A^*A$ and $B^*B$. Using the fact that $A$ and $A^*A$ have the same rank, show that $A|_U$ defines an isomorphism between $U$ and the image of $A$, and $B|_U$ defines an isomorphism between $U$ and the image of $B$.

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Another approach: with $A,B$ as above, note that $\|Ax\|^2 = \|Bx\|^2$ for all $x \in \Bbb R^n$. It follows that $\ker(A) = \ker(B)$, so that $\dim \ker(A) = \dim \ker(B)$, so that $\dim\operatorname{im}(A) = \dim \operatorname{im}(B)$.

Answer 2

Without loss of generality, let $\alpha_1, ..., \alpha_r$ and $\beta_1, ..., \beta_s$ be basises of $L(\alpha_1, ..., \alpha_n)$ and $L(\beta_1,...\beta_n)$ respectively, then $r\leq s$ and $s\leq r$ by the theroy of Gram matrix. Therefore $L(\alpha_1, ..., \alpha_n)\cong L(\beta_1, ..., \beta_n)$.