Proving that there is an inner product such that a linearly independent subset of an inner product space is an orthonormal basis

Question

I encountered a question that asked me to proof that if $V$ is a real inner product space and there is a subset $\{v_1, \dots, v_n\}$ of linearly independent vectors in $V$, then there exist an inner product such that the subset is an orthonormal basis for $V$.

I know that in order for the subset to be an orthonormal basis, it must be an ordered basis and orthonormal. However I am unsure on how to approach this question.

Answer 1

Let $A$ be a linear transformation where $A(v_i) = e_i$ for each $i$, where $e_1,\dots,e_n$ is the standard basis (verify that such an $A$ exists and must be invertible).

Define the inner product $\langle \cdot,\cdot \rangle_A$ by $$ \langle x,y \rangle_A = \langle A x,Ay\rangle $$ verify that $\langle \cdot,\cdot \rangle_A$ satisfies the definition of an inner product.

Answer 2

An inner product is completely determined by its values on basis elements. Your linearly independent set $\{v_1,\dots,v_n\}$ can only be a basis for $V$ if $\dim V=n$. Otherwise it is too small.

However, if $\dim V>n$ you can extend the set to a basis and define the inner product so that each of those original vectors is orthonormal.