Show that the basis for an inner product space $V$ is an orthonormal basis

Question

Suppose that $\{v_1,...,v_n\}$ be a basis for an inner product space $V$, and $v=a_1v_1+...+a_nv_n$ implies $||v||^2=a_1^2+...+a_n^2$, then can we say that the given basis is orthonormal? If so how?

I tried that $<a_1v_1+...+a_nv_n, a_1v_1+...+a_nv_n>=\sum_{i,j}a_ia_j<v_i,v_j>$ given that it is equal to $a_1^2+...+a_n^2$, thus can we conclude?

Answer 1

The hypothesis implies $\|v_i\|^{2}=1$ for all $i$ so each $v_i$ has norm $1$. Now consider $\|v_i+v_j\|^{2}$ where $i \neq j$. We get $\|v_i+v_j\|^{2}=1^{2}+1^{2}=2$. Expanding LHS in terms of the inner product we get $\|v_i\|^{2}+\|v_i\|^{2}+2\langle v_i,v_j \rangle =2$ which gives $\langle v_i,v_j \rangle=0$. This is the proof when teh scalare are real . I will leave the complex case to you.

Answer 2

Hint: Note that $\left\| v\right\|^2 = \langle v , v\rangle$ for all $v\in V$. Therefore, you can show using inner product properties that $\left\| v + w\right\|^2 = \left\|v \right\|^2 + \left\|w \right\|^2 + 2 \langle v, w\rangle $ in a real inner product space, so

$$ \langle v, w\rangle = \frac{\left\| v + w\right\|^2 - \left\|v \right\|^2 - \left\|w \right\|^2}{2},$$

for all $v,w\in V$.

Try to use this to show that $\langle v_i , v_i\rangle = 1$ for all $i$ and $\langle v_i , v_j\rangle = 0$ for all $i\ne j$.