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

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 $$=\sum_{i,j}a_ia_j$$ given that it is equal to $$a_1^2+...+a_n^2$$, thus can we conclude?

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 $$i\ne j$$.

• I am not getting how to apply this concept – RIYASUDHEEN TK 9747408592 Jun 8 at 5:25
• OK. Another hint: Note that $v_i + v_i = 2v_i$. Hence $\left\| v_i + v_i \right\|^2 = 2^2 = 4$ (using the assumption in the question). Similarly, find $\left\| v_i\right\|^2$ using the question's assumption, and then you have everything you need to find $\langle v_i, v_i \rangle$ using the equation in my answer. Then do a similar thing to find $\langle v_i, v_j\rangle$. – Minus One-Twelfth Jun 8 at 5:27

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.

• Ok i got it, thnk u sir – RIYASUDHEEN TK 9747408592 Jun 8 at 5:34
• In that case you can consider approving the answer by clicking on the tick mark. – Kabo Murphy Jun 8 at 5:46