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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?

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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 $i\ne j$.

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  • $\begingroup$ I am not getting how to apply this concept $\endgroup$ – RIYASUDHEEN TK 9747408592 Jun 8 at 5:25
  • $\begingroup$ 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 $\langle v_i, v_j\rangle$. $\endgroup$ – Minus One-Twelfth Jun 8 at 5:27
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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.

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

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