Suppose that $\{v_1,v_2,\ldots v_n\}$ are unit vectors in $\Bbb R^n$ such that
$||v||^2=\sum _{i=1}^n |\langle v_i,v\rangle |^2\;,\;\;\forall\, v\in \Bbb R^n$.
Which are correct?
- $\{v_i:1\le i\le n\}$ are mutually orthogonal
- $\{v_i\}$ is a basis for $\Bbb R^n$.
- $\{v_i\}$ are not mutually orthogonal.
- At-most $n-1$ of the elements in the set $\{v_i\}$ can be orthogonal.
1 is false take $\{(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}),(\frac{-1}{\sqrt 3},\frac{1}{\sqrt 3},\frac{1}{\sqrt 3}),(\frac{1}{\sqrt 3},\frac{-1}{\sqrt 3},\frac{1}{\sqrt 3})\}$
2 is true since if we take $\sum c_iv_i=0$ then $\langle \sum c_iv_i,v_i\rangle =0\implies c_i=0\forall i$
- is false take $(1,0,0),(0,1,0),(0,0,1)\}$
4.false ;take the example of 3
But the answer says correct options are 1,2 .Please help.
For any fixed $\;i\,,\,\,1\le i\le n\;$ , and for any $\;k\,,\,\,1\le k\neq i\le n\;$ , we have that:
$$\left\|v_i\right\|^2\stackrel{\text{given}}=\sum_{k=1}^n\left|\langle v_k,\,v_i\rangle\right|^2=\sum_{ k=1,\,k\neq i}^n\left|\langle v_k,\,v_i\rangle\right|^2+\overbrace{\left|\langle v_i,\,v_i\rangle\right|^2}^{=\left\|v_i\right\|^2}\implies\langle v_k,\,v_i\rangle=0\implies$$
and (1) is true, and then (2) is true from (1), as orthogonal non-zero vectors are linearly independent. Thus, (3)-(4) are false.