Prove that $\|v \|^2= |\langle v, e_1 \rangle |^2 + \cdots + | \langle v, e_m\rangle |^2$

Question

Suppose $(e_1,\cdots, e_m)$ is an orthonormal basis in $V$. Let $v \in V$ . Prove that $\|v\|^2= |\langle v, e_1 \rangle |^2 + \cdots + | \langle v, e_m\rangle |^2$

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Let $v\in V$ and $(e_1,e_2,\cdots,e_m )$ be an orthonormal basis in $V$. Then $v\in \operatorname{span} (e_1,e_2,\cdots,e_m )$. So, there exist $a_i\in F$ where $i=1,2,\cdots,m$ such that $$v=a_1e_1+a_2e_2+\cdots+a_me_m$$ Since $(e_1,e_2,\cdots,e_m )$ is linear independent, this implies each $a_i=0$, thus $$\|v\|^2=\sum_{i=1}^m |\langle v,e_i\rangle |^2 $$

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I don't understand why: $(e_1,e_2,\cdots,e_m )$ is linearly independent implies $$\|v\|^2=\sum_{i=1}^m |\langle v,e_i\rangle |^2 $$

can anyone why explain it to me? thanks

Answer 1

Hint : The scalar products $e_ie_j$ vanish for $i\ne j$, if $e_1,...e_n$ is a orthogonal basis.

The linear independence is not sufficient for the given equality.

Answer 2

I don't think this makes much sense. However, we can write: $$v = \sum_{i=1}^na_ie_i \implies \langle v, e_j \rangle = \left\langle \sum_{i=1}^na_ie_i,e_j\right\rangle = \sum_{i=1}^na_i\delta_{ij}=a_j,$$ so that $$v = \sum_{i=1}^n\langle v,e_i\rangle e_i.$$Hence: $$\|v\|^2 = \langle v,v\rangle = \left\langle \sum_{i=1}^n \langle v, e_i\rangle e_i, \sum_{j=1}^n \langle v,e_j\rangle e_j\right\rangle = \sum_{i,j=1}^n \langle v,e_i\rangle\langle v,e_j\rangle \delta_{ij} = \sum_{i=1}^n |\langle v, e_i\rangle|^2.$$

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In any case... $$\delta_{ij} = \begin{cases} 1, \text{ if }i=j \\ 0, \text{ otherwise}\end{cases}$$ Saying that $\{e_i\}_{i=1}^n$ is an orthonormal set means that $\langle e_i,e_j\rangle = \delta_{ij}$.

Answer 3

Othonormality implies that $\left \langle e_{i},v \right \rangle=a_{i}$. Doing the calculation, you get

$\left \langle e_{i},v \right \rangle =\left \langle e_{i}, (a_{1}e_{1},\cdots ,a_{n}e_{n} )\right \rangle=0+0\cdots +a_{i}\left \langle e_{i},e_{i} \right \rangle+\cdots 0+0=a_{i}$.

Can you take it from here?

Answer 4

Since $(e_i)_{1\le i\le n}$ is an orthogonal basis, $$v=\sum_{i=1}^n\langle v,e_i\rangle e_i\quad\text{and}\quad \lVert v\rVert^2=\langle v,v\rangle=\sum_{i=1}^n\langle v,e_i\rangle^2\langle e_i, e_i\rangle$$ Now $ \lVert e_i\rVert^2=\langle e_i, e_i\rangle=1$, so $$\lVert v\rVert^2=\sum_{i=1}^n\langle v,e_i\rangle^2.$$