Let $V$ be a finite dimensional vector space with an inner product $\langle . \rangle$.

Let $v_1,v_2,...,v_n\in V$, $v_i\ne 0 \quad \forall i \quad$, so that $ \forall w\in V$ we have that: $\sum_{i=1}^n \langle v_i, w\rangle^{2} = \|w\|^2$

I have already proved that $v_1,v_2,...,v_n$ form a basis for $V$, because if we let $W=span(v_1,v_2,...,v_n)$, we see that the orthogonal complement of $W$ is $0_v$ (let $x$ be in the orthogonal complement of $W$, then $\langle v_i, x\rangle=0 \quad \forall i$. Then $\sum_{i=1}^n \langle v_i, x\rangle^{2} = \|x\|^2=0$, and so $x=0$. We can see now that dimV=dimW)

Now I have to prove that if we let $w=v_i$ , then $||v_i||\le1$ $\forall i\in{1,...,n}$ , and then conclude that $v_1,v_2,...,v_n$ form an orthonormal basis for $V$, but I'm stuck in showing that $||v_i||\le1$

I've tried using the C.S inequality, the Bessel inequality, using continuity, but I can't get to anything.

Any help will be appreciated.

  • $\begingroup$ your question is not quite clear.... Are you trying to prove given any vectors $v_1,\cdots,v_n$ with $v_i\neq 0$ for all $w\in V$ we have the equality as you have written or given a basis $v_1,\cdots,v_n$ of $V$, for all $w\in V$ we have the equality as you have written.. It is not quite clear why orthogonal complement of some random $n$ non zero vectors is zero... $\endgroup$ – user87543 Nov 9 '15 at 4:05
  • $\begingroup$ Can you state the statement to be proved clearly? $\endgroup$ – Megadeth Nov 9 '15 at 4:13
  • $\begingroup$ @PraphullaKoushik - it seems clear to me that the question is: given a set of vectors for which the summation formula holds for all $w$, prove that the set of vectors is an orthonormal basis. $\endgroup$ – Paul Sinclair Nov 9 '15 at 4:14
  • $\begingroup$ @PaulSinclair : where did he say that? $\endgroup$ – user87543 Nov 9 '15 at 6:37

Orthogonality: with $w:= v_i$, we have \begin{align}\|v_i\|^2 = \sum_{j=1}^n \langle v_j,v_i\rangle^2 &= \|v_i\|^2 + \sum_{j \ne i} \langle v_j,v_i \rangle^2 \\ \sum_{j \ne i} \langle v_j,v_i \rangle^2 &= 0 \\ \langle v_j,v_i\rangle &= 0 & \forall j \ne i \end{align}

Normality(?): with $w=v_i/\|v_i\|$, (and noting the orthogonality above) we have \begin{align} 1 = \langle v_i, v_i/\|v_i\| \rangle = \langle v_i, v_i \rangle / \|v_i\| = \|v_i\|. \end{align}


Are you sure of the equality $\sum_{i=1}^{n} <v_i, w>^2 = \|{w\|^2}$ ?

As the norm is usually defined by $\|{u\|} = \sqrt{<u,u>}$, there seems to be a homogeneity problem...

For this reason, if we take $w=v_i$, we have:

$\|{v_i\|}^2 = \sum_{j=1}^{n} <v_j, v_i>^2 = \|{v_i\|}^4 + \sum_{j\neq i}<v_j, v_i>^2 $

And we can't conclude (easily) about the orthogonality of the basis.

However, with $\sum_{i=1}^{n} <v_i, w>^2 = \|{w\|^4}~~~~ \forall w \in V$, the proof of the set being a basis holds, and by taking $w=v_i$, we have:

$\|{v_i\|}^4 = \sum_{j=1}^{n} <v_j, v_i> ^2= \|{v_i\|}^4 + \sum_{j\neq i}<v_j, v_i>^2 $

As $\|{v_i\|}^4 \neq 0$ , we obtain :

$\sum_{j\neq i}<v_j, v_i>^2 = 0$

If the sum of positive terms is zero, then each term is zero, giving us orthogonality :

$<v_j, v_i>= 0 ~~ \forall j \neq i$

Then we have normality by using $w = \frac{v_i}{\|{v_i\|}}$:

$1 = \|{ \frac{v_i}{\|{v_i\|}} \|}^4 = \sum_{j=1}^{n} <v_j, \frac{v_i}{\|{v_i\|}}> ^2= \frac{<v_i, v_i>^2}{\|{v_i\|}^2} = \|{v_i\|}^2 $

And so : $ \|{v_i\|} = 1 $, and the basis is othonormal.


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