Gram-Schmidt Proof

**Anthonny** · November 18th 2010, 08:30 PM

I'm stuck on this one part for the proof of the Gram-Schmidt which states if $(v_1,\dots,v_m_)$ is a linearly independent list (my text uses the term list, but I believe the term set is used more often) of vectors in V, then there exist an orthonormal list $(e_1,\dots,e_m)$ of vectors in V such that

$span(v_1,\dots,v_j)=span(e_1,\dots,e_j)$

for j=1,...,m

The proofs follows:

Suppose $(v_1,\dots,v_j)$ is a linearly independent list of vectors in V. To construct the e's, start by setting $e_1=\frac{v_1}{||v_1||}$ . This then satisfies the above equation.

So far in the proof I understand how it works for $e_1$ .

We will choose $e_2,\dots,e_m$ inductively, as follows. Suppose j>1 and an orthonormal list $(e_1,\dots,e_{j-1})$ has been chosen so that

$span(v_1,\dots,v_{j-1})=span(e_1,\dots,e_{j-1})$ .

Here is where I am confused. I don't understand how an orthonormal list can be chosen such that the above is true. Isn't that similar to what we're trying to prove? And why can't the index j-1 just be j?

I did not post the complete proof; if this information is not sufficient enough I can post the rest.

Thank you.

**tonio** · November 18th 2010, 09:11 PM

Originally Posted by Anthonny

I'm stuck on this one part for the proof of the Gram-Schmidt which states if $(v_1,\dots,v_m_)$ is a linearly independent list (my text uses the term list, but I believe the term set is used more often) of vectors in V, then there exist an orthonormal list $(e_1,\dots,e_m)$ of vectors in V such that

$span(v_1,\dots,v_j)=span(e_1,\dots,e_j)$
for j=1,...,m

The proofs follows:

Suppose $(v_1,\dots,v_j)$ is a linearly independent list of vectors in V. To construct the e's, start by setting $e_1=\frac{v_1}{||v_1||}$ . This then satisfies the above equation.

So far in the proof I understand how it works for $e_1$ .

We will choose $e_2,\dots,e_m$ inductively, as follows. Suppose j>1 and an orthonormal list $(e_1,\dots,e_{j-1})$ has been chosen so that

$span(v_1,\dots,v_{j-1})=span(e_1,\dots,e_{j-1})$ .

Here is where I am confused. I don't understand how an orthonormal list can be chosen such that the above is true. Isn't that similar to what we're trying to prove? And why can't the index j-1 just be j?

I did not post the complete proof; if this information is not sufficient enough I can post the rest.

Thank you.

Well, how's $e_j$ constructed? First, you define $w_j:=\frac{v_j}{||v_j||}$ , and then you take

$e_j:=\sum\limits^j_{k=1}\langle w_j,e_k\rangle e_k$ . The inductive problem is to show that $e_j\neq 0$ and that it is

lin. indep. from $\{e_1,\ldots ,e_{j-1}\}$ ...and this is your job.

Tonio

Thread: Gram-Schmidt Proof

LinkBack

Thread Tools

Display

Gram-Schmidt Proof

Similar Math Help Forum Discussions

Gram-Schmidt procedure

Gram-Schmidt

[SOLVED] Gram-Schmidt

gram schmidt orthogonalisation

Gram-Schmidt

Search Tags