1. ## Gram-Schmidt Proof

I'm stuck on this one part for the proof of the Gram-Schmidt which states if $\displaystyle (v_1,\dots,v_m_)$ 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 $\displaystyle (e_1,\dots,e_m)$ of vectors in V such that

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

for j=1,...,m

The proofs follows:

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

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

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

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

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.

2. Originally Posted by Anthonny
I'm stuck on this one part for the proof of the Gram-Schmidt which states if $\displaystyle (v_1,\dots,v_m_)$ 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 $\displaystyle (e_1,\dots,e_m)$ of vectors in V such that

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

The proofs follows:

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

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

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

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

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

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

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

Tonio