# Gram-Schmidt Procedure

1. Oct 19, 2009

### boneill3

1. The problem statement, all variables and given/known data

We consider P2 the vector space of all real polynomials of degree at most 2.

<f,g> = $f(-1)g(-1)+f(0)g(0)+f(1)g(1)$

Use the Gram-Schmidt procedure to construct an orthonormal basis for P2 from the basis {1,t,t2}

2. Relevant equations

$v_{j+1}:=u_{j+1}-\sum_{i=1}^{j}<<u_{j+1},e_{i}>>e_{i}$

$e_1 = \frac{u_1}{||u_{1}|| }$
3. The attempt at a solution

I have a basis $u_1 = 1, u_2 = t, u_3 = t^2$

so

$e_1 = \frac{u_1}{||u_{1}|| }$

$e_1 = \frac{1}{\sqrt{2}}$

is the next step

$v_{2}:=u_{2}-\sum_{i=1}^{j}<<u_{2},e_{i}>>e_{i}$

$= t - << t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}}$

My question is how do I calculate the inner product $<< t,\frac{1}{\sqrt{2}} >>$
do I need to plug in the value of f(t) into

<f,g> = $f(-1)g(-1)+f(0)g(0)+f(1)g(1)$

and does g() become $g(e_1) = g(\frac{1}{\sqrt{2}})$

regards

2. Oct 19, 2009

### Dick

The inner product <<t,1/sqrt(2)>>=(-1)/sqrt(2)+0/sqrt(2)+(1)/sqrt(2). That's what you definition says, isn't it? g is a constant.

3. Oct 20, 2009

### boneill3

So

$v_{2}:=u_{2}-\sum_{i=1}^{j}<u_{j+1},e_{i}>e_{i}$

$= t - << t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}}$

$= t - (\frac{-1}{\sqrt(2)}+\frac{0}{\sqrt(2)}+\frac{1}{\sqrt(2)}) \frac{1}{\sqrt{2}}$

$= t - (0) \frac{1}{\sqrt{2}}$

$v_{2}:= t$

for v3: do I substitute $v_{2}:= t$ from above for $e_{2}$
in the next equation? or do I need to nomalise it first

$v_{3}:=u_{3}-\sum_{i=1}^{j}<u_{j+1},e_{i}>e_{i}$

$= t^2 - (<< t^2,e_{2} >>e_{2}) + (<< t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}})$

$= t^2 - (<< t^2,t >>t) + (<< t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}})$

or

$= t^2 - (<< t^2,1 >>1) + (<< t,\frac{1}{\sqrt{2}} >>\frac{1}{\sqrt{2}})$

regards

Last edited: Oct 20, 2009
4. Oct 20, 2009

### boneill3

That last equation was using $(\frac{1}{t})t= 1$
as $e_{2}$

5. Oct 20, 2009

### Dick

t is orthogonal to e1=1/sqrt(2) (i.e. <t,1/sqrt(2)>=0). But it's not normalized. <t,t> is not one. Normalize it. Then that becomes e2. Then find u3=t^2-(<t^2,e1>e1+<t^2,e2>e2) and normalize u3. Your are getting your functions all mixed up in the Gram-Schmidt process.

6. Oct 20, 2009

### Dick

BTW before you go any further <1,1>=3. So the normalization of 1 isn't 1/sqrt(2). It's 1/sqrt(3). Sorry, I missed that.

7. Oct 20, 2009

### beetle2

Thanks

So

$v_{2}:= t$
therefore

$e_{2} = \frac{t}{||v_2|| }v_2$
$e_{2} = \frac{t}{\sqrt{-1^2+0^2+1^2}}$
$e_{2} =\frac{1}{\sqrt{2}}$

is that right ?

8. Oct 20, 2009

### Dick

Is boneill3 the same person as beetle2? If so the Forum doesn't allow really allow you to use two different user pseudonyms. If you are, please stick with one, ok? Otherwise I should report this. If you aren't you are doing exactly the same kind of notational confusion as boneill3. If v2=t, then e2=v2/||v2||=t/||t|| which is t/sqrt(2). Not 1/sqrt(2).

9. Oct 20, 2009

### beetle2

Hi,

I found u3=t^2-(<t^2,e1>e1+<t^2,e2>e2) to equal

t^2-(1/sqrt(3)(1/sqrt(3)+1/sqrt(3)) + t/sqrt(2) (t/sqrt(2)+ t/sqrt(2))

= t^2 - (2/3+t^2)

u3 = -2/3

does normalising make it e3= -(2/3) / sqrt((-2/3)^2 +(-2/3)^2+(-2/3)^2)

e3 = -1/sqrt(3)

10. Oct 21, 2009

### boneill3

I think i'll make it clearer
I found
$u3=t^2-(<t^2,e_1>e_1+<t^2,e_2>e_2)$
to equal

$t^2-(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}) + (\frac{t}{\sqrt{2}}) (\frac{t}{\sqrt{2}}+\frac{t}{\sqrt{2}}) = t^2 - (\frac{2}{3}+t^2) = \frac{-2}{3}$
does normalising make it

$e_{3}= \frac{\frac{-2}{3}}{{\sqrt{(\frac{-2}{3})^2 +(\frac{-2}{3})^2+(\frac{-2}{3})^2}}}$

$e_{3} = \frac{-1}{\sqrt{3}}$

11. Oct 21, 2009

### Dick

If that whole thing had worked then you should be able to check that <e1,e3>=0. It's not zero. The inner product of two functions of t, <f(t),g(t)>, should always be a number, not another function of t. Try computing <t^2,t/sqrt(2)> again.

12. Oct 21, 2009

### boneill3

So,

to compute <t^2,t/sqrt(2)>

we have

$\left[(-1)^2(\frac{-1}{\sqrt{2}})+](0)^2(\frac{0}{\sqrt{2}})+](1)^2(\frac{1}{\sqrt{2}})\right]$

therfore

$t^2-(\frac{1}{\sqrt{3}})(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}) + (\frac{t}{\sqrt{2}}) (0) = t^2 - (\frac{2}{3})$

13. Oct 21, 2009

### Dick

Ok, now you just have to normalize t^2-(2/3).

14. Oct 21, 2009

### beetle2

I normalised u3 and got

$e_{3}= \frac{t^2-\frac{-2}{3}}{\sqrt{\frac{2}{3}}}$

I checked that <e1,e3>=0 which it does.

Thanks so much for your help