Question

Let R^3 have the Euclidean inner product = u1v1+2u2v2+3u3v3 Use the Gram-Schmidt process to transform u1 =(1,1,1), u2=(1,1.0), u3=(1,0,0) into an orthonormal basis. I did as usual: let v1= u1, and continue, but the answer in book is for orthonormal v1 = (1/sqr (6), 1/sqr(6), 1/sqr(6)) . it's not the same with mine. Anyone explains me, please.

Answer 1

I checked after each step to make sure that I am on the right track. So, I know I am wrong and stop to get explanation

Answer 2

\[v_2 = u_2-proj_{v_1}u_2 = \frac{ <u_2,v_1> }{ ||v_1||^2 }\]

Answer 3

to Gram- we must let v1 =u1. then everything come to that v1. if it's not ok, everything then ruin. we cannot ignore it.

Answer 4

@Hoa
r u getting orthonormal v1= (1/sqrt(3) , 1/sqrt(3) , 1/sqrt(3)) ??????

Answer 5

yes

Answer 6

do you want me attach the problem and the answer from the book to compare?

Answer 7

i think we are doing something wrong in taking the norm

Answer 8

do attach the problem
no need for solution something wrong with the inner product given in the question

Answer 9

yes, I am doing

Answer 10

Problem 25

Answer 11

the inner product is given as <u,v> = u1v1+2u2v2+3u3v3
we need to find the norm w.r.t this inner product

Answer 12

@Hoa if i ask u to normalize u1 w.r.t the inner product <u,v> = u1v1+2u2v2+3u3v3
how r u going to normalize it?

Answer 13

I don't know. everything starts at let v1=u1, but if you ask me normalize u1, i just take u1'= <u1>/||u1||. done.

Answer 14

yeah we use this concept when we find norm w.r.t Eucledian inner product
but in this problem inner product is given
for comparison see problem 22 and 23
compare it with 25

Answer 15

I think you have to use the following definition of the norm: For some vector \(u\),
\[||u||=\sqrt{\langle u,u\rangle} = \sqrt{u_1^2+2u_2^2+3u_3^2}\]
(The above is obtained from the given inner product.)
So if \(u_1=v_1,\) the unit vector is the following:
\[\frac{v_1}{||v_1||}=\frac{(1,1,1)}{\sqrt{1+2+3}}=\left(\frac{1}{\sqrt6},\frac{1}{\sqrt6},\frac{1}{\sqrt6}\right)\]

Answer 16

thanks a lot. I got it..

Answer 17

You're welcome!