Question

let R^2 have the Euclidean inner product. Use the Gram-Schmidt process to transform the basis {u1,u2} into an orthonormal basis.
u1= (1,-3), u2= (2,2)

Answer 1

so, if we consider u1 as one basis, we want one that is orthonormal to u1

Answer 2

i.e., project u2 on a perpendicular vector.

Answer 3

yea. but i confuse when the answer is not that.

Answer 4

you can start with u2 as your original vector too

Answer 5

and find the projection of u1 on an orthogonal verctor

Answer 6

sometimes, the process tells me keep v1 =u1, and count v2= ......
but in this case, they turn v1 to orthonormal vector, too

Answer 7

oh ya, first count them by using Gram, second, turn to orthonomal. sorry kid.

Answer 8

is it right?

Answer 9

let me see, did they take
\[u_1=\left[{1\over2},{-3\over2}\right]^T\]

Answer 10

don't get!!

Answer 11

normalized the vector!

Answer 12

yea.. my bad
\[
u_1=\left[{1\over\sqrt{10}},-{3\over\sqrt{10}}\right]^T
\]

Answer 13

yeap

Answer 14

but not T, just in brake

Answer 15

ok

Answer 16

what do you mean by that ok? I 'm not done yet. my question is: getting v1 and count v2 base on that v1 or taking v1 , orthonormal v1 and then count v2 base on the orthonormal v1? which one?

Answer 17

so, starting with v1, we find v2:
\[
v_2=(2,2)^T-\frac{(2,2)\cdot(1,-3)}{(1,-3)\cdot(1,-3)}(1,-3)^T\\
v_2=(2,2)^T-\frac{-4}{10}(1,-3)^T=\left(2+{2\over5},2-{6\over 5}\right)^T\\
v_2=\left({12\over5},{4\over5}\right)^T
\]

Answer 18

now, normalize it

Answer 19

normalize both them is the last step?

Answer 20

yes, got it. thank you, sir

Answer 21

yep. you cannot normalize v1 before coz you will loose the length information.

Answer 22

now, i am done with it.

Answer 23

you are welcome madam.