Question

Let R^3 have the Euclidean inner product. Find an orthonormal basis for the subspace spanned by (0,1,2) (-1,0,1) (-1,1,3). Show the steps obviously. Thanks!

Answer 1

wouldnt that ne the nulspace of that span?

Answer 2

we need to turn this into an orthogonal set first I believe, which requires me looking at my notes

Answer 3

http://www.wolframalpha.com/input/?i=rref%7B%280%2C1%2C2%29+%2C%28-1%2C0%2C1%29%2C+%28-1%2C1%2C3%29%7D

if so i rref the matrix to get; (1,-2,1)

Answer 4

we gotta use the Gram-Schmidt process I believe

Answer 5

Gram Smidt is for weiners lol

Answer 6

I know, that's why I always forget it and have to look at my notes

Answer 7

the set given is not linearly independant, if that helps

Answer 8

\[\vec u_1=\vec v_1=<0,1,2>\]\[\vec u_2=\vec v_2-{<\vec v_2,\vec u_1>\over||\vec u_1||^2}\vec u_1\]oh this is gonna take forever

Answer 9

Use words instead then :) I know GS method, so it's fine.

Answer 10

well if you continue along with GS you get an orthogonal basis
you want an \(orthonormal\) basis, so just divide each resultant vector by its norm at the end.

Answer 11

Hmm, so it's that easy is it.. Thought it would be more work.

Answer 12

\[\vec u_1=\vec v_1\]\[\vec u_2=\vec v_2-{<\vec v_2,\vec u_1>\over||\vec u_1||^2}\vec u_1\]\[\vec u_2=\vec v_3-{<\vec v_3,\vec u_1>\over||\vec u_1||^2}\vec u_1-{<\vec v_3,\vec u_2>\over||\vec u_2||^2}\vec u_2\]not that's all according to my notes

Answer 13

(0,1,2) (-1,1,3) (0,1,2)
(-1,0,1) (-1,0,1) (-1,1,3)
-------------------------
0+0+2 1+0+3 0+1+6

none of the vectors in the set ore orthogonal to each other ... is that a problem?

Answer 14

Are there any sweeter alternatives then GS to convert the set to a orthogonal base?

Answer 15

I really wish there were, I always forget the GS formula
if you find one let me know

Answer 16

@amistre64 yes it is a problem, that's why we use the GS process first to make it an orthogonal set

Answer 17

I see, well, crackin on the GS then, thanks for the help. Do you know any page where all the spaces are geometrically explained, so it's easier to imagine how they all relate to each others?

Answer 18

I almost always refer people here for linear algebra
http://tutorial.math.lamar.edu/Classes/LinAlg/OrthonormalBasis.aspx
check out the whole site, it's pretty cool

Answer 19

http://tutorial.math.lamar.edu/Classes/LinAlg/OrthonormalBasis.aspx

example 2 says: Solution - You should verify that the set of vectors above is in fact a basis for R^3

Answer 20

echo echo... lol

Answer 21

your set is NOT a basis for R^3

Answer 22

oh I didn't check, good eye amistre

Answer 23

Weird, since the book has nothing to say about that when I check the answer. Just shows the 2 answer-vectors.

Answer 24

hmf... well now I have no idea what to do
if it doesn't span \(\mathbb R^3\) how can we make it do so?

Answer 25

then example 2 is finding an orthoGONAL basis and not really an orthoNORMAL basis for othro vectors of orthodontists of orthopedic institutions of orthogonality on the book of orthometrical identites of othro ortho orthos

Answer 26

lol yeah that's how I read it too

Answer 27

:)

Answer 28

@Nightie are you \(sure\) those are the three vectors you are given? not typos?
'cuz if those are the vectors, they don't form a basis for \(\mathbb R^3\), so I got no idea what to do
maybe amistre does

Answer 29

"First, note that this is almost the same problem as the previous one except this time we’re looking for an orthonormal basis instead of an orthogonal basis. There are two ways to approach this. The first is often the easiest way and that is to acknowledge that we’ve got a orthogonal basis and we can turn that into an orthonormal basis simply by dividing by the norms of each of the vectors. " But we dont have an orthoG basis to work with

Answer 30

yeah, that's why I suggested GS to get an othogonal basis, but how do we get around the fact that this is not a basis for R^3 ?

Answer 31

Well in example 2 he doesnt convert them to a orthonormal basis, he keeps them as orthogonal, afaik. And yes, I've allready checked the vectors 4 times, just in case...

Answer 32

example 2 he makes the set orthogonal
example 3 he makes it orthonormal

Answer 33

this website is acting wierd lately; keeps booting me out

Answer 34

me too at times...

Answer 35

you say your answer book gives 2 vectors?

Answer 36

Indeed

Answer 37

then i believe that are omiting the last vector since its not linearly independant; and using the first 2 in some process

Answer 38

Thought: Does a subspace actually have to be a base? Maybe that's in a defintion somewhere.. But yeah, does it?

Answer 39

what do you mean "be a base"?
a set of vectors can form a base for a susbspace
I think your terminology is a bit confused

Answer 40

a subspace of R^3 that spans R^3 has to have 3 linearly independant vectors

Answer 41

I think so as well. Thanks amistre

Answer 42

otherwise its a subspace in R^2 i think

Answer 43

right, the vectors have to be linearly independent to form a basis for anything

Answer 44

ah, find an orthoN for the subspace spanned by .... yeah, so drop that last vector and try your luck

Answer 45

teh subspace is not being defined as R^3

Answer 46

I agree, there seems no better option

Answer 47

Sneaky bastards, will see if that works out.

Answer 48

good luck!

Answer 49

the only caveat i see there is in that the first 2 vectors still aint orthoG

Answer 50

do you have to make them orthoG? by finding 2 vectors in the plane that are orthoG?

Answer 51

but you can make them orthoG with GS
but that makes no sense because they aren't a basis because they have 3 components
oh jeez, this makes no sense

Answer 52

lol, they are still a basis; they just dont exists as the basis for the xy plane; its tilted

Answer 53

so then the z-components should be linearly dependent ?

Answer 54

and the other two no?
I'm confusiddddd

Answer 55

this row reduces to:

1 0 -1
0 1 2
0 0 0

so its a plane with a normal vector of (1,-2,1) if i see it right

Answer 56

and you would just do GS on... what now?

Answer 57

x 1
y = z -2 ; and the point (0,0,0)
z 1

x-2y+z = 0

Answer 58

http://www.wolframalpha.com/input/?i=z+%3D+2y-x

this is the subspace we are looking at

Answer 59

so I was right that z is linearly dependent on x and/or y
but I still have no idea what to do from here to get an otho-anything basis

Answer 60

well, if all that is needed is to orthoG vectors from the plane; we have a normal and 2 other vectors to choose from; id say we could cross the normal and another vector to produce the orthoG (binormal) that sits in the plane to play with

Answer 61

ok, that's a good idea :)
then orthoN just comes from dividing each vector by it's norm ?

Answer 62

thats my thought; but then im just taking a blind stab at it :)

Answer 63

yeah, I doubt this is how they want the problem solved, but since the vectors given do not span R^3 I think your way should work

Answer 64

any advice from nightie would be good; since they are actually invloved in the subject material to guide this idea into the gutter or not lol

Answer 65

I'm just trying to pass the test... Scratching surfaces of the real material, my terminology even sucks.. :P So you're probably better off guiding this, by far :)

Answer 66

what are the answers so that we know where to go from here?

Answer 67

im assuming youve got 2 answer vectors

Answer 68

Fock, annoying equation system, I'll just use root as root

Answer 69

(0, 1/r5, 2/r5)
(-(r5/r6), -(2/r30), 1/r30)

Answer 70

Enjoy, btw, if you reached 3 vectors... and the answer is still 2?

Answer 71

so they started with the vector (0,1,2) and applied GS
than divided by the norm it looks like

Answer 72

x 1 0 x= -5
y -2 1 y= -2
z 1 2 z= 1

x 1 -1 x= -2
y -2 0 y= -2
z 1 1 z= -2

well, I crossed the normal to each vector just for kicks

Answer 73

the first one is the v1 x n

Answer 74

lets call the first cross b1 :) |b1| = r30

Answer 75

yep, first one seems like that is done at least.

Answer 76

-5/r30, -2/r30, 1/r30 for the binormal i found; you sure your secind is -r5 as a numerator?

Answer 77

yes, in the solution it's -r5/r6, -2/r30, 1/r30

Answer 78

I also got your answer btw, using the GS method.

Answer 79

i think thats too much of a coincedence to be wrong; maybe the answer key has a typo?

Answer 80

Yeah, should be safe to assume it is. I'll see what happens with the 3rd vector using the GS method, would be interesting.