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!

- anonymous

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- amistre64

wouldnt that ne the nulspace of that span?

- TuringTest

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

- amistre64

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)

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## More answers

- TuringTest

we gotta use the Gram-Schmidt process I believe

- amistre64

Gram Smidt is for weiners lol

- TuringTest

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

- amistre64

the set given is not linearly independant, if that helps

- TuringTest

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

- anonymous

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

- TuringTest

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.

- anonymous

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

- TuringTest

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

- amistre64

(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?

- anonymous

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

- TuringTest

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

- TuringTest

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

- anonymous

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?

- TuringTest

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

- amistre64

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

- TuringTest

echo echo... lol

- amistre64

your set is NOT a basis for R^3

- TuringTest

oh I didn't check, good eye amistre

- anonymous

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

- TuringTest

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?

- amistre64

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

- TuringTest

lol yeah that's how I read it too

- amistre64

:)

- TuringTest

@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

- amistre64

"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

- TuringTest

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 ?

- anonymous

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...

- TuringTest

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

- amistre64

this website is acting wierd lately; keeps booting me out

- TuringTest

me too at times...

- amistre64

you say your answer book gives 2 vectors?

- anonymous

Indeed

- amistre64

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

- anonymous

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

- TuringTest

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

- amistre64

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

- anonymous

I think so as well. Thanks amistre

- amistre64

otherwise its a subspace in R^2 i think

- TuringTest

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

- amistre64

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

- amistre64

teh subspace is not being defined as R^3

- TuringTest

I agree, there seems no better option

- anonymous

Sneaky bastards, will see if that works out.

- TuringTest

good luck!

- amistre64

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

- amistre64

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

- TuringTest

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

- amistre64

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

- amistre64

|dw:1333376644461:dw|

- TuringTest

so then the z-components should be linearly dependent ?

- TuringTest

and the other two no?
I'm confusiddddd

- amistre64

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

- TuringTest

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

- amistre64

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

- amistre64

http://www.wolframalpha.com/input/?i=z+%3D+2y-x
this is the subspace we are looking at

- TuringTest

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

- amistre64

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

- TuringTest

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

- amistre64

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

- TuringTest

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

- amistre64

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

- anonymous

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 :)

- amistre64

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

- amistre64

im assuming youve got 2 answer vectors

- anonymous

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

- anonymous

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

- anonymous

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

- TuringTest

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

- amistre64

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

- amistre64

the first one is the v1 x n

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

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

- amistre64

well, thats all the crazy ideas I can make for this, so good luck ;)

- anonymous

Thank you for your time :) you seem to know your planes in and out!

- TuringTest

likewise
I'd stick with GS, but amistre's logic seems good too
I still suspect that something is amiss here....
good luck!

- amistre64

got another crazy idea
does sqrt(5/6) = 5/sqrt(30)?

- TuringTest

no

- TuringTest

haha whaddya think?

- amistre64

sqrt(5/6) = sqrt(25/36) = 5/sqrt(36) lol ... like i said, crazy

- anonymous

yes, it does :)

- anonymous

It's the same actually, got my calc right here :P

- amistre64

http://www.wolframalpha.com/input/?i=5%2Fsqrt%2830%29
LOL

- amistre64

one of my personalities is a genius :)

- TuringTest

\[\sqrt{\frac56}\cdot\sqrt{\frac55}=\frac5{\sqrt30}\]

- TuringTest

so whoopie!

- amistre64

what math course is this?

- anonymous

haha, this damn book is doing the most amazing factors ever.. Who on earth knows 5/r30 = r5/6 without checking properly?

- TuringTest

rationalizing irrational problems

- anonymous

Linear algebra, basic course, 1/4th of a term.

- amistre64

hmm, im in linear now; elementary stuff. havent seen nothing like this tho.
grad or undergrad?

- anonymous

undergrad, it's in the end of the course, getting a bit more deep towards the end.

- amistre64

we got like 2 weeks left over here; and we are just getting to the eugene stuff

- anonymous

Guess it's different depending on uni, I suppose. Eigenvectors is the last thing in this course.

- amistre64

:) goog luck with it all

- amistre64

good luck even lol

- anonymous

Likewise, doubt you need it though :P

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