anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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amistre64
  • amistre64
wouldnt that ne the nulspace of that span?
TuringTest
  • TuringTest
we need to turn this into an orthogonal set first I believe, which requires me looking at my notes
amistre64
  • 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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TuringTest
  • TuringTest
we gotta use the Gram-Schmidt process I believe
amistre64
  • amistre64
Gram Smidt is for weiners lol
TuringTest
  • TuringTest
I know, that's why I always forget it and have to look at my notes
amistre64
  • amistre64
the set given is not linearly independant, if that helps
TuringTest
  • 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
  • anonymous
Use words instead then :) I know GS method, so it's fine.
TuringTest
  • 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
  • anonymous
Hmm, so it's that easy is it.. Thought it would be more work.
TuringTest
  • 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
  • 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
  • anonymous
Are there any sweeter alternatives then GS to convert the set to a orthogonal base?
TuringTest
  • TuringTest
I really wish there were, I always forget the GS formula if you find one let me know
TuringTest
  • TuringTest
@amistre64 yes it is a problem, that's why we use the GS process first to make it an orthogonal set
anonymous
  • 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
  • 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
  • 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
  • TuringTest
echo echo... lol
amistre64
  • amistre64
your set is NOT a basis for R^3
TuringTest
  • TuringTest
oh I didn't check, good eye amistre
anonymous
  • anonymous
Weird, since the book has nothing to say about that when I check the answer. Just shows the 2 answer-vectors.
TuringTest
  • 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
  • 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
  • TuringTest
lol yeah that's how I read it too
amistre64
  • amistre64
:)
TuringTest
  • 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
  • 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
  • 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
  • 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
  • TuringTest
example 2 he makes the set orthogonal example 3 he makes it orthonormal
amistre64
  • amistre64
this website is acting wierd lately; keeps booting me out
TuringTest
  • TuringTest
me too at times...
amistre64
  • amistre64
you say your answer book gives 2 vectors?
anonymous
  • anonymous
Indeed
amistre64
  • amistre64
then i believe that are omiting the last vector since its not linearly independant; and using the first 2 in some process
anonymous
  • anonymous
Thought: Does a subspace actually have to be a base? Maybe that's in a defintion somewhere.. But yeah, does it?
TuringTest
  • 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
  • amistre64
a subspace of R^3 that spans R^3 has to have 3 linearly independant vectors
anonymous
  • anonymous
I think so as well. Thanks amistre
amistre64
  • amistre64
otherwise its a subspace in R^2 i think
TuringTest
  • TuringTest
right, the vectors have to be linearly independent to form a basis for anything
amistre64
  • amistre64
ah, find an orthoN for the subspace spanned by .... yeah, so drop that last vector and try your luck
amistre64
  • amistre64
teh subspace is not being defined as R^3
TuringTest
  • TuringTest
I agree, there seems no better option
anonymous
  • anonymous
Sneaky bastards, will see if that works out.
TuringTest
  • TuringTest
good luck!
amistre64
  • amistre64
the only caveat i see there is in that the first 2 vectors still aint orthoG
amistre64
  • amistre64
do you have to make them orthoG? by finding 2 vectors in the plane that are orthoG?
TuringTest
  • 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
  • amistre64
lol, they are still a basis; they just dont exists as the basis for the xy plane; its tilted
amistre64
  • amistre64
|dw:1333376644461:dw|
TuringTest
  • TuringTest
so then the z-components should be linearly dependent ?
TuringTest
  • TuringTest
and the other two no? I'm confusiddddd
amistre64
  • 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
  • TuringTest
and you would just do GS on... what now?
amistre64
  • amistre64
x 1 y = z -2 ; and the point (0,0,0) z 1 x-2y+z = 0
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=z+%3D+2y-x this is the subspace we are looking at
TuringTest
  • 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
  • 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
  • TuringTest
ok, that's a good idea :) then orthoN just comes from dividing each vector by it's norm ?
amistre64
  • amistre64
thats my thought; but then im just taking a blind stab at it :)
TuringTest
  • 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
  • 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
  • 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
  • amistre64
what are the answers so that we know where to go from here?
amistre64
  • amistre64
im assuming youve got 2 answer vectors
anonymous
  • anonymous
Fock, annoying equation system, I'll just use root as root
anonymous
  • anonymous
(0, 1/r5, 2/r5) (-(r5/r6), -(2/r30), 1/r30)
anonymous
  • anonymous
Enjoy, btw, if you reached 3 vectors... and the answer is still 2?
TuringTest
  • TuringTest
so they started with the vector (0,1,2) and applied GS than divided by the norm it looks like
amistre64
  • 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
  • amistre64
the first one is the v1 x n
amistre64
  • amistre64
lets call the first cross b1 :) |b1| = r30
anonymous
  • anonymous
yep, first one seems like that is done at least.
amistre64
  • amistre64
-5/r30, -2/r30, 1/r30 for the binormal i found; you sure your secind is -r5 as a numerator?
anonymous
  • anonymous
yes, in the solution it's -r5/r6, -2/r30, 1/r30
anonymous
  • anonymous
I also got your answer btw, using the GS method.
amistre64
  • amistre64
i think thats too much of a coincedence to be wrong; maybe the answer key has a typo?
anonymous
  • 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
  • amistre64
well, thats all the crazy ideas I can make for this, so good luck ;)
anonymous
  • anonymous
Thank you for your time :) you seem to know your planes in and out!
TuringTest
  • 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
  • amistre64
got another crazy idea does sqrt(5/6) = 5/sqrt(30)?
TuringTest
  • TuringTest
no
TuringTest
  • TuringTest
haha whaddya think?
amistre64
  • amistre64
sqrt(5/6) = sqrt(25/36) = 5/sqrt(36) lol ... like i said, crazy
anonymous
  • anonymous
yes, it does :)
anonymous
  • anonymous
It's the same actually, got my calc right here :P
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=5%2Fsqrt%2830%29 LOL
amistre64
  • amistre64
one of my personalities is a genius :)
TuringTest
  • TuringTest
\[\sqrt{\frac56}\cdot\sqrt{\frac55}=\frac5{\sqrt30}\]
TuringTest
  • TuringTest
so whoopie!
amistre64
  • amistre64
what math course is this?
anonymous
  • anonymous
haha, this damn book is doing the most amazing factors ever.. Who on earth knows 5/r30 = r5/6 without checking properly?
TuringTest
  • TuringTest
rationalizing irrational problems
anonymous
  • anonymous
Linear algebra, basic course, 1/4th of a term.
amistre64
  • amistre64
hmm, im in linear now; elementary stuff. havent seen nothing like this tho. grad or undergrad?
anonymous
  • anonymous
undergrad, it's in the end of the course, getting a bit more deep towards the end.
amistre64
  • amistre64
we got like 2 weeks left over here; and we are just getting to the eugene stuff
anonymous
  • anonymous
Guess it's different depending on uni, I suppose. Eigenvectors is the last thing in this course.
amistre64
  • amistre64
:) goog luck with it all
amistre64
  • amistre64
good luck even lol
anonymous
  • anonymous
Likewise, doubt you need it though :P

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