KonradZuse
  • KonradZuse
6.3 #2 Which of the sets in Exercise 1 are orthonormal with respect to the Euclidean inner product on R^2 a = (0, 1), 2, 0; b = (-1/sqrt(2), 1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); c = (-1/sqrt(2), -1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); d = (0, 0), 0, 1;
Linear Algebra
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
KonradZuse
  • KonradZuse
Now it says in the book that A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthogonal.
KonradZuse
  • KonradZuse
Now in the problem before this I found that a,b, and d were orthogonal vectors. So would I use those 3 only, or all 4? I'm not sure what the latter part means "An orthogonal set in which each vector has norm 1 is said to be orthogonal." I would think we wouldn't know it was orthogonal, and isn't saying an orthogonal set is orthogonal kind of redundant??? If it meant that we know the vectors are already orthogonal then I would assume the 3 a b, and d.
KonradZuse
  • KonradZuse
@mahmit2012 @kropot72 @asnaseer

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

KonradZuse
  • KonradZuse
1 Attachment
anonymous
  • anonymous
b & c is a orthonormal set. a with (1,0) also is a orthonormal, but d is a (0,0) and it can not be in a orthonormal set.
KonradZuse
  • KonradZuse
so I basically compare 1 at a time to each other...?
KonradZuse
  • KonradZuse
in the book's example they compare each of them to each other it seems.(u1,u3) = (u2,u1) = (u2,u3) = 0
KonradZuse
  • KonradZuse
I guess they do 2 at a time though.....
KonradZuse
  • KonradZuse
@mahmit2012 should I try a and C and b and D? I'm also not 100% exact what you were saying about (0,0) for d?
KonradZuse
  • KonradZuse
Also wouldn't a b and D be an orthogonal set since each individual is orthogonal?
KonradZuse
  • KonradZuse
@UnkleRhaukus
KonradZuse
  • KonradZuse
I think that according to the definition I would have to just use the ones that make sense... C shouldn't be anything since it's not orthogonal, and according to definition 1 that states: A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal.
KonradZuse
  • KonradZuse
@Outkast3r09 save me :)
KonradZuse
  • KonradZuse
Definition 1 stating "A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal." So we also should concluded that set a and d are orthogonal, a and b are orthogonal, b and d are orthogonal to each other.
KonradZuse
  • KonradZuse
Now it also says that orthogonal sets have to have a norm = 1 to be orthogonal, so should I check for that also? a's norm is 2, but by normalizing it you get 1 as the new norm.
KonradZuse
  • KonradZuse
@jim_thompson5910 save me ;)
UnkleRhaukus
  • UnkleRhaukus
you could plot the point on a graph
UnkleRhaukus
  • UnkleRhaukus
orthogonal points will be separated by a 90° angle at the origin
UnkleRhaukus
  • UnkleRhaukus
orthogonal sets do not have to have a norm = 1. orthonormal sets do
KonradZuse
  • KonradZuse
DEFINITION 1 A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthogonal. Wtf is ortonormal? @UnkleRhaukus
KonradZuse
  • KonradZuse
@tkhunny
tkhunny
  • tkhunny
OrthoNormal = Orthogonal + Normal = Orthogonal + (Unit Length) Orthogonal -- Calculate the Inner Product. If you get "0", you are done. They are Orthogonal. Normal -- Calculate the Length of each Basis Vector. If all lengths are "1", you are done - Normal it is.
KonradZuse
  • KonradZuse
so did I do this correctly since the vectors a b and d are orthogonal, that the sets a and b, b and d, d and a are arthogonal?
tkhunny
  • tkhunny
You will need to test all basis vectors, pairwise, for orthogonality. You will need to test all basis vecors individually for length. Just question on notation: a = (0, 1), 2, 0; Is that one basis vector (0,1) and another basis vector (2,0), or is it something else. I'm hoping you just failed to use the parentheses on the second vector. Otherwise, I don't know what it means.
KonradZuse
  • KonradZuse
yeah I failed haha.
KonradZuse
  • KonradZuse
well I tested each pair of vectors to see if they were orthogonal to each other, and A, B, and D are the ones that are orthogonal.
KonradZuse
  • KonradZuse
definition 1 states that if each are orthogonal sets, then they will be orthogonal to each other.
tkhunny
  • tkhunny
a = (0, 1), 2, 0; b = (-1/sqrt(2), 1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); c = (-1/sqrt(2), -1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); d = (0, 0), 0, 1; a --- Orthogonal, but not Normal d --- Definition Question,is EVERTHING considered Orthogonal to the zero-vector? Are you SURE that using the zeo-vector constitutes a Basis?
KonradZuse
  • KonradZuse
Idk I forgot :P
tkhunny
  • tkhunny
Well, check that definition and be sure. You'll be the only one in your class to notice the technicality.
KonradZuse
  • KonradZuse
BLEHEHEH
KonradZuse
  • KonradZuse
This one?? I"t is not true that every vector space has a basis in the sense of Definition 1. The simplest example is the zero vector space, which contains no linearly independent sets and hence no basis. The following is an example of a nonzero vector space that has no basis in the sense of Definition 1 because it cannot be spanned by finitely many vectors."
KonradZuse
  • KonradZuse
so does that mean the only orthagonal sets are a and b?
tkhunny
  • tkhunny
Okay, we threw out 'c' as not being orthogonal. That leaves ONLY the length of the b-vectors to consider. Are they length one (1)? I think we have a winner.
KonradZuse
  • KonradZuse
I did't think we needed to use basis on this one... :(
tkhunny
  • tkhunny
As far as the zero vector goes, it cannot be a Basis Vector, since it cannot EVER be linerly independent from other Basis Vectors. Example, If I have two basis vectors (1,0) and (0,1), I can get (0,0) from 0*(1,0) or from 0*(0,1). Definitely not a valid basis vector, zero.
KonradZuse
  • KonradZuse
mhm
tkhunny
  • tkhunny
You always need to consider Basis Vectors, whether it is stated explicitly in the problem statement or not. The correct answer to 'd' is , "Cannot be an orthonormal basis, since it isn't a basis at all." We could also argue that the zero vector has length zero, which clearly is not one (1).
KonradZuse
  • KonradZuse
so a is gone too?
tkhunny
  • tkhunny
Well, how are we doing? Are we getting to the bottom of this one, too?
KonradZuse
  • KonradZuse
I thought we were comparing based on the definition.. Guess not :P
tkhunny
  • tkhunny
'a' is gone by inspections. (2,0) has length 2, not 1.
KonradZuse
  • KonradZuse
kk
KonradZuse
  • KonradZuse
my book says that you can first take the norm of a vector, then normalize it based on that norm, then take the norm again to see.... In that case a would = 1.
tkhunny
  • tkhunny
Maybe you're right. It does not ask, "Which form an orthonormal basis?" That is how I was thinking about the problem. It never hurts to go back and read the problem statement. Recap: a) No good. (2,0) has length 2, not 1 d) No good. (0,0) has length 0, not 1 c) No Good. Inner Product is -1, not 0 b) Got it!
KonradZuse
  • KonradZuse
Which of the sets in Exercise 1 are orthonormal with respect to the Euclidean inner product on R^2
KonradZuse
  • KonradZuse
which from the book def 1 says what i've been saying above.
KonradZuse
  • KonradZuse
EFINITION 1 A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthogonal.
tkhunny
  • tkhunny
a) That makes no sense. It's not a very interesting question if we can just normalize the ones that aren't normalized. d) Of course, I dare you to normalize this one!
KonradZuse
  • KonradZuse
maybe that means something else.... But liek you were saying above norm = 1.
KonradZuse
  • KonradZuse
hmm?
KonradZuse
  • KonradZuse
not sure what you mean...?
tkhunny
  • tkhunny
I think the last word in Definition 1 should be "orthonormal".
KonradZuse
  • KonradZuse
meh stupid book I hate thee....
KonradZuse
  • KonradZuse
This is what I was saying about normalizing.... It shows v2 and v2 = sqrt(2) but then normalizing it they got it = 1?
1 Attachment
KonradZuse
  • KonradZuse
it says we should verify it tho.
tkhunny
  • tkhunny
DEFINITION 1 A set of two or more vectors in a real inner product space is said to be orthogonal if all pairs of distinct vectors in the set are orthogonal. An orthogonal set in which each vector has norm 1 is said to be orthonormal. I stand by this answer: a) No good. (2,0) has length 2, not 1 d) No good. (0,0) has length 0, not 1 c) No Good. Inner Product is -1, not 0 b) Got it! This seems to be most consistent with the problem statement and Definition #1.
KonradZuse
  • KonradZuse
by the way I get an inner product of 1 for c.
tkhunny
  • tkhunny
Verify what? We did all the verification in order to answer the question. Show the inner product (0) and the Norms (1). Done.
KonradZuse
  • KonradZuse
look at the picture i posted.....
KonradZuse
  • KonradZuse
it takes the norm of 3 vectors, and then normalizes the vectors which then sets them = 1?
KonradZuse
  • KonradZuse
a also = 1 if done this way....
tkhunny
  • tkhunny
?? Both pieces are negative. Can't be +1.
KonradZuse
  • KonradZuse
neg * neg + pos * pos...
tkhunny
  • tkhunny
?? \(() = a*c + b*d\) You seem to be doing a*b + c*d.
KonradZuse
  • KonradZuse
Oh yeah :P good call.
tkhunny
  • tkhunny
Your picture has two sections. One side is determining orthogonality. The other side is concerned with CONSTRUCTING an orthonormal set. We are not doing any CONSTRUCTING in this problem. We are just determining if it has already been constructed.
KonradZuse
  • KonradZuse
so for the norm of a do we just do each individual section? sqrt(0^2 + 2^2) = 2
KonradZuse
  • KonradZuse
kk
KonradZuse
  • KonradZuse
norm =length?
KonradZuse
  • KonradZuse
so we take each individual vector than to see? So all vectors have to = 1 then?
tkhunny
  • tkhunny
That's it. In this case, if we don't get one (1), we throw our hands in the air and exclaim, "Not Normal!" We can make it Normal, but that is not part of this problem statement.
KonradZuse
  • KonradZuse
I just wanna write it out all nice :).
tkhunny
  • tkhunny
I'm all for exploration and playing around to see what works. This answer is my actual thought process for answering this problem specifically. First, I rule out the obvious ones: a) No good. (2,0) has length 2, not 1 d) No good. (0,0) has length 0, not 1 Then, with a little work, I narrow it down some more. c) No Good. Inner Product is -1, not 0 b) Got it! If you want to display it very well, I would do this a = (0, 1), (2, 0); \(\sqrt{0^{2} + 1^{2}} = \sqrt{1} = 1\) - Normal \(\sqrt{2^{2} + 0^{2}} = \sqrt{4} = 2\) - Not Normal \((<0,1>,<2,0>) = 0*2 + 1*0 = 0\) - Orthogonal This set is NOT orthonormal And then proceed similarly with the other three. b = (-1/sqrt(2), 1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); c = (-1/sqrt(2), -1/sqrt(2)), 1/sqrt(2), 1/sqrt(2); d = (0, 0), 0, 1;
KonradZuse
  • KonradZuse
This is what I wrote out... Should be good :). Thanks for the help!
1 Attachment
tkhunny
  • tkhunny
Always a pleasure to see a good student working hard to get it right! Let the definitions soak into your brain. Oh, and don't forget to fix that error in your book's Definition 1.
KonradZuse
  • KonradZuse
So it looks good? :) "An orthogonal set in which each vector has norm 1 is said to be orthognormal????"
KonradZuse
  • KonradZuse
I added that to my homework, that isn't in that pic :P
tkhunny
  • tkhunny
orthonormal. No extra 'g' in there. And yes, the "Norm" is essentially the Length. Ortholengthal just looks dumb, so we go with orthonormal. (Not really. I just made that up. MATH JOKES!!)
KonradZuse
  • KonradZuse
By the way we have a discussion post where it asks us to use 6.5 to post an example of why we should use this... Normally I go by what I know, but I'm curious if this will wokr... They talk about least square regression lines, and I saw one y = a + bx which I've used in Physics for finding % error and such in our data.
KonradZuse
  • KonradZuse
I'm not sure if that's a good example tho....
KonradZuse
  • KonradZuse
Someone posted water flow, health care cost, population, NFL wins... :(
tkhunny
  • tkhunny
Interest Rate / Investment Earnings Prognostication.

Looking for something else?

Not the answer you are looking for? Search for more explanations.