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 2 years ago
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;
 2 years ago
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;

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KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Now 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
 2 years ago
Best ResponseYou've already chosen the best response.0Now 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
 2 years ago
Best ResponseYou've already chosen the best response.0@mahmit2012 @kropot72 @asnaseer

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.1b & 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
 2 years ago
Best ResponseYou've already chosen the best response.0so I basically compare 1 at a time to each other...?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0in the book's example they compare each of them to each other it seems.(u1,u3) = (u2,u1) = (u2,u3) = 0

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I guess they do 2 at a time though.....

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0@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
 2 years ago
Best ResponseYou've already chosen the best response.0Also wouldn't a b and D be an orthogonal set since each individual is orthogonal?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.0@Outkast3r09 save me :)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Definition 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
 2 years ago
Best ResponseYou've already chosen the best response.0Now 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
 2 years ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 save me ;)

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0you could plot the point on a graph

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0orthogonal points will be separated by a 90° angle at the origin

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0orthogonal sets do not have to have a norm = 1. orthonormal sets do

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0DEFINITION 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

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1OrthoNormal = 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
 2 years ago
Best ResponseYou've already chosen the best response.0so 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
 2 years ago
Best ResponseYou've already chosen the best response.1You 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
 2 years ago
Best ResponseYou've already chosen the best response.0well 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
 2 years ago
Best ResponseYou've already chosen the best response.0definition 1 states that if each are orthogonal sets, then they will be orthogonal to each other.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1a = (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 zerovector? Are you SURE that using the zeovector constitutes a Basis?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Well, check that definition and be sure. You'll be the only one in your class to notice the technicality.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0This 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
 2 years ago
Best ResponseYou've already chosen the best response.0so does that mean the only orthagonal sets are a and b?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, we threw out 'c' as not being orthogonal. That leaves ONLY the length of the bvectors to consider. Are they length one (1)? I think we have a winner.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I did't think we needed to use basis on this one... :(

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1As 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.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1You 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).

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Well, how are we doing? Are we getting to the bottom of this one, too?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I thought we were comparing based on the definition.. Guess not :P

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1'a' is gone by inspections. (2,0) has length 2, not 1.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0my 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
 2 years ago
Best ResponseYou've already chosen the best response.1Maybe 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
 2 years ago
Best ResponseYou've already chosen the best response.0Which of the sets in Exercise 1 are orthonormal with respect to the Euclidean inner product on R^2

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0which from the book def 1 says what i've been saying above.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0EFINITION 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
 2 years ago
Best ResponseYou've already chosen the best response.1a) 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
 2 years ago
Best ResponseYou've already chosen the best response.0maybe that means something else.... But liek you were saying above norm = 1.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0not sure what you mean...?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1I think the last word in Definition 1 should be "orthonormal".

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0meh stupid book I hate thee....

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0This is what I was saying about normalizing.... It shows v2 and v2 = sqrt(2) but then normalizing it they got it = 1?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0it says we should verify it tho.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1DEFINITION 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
 2 years ago
Best ResponseYou've already chosen the best response.0by the way I get an inner product of 1 for c.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Verify what? We did all the verification in order to answer the question. Show the inner product (0) and the Norms (1). Done.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0look at the picture i posted.....

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0it takes the norm of 3 vectors, and then normalizes the vectors which then sets them = 1?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0a also = 1 if done this way....

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1?? Both pieces are negative. Can't be +1.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0neg * neg + pos * pos...

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1?? \((<a,b><c,d>) = a*c + b*d\) You seem to be doing a*b + c*d.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Oh yeah :P good call.

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Your 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
 2 years ago
Best ResponseYou've already chosen the best response.0so for the norm of a do we just do each individual section? sqrt(0^2 + 2^2) = 2

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0so we take each individual vector than to see? So all vectors have to = 1 then?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1That'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
 2 years ago
Best ResponseYou've already chosen the best response.0I just wanna write it out all nice :).

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1I'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
 2 years ago
Best ResponseYou've already chosen the best response.0This is what I wrote out... Should be good :). Thanks for the help!

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Always 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
 2 years ago
Best ResponseYou've already chosen the best response.0So it looks good? :) "An orthogonal set in which each vector has norm 1 is said to be orthognormal????"

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I added that to my homework, that isn't in that pic :P

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1orthonormal. 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
 2 years ago
Best ResponseYou've already chosen the best response.0By 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
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not sure if that's a good example tho....

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Someone posted water flow, health care cost, population, NFL wins... :(

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Interest Rate / Investment Earnings Prognostication.
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