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;

- KonradZuse

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

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

@mahmit2012 @kropot72 @asnaseer

Looking for something else?

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

## More answers

- KonradZuse

##### 1 Attachment

- 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

so I basically compare 1 at a time to each other...?

- KonradZuse

in the book's example they compare each of them to each other it seems.(u1,u3) = (u2,u1) = (u2,u3) = 0

- KonradZuse

I guess they do 2 at a time though.....

- 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

Also wouldn't a b and D be an orthogonal set since each individual is orthogonal?

- KonradZuse

@UnkleRhaukus

- 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

@Outkast3r09 save me :)

- 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

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

@jim_thompson5910 save me ;)

- UnkleRhaukus

you could plot the point on a graph

- UnkleRhaukus

orthogonal points will be separated by a 90° angle at the origin

- UnkleRhaukus

orthogonal sets do not have to have a norm = 1.
orthonormal sets do

- 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

@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

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

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

yeah I failed haha.

- 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

definition 1 states that if each are orthogonal sets, then they will be orthogonal to each other.

- 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

Idk I forgot :P

- tkhunny

Well, check that definition and be sure. You'll be the only one in your class to notice the technicality.

- KonradZuse

BLEHEHEH

- 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

so does that mean the only orthagonal sets are a and b?

- 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

I did't think we needed to use basis on this one... :(

- 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

mhm

- 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

so a is gone too?

- tkhunny

Well, how are we doing? Are we getting to the bottom of this one, too?

- KonradZuse

I thought we were comparing based on the definition.. Guess not :P

- tkhunny

'a' is gone by inspections. (2,0) has length 2, not 1.

- KonradZuse

kk

- 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

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

Which of the sets in Exercise 1 are orthonormal with respect to the Euclidean inner product on R^2

- KonradZuse

which from the book def 1 says what i've been saying above.

- 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

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

maybe that means something else.... But liek you were saying above norm = 1.

- KonradZuse

hmm?

- KonradZuse

not sure what you mean...?

- tkhunny

I think the last word in Definition 1 should be "orthonormal".

- KonradZuse

meh stupid book I hate thee....

- 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

it says we should verify it tho.

- 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

by the way I get an inner product of 1 for c.

- 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

look at the picture i posted.....

- KonradZuse

it takes the norm of 3 vectors, and then normalizes the vectors which then sets them = 1?

- KonradZuse

a also = 1 if done this way....

- tkhunny

?? Both pieces are negative. Can't be +1.

- KonradZuse

neg * neg + pos * pos...

- tkhunny

- KonradZuse

Oh yeah :P good call.

- 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

so for the norm of a do we just do each individual section? sqrt(0^2 + 2^2) = 2

- KonradZuse

kk

- KonradZuse

norm =length?

- KonradZuse

so we take each individual vector than to see? So all vectors have to = 1 then?

- 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

I just wanna write it out all nice :).

- 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

This is what I wrote out... Should be good :). Thanks for the help!

##### 1 Attachment

- 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

So it looks good? :)
"An orthogonal set in which each vector has norm 1 is said to be orthognormal????"

- KonradZuse

I added that to my homework, that isn't in that pic :P

- 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

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

I'm not sure if that's a good example tho....

- KonradZuse

Someone posted water flow, health care cost, population, NFL wins... :(

- tkhunny

Interest Rate / Investment Earnings Prognostication.

Looking for something else?

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