Solving Inner Product Equations for Algebra Students

[playboy]

Oh, its algebra time again!

A Question reads:
let ||u|| =1 ||v|| = 2 ||w|| = 3^0.5 (or root 3) <u,v>=-1
<u,w>=0 and <v,w>=3

Given this information, who that u + v = w

I gave it my best show.

i know that ||u|| (im going to write |u| for slimpicity) ... i know that |u| = <u,u> = 1 and |v| = <v,v> = 2 and that |w| = <w,w> = root 3.

I did some arithmatic with the given data, but i just cannot seem to isolate u and v.

What I am now trying to do is work from converting an inner products to the vectors, however, I don't even think that is possible. Infact, I looked through two different textbooks and I still couldn't find it.

Could somebody give me a head start, for where to being. Dont give me too much because I want to work this out on my own.

Thanks

 

[Gokul43201]

Consider the vector u+v-w. Being a linear combination of these three vectors, in must lie in any subspace that contains these vectors. Now what can you infer, if you find that the inner product of this vector with each of u, v, and w turns out to be 0 ?

 

[AKG]

Oh, its algebra time again!
A Question reads:
let ||u|| =1 ||v|| = 2 ||w|| = 3^0.5 (or root 3) <u,v>=-1
<u,w>=0 and <v,w>=3
Given this information, who that u + v = w
I gave it my best show.
i know that ||u|| (im going to write |u| for slimpicity) ... i know that |u| = <u,u> = 1 and |v| = <v,v> = 2 and that |w| = <w,w> = root 3.

This is wrong. Given an inner product, the norm is usually defined:

\|v\| \equiv \sqrt{\langle v,\, v\rangle }

 

[Gokul43201]

Also, in terms of vectors in Euclidean space, the inner product <u,v> = |u| |v| cos(A), where A is the angle between u and v. In the above case, u and w are orthogonal vectors whose lengths are given. The direction and magnitude of v can be easily found from the given data and the above definition. Knowing all three vectors (in polar representation), you can show that the required expression holds.

 

[playboy]

Ahh Gokul43201! I showed it by the second method you told me. Thats geometry! but it works!

so...

1. <u w> = |u| |w| cos(A) A = 90deg
2. <v w> = |v| |w| cos(A) A = 30deg
3. <u v> = |u| |v| cos(A) A = 120deg

So the angle between u and v is 120 deg.
And so, u+v = w.. and the angle between w and v is 30deg, and the angle between w and u is 90deg, and this adds to 120deg.

Therefore, u+v =w!

and AKG... yes, that was a very silly mistake that I made. I don't know how i missed that.

Furthermore, Gokul43201... on your first post, you were speaking about linear combinations.. " Now what can you infer, if you find that the inner product of this vector with each of u, v, and w turns out to be 0 ? " You mean that it is perpendicular right!

 

[AKG]

I don't understand how your argument works playboy. Gokul, does what he wrote make sense? Anyways, you know that <x,x> = 0 if and only if x = 0. And you want to prove u+v = w, i.e. u+v-w = 0.

 

[Gokul43201]

No playboy : just showing that the angles add up correctly means nothing...they have to do that. If you take any 3 vectors u,v,w with A being the angle between u and v and B being the angle between v and w, then the angle between u and w can only be one of A+B or A-B. This is true for any 3 vectors !

The best way to prove it (ignore my first post, that is quite non-optimal - but no, the inference is not that it is perpendicular^*) is as AKG says above. There's only one vector which has norm=0.

If you have to go the geometrical way, it's a little more elaborate. If you've done Newtonian Mechanics and vector addition, you'll easily recall the formulas for the magnitude and direction of a resultant vector - which is nothing but the sum of the two vectors. You can find the resultant of u,v and show that it is w.

* If <a,b> = 0, then a is orthogonal to b or one of the two vectors is a zero-vector. If some vector x satisfies <x,a> = <x,b> = <x,c> = 0 (with a,b,c non-zero) then either x must be orthogonal to the minimal space containing a,b,c or x=0. But the first is impossible since x lies in the minimal space containing a,b,c.

 

[playboy]

oh that's right!

Its to hard to show my geometry on here, but when you draw out the vectors, it does make sense.

so i just solved for the direction, now i need the magnitude.

i used the angel that I found between v and w, 30deg and the angel outside u and v which is 60deg... and used that sin law...

sin(u) = sin(W)
u w

sin(30deg) = sin(60deg) and solve for w
1 w

and i get root 3!

So, I found all the angels and showed the directions of the vectors, and now i just showed the magnitude! So this has to be right!

 

[AKG]

No playboy : just showing that the angles add up correctly means nothing...they have to do that. If you take any 3 vectors u,v,w with A being the angle between u and v and B being the angle between v and w, then the angle between u and w can only be one of A+B or A-B. This is true for any 3 vectors !

Are you sure? What if the vectors are not coplanar?

 

[playboy]

Yes, perhaps. But I still think that showing the magnitude of u+v using trig proves the question!

I have another question that comes from the textbook...

b) If < , > and < , >' are two inner products on V that have equal associated norm functions, show that <u,v>=<u,v>' for all u and v.

This is how i understand the question:

< , > and < , >' are two DIFFERENT inner products in V.
they have equal associated norm funtions: ||< , >|| = ||< , >'||
Does that sound right?

i mean, commen sense would tell me that <a,b> = <a,b>'

can somebody give me a little push.

Thanks.

 

[Gokul43201]

Are you sure? What if the vectors are not coplanar?

Oops...of course !

 

[Gokul43201]

This is how i understand the question:
< , > and < , >' are two DIFFERENT inner products in V.
they have equal associated norm funtions: ||< , >|| = ||< , >'||
Does that sound right?

No, how can that be ? The inner product is a csalar. What's the norm of a scalar ?

What that line says is that given the two inner products <u,v> and <u,v>', you have ||u|| = sqrt(<u,u>) = sqrt(<u,u>') = ||u||' for all u.

 

[playboy]

Isn't that still commen sense though? that if they have equal norm funtions, then obviously <u,v> and <u,v>' are going to be equal?
I mean, further the question gives
||u|| = ||u||'
||v|| = ||v||'
So wouldn't one conculde that <u,v> = <u,v>'

 

[Gokul43201]

Isn't that still commen sense though?

In math, you do not believe that something is true even if it makes sense that it should. Nothing is taken as a truth unless it can be proved from the set of axioms.

Try finding the two norms of a vector like u+v and set them equal to each other...

 

[playboy]

In math, you do not believe that something is true even if it makes sense that it should. Nothing is taken as a truth unless it can be proved from the set of axioms.

...That is almost the same thing the prof told us on the first week of university... and the last word we ever want to hear is "prove" or "show"

 

[playboy]

In the inner product section of the textbook...

the norm (length) is... ||u|| = root <u,u>
the distance is... d(u,v) = ||u-v||

if you have u+v, then the norm would be... root u^2 + v^2 right...

So what do you mean by finding TWO norms of a vector like u+v?

 

[Gokul43201]

Look into the properties of inner products, specifically <a, b+c> = <a,b> + <a,c>

Exploit this property with the suggestion I made. Calculate <u+v,u+v> and <u+v,u+v>' and compare terms.

 

[HallsofIvy]

In the inner product section of the textbook...
the norm (length) is... ||u|| = root <u,u>
the distance is... d(u,v) = ||u-v||
if you have u+v, then the norm would be... root u^2 + v^2 right...
So what do you mean by finding TWO norms of a vector like u+v?

Are you under the impression that (u+v).(u+v)= u^2+ v^2? I would hope that a beginning algebra student would know that that is not true for u and v numbers, much less a person who is taking linear algebra! What exactly do you mean by "u^2"? the dot product of u with itself? if so then
the norm would be the root of (u^2+ 2u.v+ v^2).

 

[HallsofIvy]

b) If < , > and < , >' are two inner products on V that have equal associated norm functions, show that <u,v>=<u,v>' for all u and v.

This is how i understand the question:

< , > and < , >' are two DIFFERENT inner products in V.
they have equal associated norm funtions: ||< , >|| = ||< , >'||
Does that sound right?

i mean, commen sense would tell me that <a,b> = <a,b>'

The fact that the norms are the same tells you that <a,a>= <a,a>' for any a in the vector space. It does not immediately tell you that <a,b>= <a,b>' for any a,b. You might try looking at <a+b,a+b>= <a+b,a+b>' and expanding.

 

[playboy]

Are you under the impression that (u+v).(u+v)= u^2+ v^2? I would hope that a beginning algebra student would know that that is not true for u and v numbers, much less a person who is taking linear algebra! What exactly do you mean by "u^2"? the dot product of u with itself? if so then
the norm would be the root of (u^2+ 2u.v+ v^2).

Hallsofivy... I should have been more clear on that! Next time I will!

I was actually trying to refer to the distance in 3 space between two points u and v which is the root of (x-x0)^2 + (y-y0)^2 + (z-z0)^2 which is completely irrelevant to this question!

Im just getting so confused with working with inner products right now. So I am just going to spend some time working with simple questions with the properties right now.

 

[playboy]

So it must look like this:

<a+b, a+b> = <a+b, a+b>'

<a,a> + <a,b> + <b,a> + <b,b> = <a,a>' + <a,b>' + <b,a>' + <b,b>'

|a|^2 + 2<a,b> + |b|^2 = |a|'^2 + 2<a,b>' + |b|'^2

and since the have equal norm funtions, they cancel out... and...

2<a,b> = 2<a,b>'

<a,b> = <a,b>'

hmmm...

 

[Gokul43201]

Yes, that works.

 

[playboy]

Okay, I didn't know if I should have started a new thread on the forum, or just continue using this one. Since we were talking about "obvious" proofs, i thought I might as well just put this one here!

the question reads: Let U ⊆ W be subspaces of V with dimU=k and dimW=m, whre k<m. If k<l<m, SHOW that a subspace X exists where U ⊆ X ⊆ W and dimX =l.

by the way... that ⊆ means "contained in and equal to" It is not coming out right on my screen, but perhaps it is on yours!

Okay, that is so obvious that is true, but i just don't have any idea where to start? This reminds of of The Mean Value theorem, could somebody give me a start please?

 

[HallsofIvy]

Since dim U= k, dim W= m, there exist bases for U and W containing k and m vectors respectively. Since m> k, there exist at least one vector in W which is not in U. What happens if you add that vector to a basis for U?

 

[playboy]

If you add that 1 vector from W to U, then its no longer a basis. however, then you may have created an entirley new subspace containing that basis which has to be U ⊆ X ⊆ W ?

does that make sense?

It sounds good to me.

 

[playboy]

Hi.

I was just wondering if their is an "algebraic" method of proving this...

I mean, i wrote a lot in words for my solution, which makes sense... but i just cannot see an algebraic method?

Anybody have any ideas?