Linear Algebra Question

[TecHNooB]

The section is orthogonal compliments. Someone explain to me what happened between the steps specified:

u = v - w

Prove that u lies in W perp by showing that u is orthogonal to every vector in S, a basis for W. For each wi in S, we have

(u, wi) = (v - w, wi) = (v, wi) - (w, wi)
= (v,wi) - ((v, w1)w1 + (v, w2)w2 + ... + (v, wm) wm, wi) <--
= (v,wi) - (v, wi)(wi, wi) <--
= 0

What happened there? O_O So confused..

u, v, and w all represent vectors. (u,v) is the dot product of vector u and vector v.

Additional info:

W is a subspace of inner product space V.
dim W = m
W has a basis of m vectors.
S = {w1, w2,... wm} is an orthonormal basis for W.
v is a vector in V.

 

[TecHNooB]

 

[JohnCU]

Originally posted by: TecHNooB

okay i haven't used this since mthsc 435 but doesn't orthogonal have something to do with the dot product being equal to zero?

 

[TecHNooB]

Originally posted by: JohnCU

Originally posted by: TecHNooB

okay i haven't used this since mthsc 435 but doesn't orthogonal have something to do with the dot product being equal to zero?

Yep. Still not sure what happened here tho

= (v,wi) - ((v, w1)w1 + (v, w2)w2 + ... + (v, wm) wm, wi) <--
= (v,wi) - (v, wi)(wi, wi) <--

WTF! Explain this madness

 

[TecHNooB]

Originally posted by: TecHNooB

 

[JohnCU]

Originally posted by: TecHNooB

Originally posted by: TecHNooB

i'm working on it but i hated this stuff and i can't find my linear algebra book.

 

[blinky8225]

How does this first step even make sense?
(u, wi) = (v - w, wi) = (v, wi) - (w, wi)

When you subtract vectors, you just subtract the respective components, no?
(v, wi) - (w, wi) = (v-w, 0) =/= (v-w, wi)?

 

[TecHNooB]

Originally posted by: JohnCU

Originally posted by: TecHNooB

Originally posted by: TecHNooB

i'm working on it but i hated this stuff and i can't find my linear algebra book.

Okay, thanks I appreciate this T_T

 

[chuckywang]

Originally posted by: TecHNooB
The section is orthogonal compliments. Someone explain to me what happened between the steps specified:

u = v - w

Prove that u lies in W perp by showing that u is orthogonal to every vector in S, a basis for W. For each wi in S, we have

(u, wi) = (v - w, wi) = (v, wi) - (w, wi)
= (v,wi) - ((v, w1)w1 + (v, w2)w2 + ... + (v, wm) wm, wi) <--
= (v,wi) - (v, wi)(wi, wi) <--
= 0

What happened there? O_O So confused..

First of all (v - w, wi) =/= (v, wi) - (w, wi)

 

[TecHNooB]

Originally posted by: blinky8225
How does this first step even make sense?
(u, wi) = (v - w, wi) = (v, wi) - (w, wi)

When you subtract vectors, you just subtract the respective components, no?
(v, wi) - (w, wi) = (v-w, 0) =/= (v-w, wi)?

(a,b) is the dot product of a and b. It is not the same as vector [a b]. God.. linear took a turn for the worst after chapter four

 

[TuxDave]

Originally posted by: TecHNooB

Originally posted by: blinky8225
How does this first step even make sense?
(u, wi) = (v - w, wi) = (v, wi) - (w, wi)

When you subtract vectors, you just subtract the respective components, no?
(v, wi) - (w, wi) = (v-w, 0) =/= (v-w, wi)?

(a,b) is the dot product of a and b. It is not the same as vector [a b]. God.. linear took a turn for the worst after chapter four

Bah I was boggled by the first couple lines too until you mentioned that it was a dot product.

 

[JohnCU]

oh damnit your notation threw me off, i don't remember (a,b) being the dot product

 

[blinky8225]

Originally posted by: TecHNooB

Originally posted by: blinky8225
How does this first step even make sense?
(u, wi) = (v - w, wi) = (v, wi) - (w, wi)

When you subtract vectors, you just subtract the respective components, no?
(v, wi) - (w, wi) = (v-w, 0) =/= (v-w, wi)?

(a,b) is the dot product of a and b. It is not the same as vector [a b]. God.. linear took a turn for the worst after chapter four

Oh, I see, everything makes so much more sense now. However, isn't <a, b> the traditional notation used for an inner product, not (a, b)?

 

[TecHNooB]

Originally posted by: blinky8225

Originally posted by: TecHNooB

Originally posted by: blinky8225
How does this first step even make sense?
(u, wi) = (v - w, wi) = (v, wi) - (w, wi)

When you subtract vectors, you just subtract the respective components, no?
(v, wi) - (w, wi) = (v-w, 0) =/= (v-w, wi)?

(a,b) is the dot product of a and b. It is not the same as vector [a b]. God.. linear took a turn for the worst after chapter four

Oh, I see, everything makes so much more sense now. However, isn't <a, b> the traditional notation used for an inner product, not (a, b)?

Not in my book. And in physics, <a,b> is how they write vectors.

 

[JohnCU]

Originally posted by: TecHNooB

Originally posted by: blinky8225

Originally posted by: TecHNooB

Originally posted by: blinky8225
How does this first step even make sense?
(u, wi) = (v - w, wi) = (v, wi) - (w, wi)

When you subtract vectors, you just subtract the respective components, no?
(v, wi) - (w, wi) = (v-w, 0) =/= (v-w, wi)?

(a,b) is the dot product of a and b. It is not the same as vector [a b]. God.. linear took a turn for the worst after chapter four

Oh, I see, everything makes so much more sense now. However, isn't <a, b> the traditional notation used for an inner product, not (a, b)?

Not in my book. And in physics, <a,b> is how they write vectors.

we wrote it a . b with the . being raised up to the middle portion..

 

[blinky8225]

Originally posted by: TecHNooB

Originally posted by: blinky8225

Originally posted by: TecHNooB

Originally posted by: blinky8225
How does this first step even make sense?
(u, wi) = (v - w, wi) = (v, wi) - (w, wi)

When you subtract vectors, you just subtract the respective components, no?
(v, wi) - (w, wi) = (v-w, 0) =/= (v-w, wi)?

(a,b) is the dot product of a and b. It is not the same as vector [a b]. God.. linear took a turn for the worst after chapter four

Oh, I see, everything makes so much more sense now. However, isn't <a, b> the traditional notation used for an inner product, not (a, b)?

Not in my book. And in physics, <a,b> is how they write vectors.

I see, maybe math majors do things differently, but wikipedia agrees with me for what it's worth.

http://en.wikipedia.org/wiki/Vector_%28spatial%29
http://en.wikipedia.org/wiki/Inner_product

 

[chuckywang]

Originally posted by: TecHNooB
The section is orthogonal compliments. Someone explain to me what happened between the steps specified:

u = v - w

Prove that u lies in W perp by showing that u is orthogonal to every vector in S, a basis for W. For each wi in S, we have

(u, wi) = (v - w, wi) = (v, wi) - (w, wi)
= (v,wi) - ((v, w1)w1 + (v, w2)w2 + ... + (v, wm) wm, wi) <--
= (v,wi) - (v, wi)(wi, wi) <--
= 0

What happened there? O_O So confused..

u, v, and w all represent vectors. (u,v) is the dot product of vector u and vector v.

Additional info:

W is a subspace of inner product space V.
dim W = m
W has a basis of m vectors.
S = {w1, w2,... wm} is an orthonormal basis for W.
v is a vector in V.

Are you missing something, cause I don't think this is true.

Counterexample:

Let W = xy plane and V = R^3

Set w = (1,1,0) and v = (2,2,1)

Therefore u = (1,1,1) which is not perpendicular to W.

 

[TecHNooB]

Originally posted by: chuckywang

Originally posted by: TecHNooB
The section is orthogonal compliments. Someone explain to me what happened between the steps specified:

u = v - w

Prove that u lies in W perp by showing that u is orthogonal to every vector in S, a basis for W. For each wi in S, we have

(u, wi) = (v - w, wi) = (v, wi) - (w, wi)
= (v,wi) - ((v, w1)w1 + (v, w2)w2 + ... + (v, wm) wm, wi) <--
= (v,wi) - (v, wi)(wi, wi) <--
= 0

What happened there? O_O So confused..

u, v, and w all represent vectors. (u,v) is the dot product of vector u and vector v.

Additional info:

W is a subspace of inner product space V.
dim W = m
W has a basis of m vectors.
S = {w1, w2,... wm} is an orthonormal basis for W.
v is a vector in V.

Are you missing something, cause I don't think this is true.

Counterexample:

Let W = xy plane and V = R^3

Set w = (1,1,0) and v = (2,2,1)

Therefore u = (1,1,1) which is not perpendicular to W.

w is the projection of vector v on subspace W.

u is the orthogonal compliment to W.

 

[TuxDave]

= (v,wi) - [ (v, w1)w1 + (v, w2)w2 + ... + (v, wm) wm, wi) <--

= (v,wi) - [ ((v, w1)w1,wi) + ((v, w2)w2,wi) ... ]

= (v,wi) - [ (v, w1)(w1,wi) + (v, w2)(w2,wi) ... ]

= (v,wi) - [ (v, w1)(w1,w1) + (v, w2)(w2,w2) ... ]

= (v,wi) - [ (v, w1) + (v,w2) .... ]

= (v,wi) - (v,w1+w2...)

= (v,wi) - (v,wi)

= 0


*** remember wi is a set of orthogonal basis so (w1,w2) = 0
*** so (w1,wi) = (w1,w1+w2+w3...) = (w1,w1) + (w1,w2) ... = (w1,w1) + 0...

Oh and my other assumption is that I think a basis vector has magnitude 1?

Here's my best shot of playing around with the dot products without really understanding what the hell I'm talking about.

Edit: Trying to clean up duplicate lines cuz I was copying and pasting like mad.

Edit #2: Sorry I just noticed I don't have an intermediate step that matches exactly the second arrow in your question. At the same time it's time to go home from work so neffing will have to wait.

 

[blinky8225]

I have it. I'm typing up the solution now. I also had to make the assumption that each basis vector has a magnitude of 1.

 

[TecHNooB]

Originally posted by: blinky8225
I have it. I'm typing up the solution now. I also had to make the assumption that each basis vector has a magnitude of 1.

Yea, that is correct. Every vector in the orthonormal basis is a unit vector. Thanks for your help guys!

I'm doing some practice problems right now but i'll be sure to decipher what you guys put up

 

[blinky8225]

I?m going to use my notation because yours really messes with me.
Okay I?ll focus on how <w, wi> becomes <v, wi> x <wi, wi>.

So we have <w, wi> and u = v-w -> v = u + w
So, w can be written as the projection of v onto w. From the projection formula, we obtain (<v, w>/ (Norm[w])^2) w.

Now since, w is in W and S forms a basis for W, w can be written with the same basis as wi. So w = Norm[wA]w1 + Norm[wB]w2 + ?+Norm[wM]wm where wA,?,wM form the components to w.

Now if we break the projection formula into components we have:
(<v, wA>/ (Norm[wA])^2) wA+?+(<v, wM>/ (Norm[wM])^2) wM. From above, wA=Norm[wA]w1.

If we make that subsitution:
(<v, Norm[wA]w1>/ (Norm[wA])^2) Norm[wA]w1. Since Norm[wA] is a scalar, it can be moved out, and all the Norm expressions cancel out and we get:
<(<v, w1>w1 +?+<v,wm>wm), wi> as in the 2nd line.

This is the same as <(<v, wi>wi), wi>. Because <v, wi> is a scalar it can be moved out and we get <y, wi> x <wi, wi> as in the third line. Because each basis vector has a magnitude of 1, <v, wi> x <wi, wi> = <v, wi>

Representing Duke Math. Feel free to ask for clarification; it's hard to type math notation.

 

[TecHNooB]

Originally posted by: blinky8225
I?m going to use my notation because yours really messes with me.
Okay I?ll focus on how <w, wi> becomes <v, wi> x <wi, wi>.

So we have <w, wi> and u = v-w -> v = u + w
So, w can be written as the projection of v onto w. From the projection formula, we obtain (<v, w>/ (Norm[w])^2) w.

Now since, w is in W and S forms a basis for W, w can be written with the same basis as wi. So w = Norm[wA]w1 + Norm[wB]w2 + ?+Norm[wM]wm where wA,?,wM form the components to w.

Now if we break the projection formula into components we have:
(<v, wA>/ (Norm[wA])^2) wA+?+(<v, wM>/ (Norm[wM])^2) wM. From above, wA=Norm[wA]w1.

If we make that subsitution:
(<v, Norm[wA]w1>/ (Norm[wA])^2) Norm[wA]w1. Since Norm[wA] is a scalar, it can be moved out, and all the Norm expressions cancel out and we get:
<(<v, w1>w1 +?+<v,wm>wm), wi> as in the 2nd line.

This is the same as <(<v, wi>wi), wi>. Because <v, wi> is a scalar it can be moved out and we get <y, wi> x <wi, wi> as in the third line. Because each basis vector has a magnitude of 1, <v, wi> x <wi, wi> = <v, wi>

Representing Duke Math. Feel free to ask for clarification; it's hard to type math notation.

Thanks, I will go over this in a bit.