Linear Algebra; Transformation of cross product

Myr73
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Pre-knowledge
A matrix is a linear transformation if, T(u+v)= T(u) +T(v) and T(cu)=cT(u).

Theorem 8.4.2 If V is a finnite dimensional vector space, and T: V-> V is a linear operator then the following are equivalent.
a) T is one to one, b) ker(T)=0, c) nuttility(T)=0 and d) the Range of T is V; R(T)=V

1- Question
Let V be a fixed vector in R^3. a)Show that the transformation defined by T( u)= v X u is a linear transformation.
b) Find the range ot the linear transformation
c) If v=i , find the matrix for this linear transformation.

2- Answer
a) I proved that T(u+v)= T(u) +T(v) and T(cu)=cT(u), through cross product properties, And therefore proved its linear transformation.

b) This one i am not entirely sure, however with the Theroem 8.4.2 d) I concluded that R(T) = R^3

c) This one is where I am unsure where to begin. Since T(u) = v X u

Then T(u) = { Det ( v2, v3 - Det ( v1, v3 Det ( v1, v2
u2, u3 ) , u1, u3) , u1, u2) }

I am not sure where to go from there, and unsure what they mean by v=i

Thank you,
 
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To answer part(c), just plug in corresponding unit basis vectors into the linear transformation formula, and you will obtain each column of the matrix.
 
I am not sure what you mean.
 
Linear transformation can be written into a matrix form. When you times a matrix with a unit basis vector, if the unit basis vector has a 1 as its first element and 0s as its other elements, then you obtain the first column of the matrix as a result. Keep doing this you can calculate all columns of the corresponding matrix, thus obtain the matrix itself.
c71eb1eb2ef2fed843a0e7dc8e5c9bf3.png
 
ok, so v= i means i=( 1
0
0) ?

But then I don't know what u is . So would I not just have u= u1
u2
u3 ?

And then T(u) = v X u = (0, -1u3, 1u2).
 
No, here i is just an example. I couldn't tell what i is from your information. Is this the whole problem?
 
yes, that is all the information I am given ?
 
Myr73 said:
A matrix is a linear transformation if, T(u+v)= T(u) +T(v) and T(cu)=cT(u).
I think you meant "function" or "map" here, not "matrix". If T is a matrix, then the map that takes x to Tx is always linear.

Myr73 said:
Theorem 8.4.2 If V is a finnite dimensional vector space, and T: V-> V is a linear operator then the following are equivalent.
a) T is one to one, b) ker(T)=0, c) nuttility(T)=0 and d) the Range of T is V; R(T)=V
nuttility?

Myr73 said:
b) This one i am not entirely sure, however with the Theroem 8.4.2 d) I concluded that R(T) = R^3
This is wrong. Is there something you can say about how T(u) is related to u?

Myr73 said:
c) This one is where I am unsure where to begin.
You will need to learn how the matrix of a linear transformation is defined. It's described in one of our FAQ posts: https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/
 
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sorry , for theroem 8.4.2 c) I meant, nullity.

If I prove that a,b or c is true, then i can state that d is true also, right?
 
  • #10
If A is an mXn matrix and TA:Rn-->Rm is multiplication by A then, nullity(TA)= nullity(A) and the rank(TA)=rank(A)
 
  • #11
Myr73 said:
sorry , for theroem 8.4.2 c) I meant, nullity.

If I prove that a,b or c is true, then i can state that d is true also, right?
Correct. But if you think about what the cross product does, you should find it easy to see that none of the statements is true for the specific T you're working with.
 
  • #12
Ok, so the range of T is not R^3? And the nullity and kernel is not zero? I had "figured" out that the ker(T)=0, but I must be miss understanding the kernel concept.
 
  • #13
The kernel is the set of vectors that are mapped to the 0 vector, i.e. the set of all x such that Tx=0. You should have no difficulty finding a non-zero vector that's mapped to 0 by this T.
 
  • #14
Oh ok, you makes sense, thanks :).. So how would I determine what the kernel and the range is here then ?
 
  • #15
Are you familiar with the geometric interpretation of the cross product? The right-hand rule and all that? If yes, then you should be able to correctly guess what the kernel and range are. Then you can try to prove that your guesses are correct.
 
  • #16
Ok, So i'll do this one at a time.

I am kinda familiar with the geometric interpration of the cross product, the right hand rule. I also just revisited it.

In order to find the Kernel I need to find when does vXu=0, correct?

If that's correct then this is the case when; vector u, equals zero; vector v equals zero; when vector v is in the exact opposite direction of u; and also when vector v is in the exact same direction of u.

What I understand is that uXv=0 when the angle between them is 0. Is this correct?

If so how would I write that?
 
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  • #17
Myr73 said:
Ok, So i'll do this one at a time.

I am kinda familiar with the geometric interpration of the cross product, the right hand rule. I also just revisited it.

In order to find the Kernel I need to find when does uXv=0, correct?

If that's correct then this is the case when; vector u, equals zero; vector v equals zero; when vector v is in the exact opposite direction of u; and also when vector v is in the exact same direction of u.

What I understand is that uXv=0 when the angle between them is 0. Is this correct?

If so how would I write that?

So your conjecture is that ##\mathbf{u}\times \mathbf{v} =\mathbf{0}## iff ##\mathbf{u} = t\mathbf{v}## for some ##t\in \mathbb{R}##. This would make the kernel ##\{t\mathbf{v}~\vert~t\in \mathbb{R}\}##. So now try to prove this. This involves two steps:
1) Each ##t\mathbf{v}## is in the kernel. I.e. for each ##t##, we have ##(t\mathbf{v})\times \mathbf{v} =\mathbf{0}##.
2) There are no other vectors in the kernel. I.e. if ##\mathbf{u}\times\mathbf{v} = \mathbf{0}##, then ##\mathbf{u} = t\mathbf{v}## for some ##t\in \mathbb{R}##.
 
  • #18
Does this make sense?
What I am understanding the kernel of T is always a subset of V. And it would be either the zero subspace,the entire vector space or a line through the origin. Ane from my understanding that I explained above, I conclude that,
The Kernel of T will be the line through the origin spanned by V? and so, Ker(T)=Span(V)

And the Range of T has to be a subspace of the new space. And the new space is an orthogonal projection of V.
Range would therefore be a plane through the origin orthogonal to v. and so R(T)= Span( VxU) ??
 
  • #19
Myr73 said:
The Kernel of T will be the line through the origin spanned by V? and so, Ker(T)=Span(V)

Yes, but what if ##\mathbf{v} = \mathbf{0}##? Is it still true then?

Range would therefore be a plane through the origin orthogonal to v.

A plane? Is there more than one such plane?

and so R(T)= Span( VxU) ??

That doesn't really make much sense. The vector ##\mathbf{u}## is the variable in the function. So I don't know what ##\text{Span}(\mathbf{v}\times \mathbf{u})## means. Whatever it means, it's not correct since ##\text{Span}(\mathbf{v}\times \mathbf{u})## is 1-dimensional, while you just said that the range is a plane, which would make it two-dimensional.
 
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  • #20
Ok- Can someone help me find what the kernel and the range of T is?
 
  • #21
Your attempt in your previous post was very good, don't give up now! Try to think about the remarks I gave you.
 
  • #22
ok...I am really confused!
I thought the line through the origin would include when v=0.
And for the second part, I really don't know what to do, I understand what you mean by span (vXu) would be a line and not a plane. But I don't know if what I said about the plane was correct and how I would express that.
 
  • #23
Myr73 said:
I thought the line through the origin would include when v=0.

Does it? It's not unusual to get separate cases for extreme situations.

. But I don't know if what I said about the plane was correct

What you said about the plane is certainly correct. The plane orthogonal to ##\mathbb{v}## is the range of ##T## (but again, is this correct if ##\mathbf{v}= \mathbf{0}##)?
 
  • #24
ok...ummm.. I don't know.. I'm not even sure where to go from there lol
 
  • #25
any chance you can you give me little more guidance? :/
 
  • #26
Where exactly are you stuck?

You will have to consider two cases separately: ##v=0## and ##v\neq 0##. The ##v=0## case is very simple. Can you find the kernel and range in that case?

Suppose that ##v\neq 0##. What you need to prove is that the kernel is the 1-dimensional subspace that contains ##v##, i.e. ##\ker T=\{tv|t\in\mathbb R\}##, and that the range is a plane that's orthogonal to ##v##. Which one of those planes is it? (Hint: As you should be able to verify for yourself, the range of a linear operator is always a subspace).
 
  • #27
for T(u)=vXu

when v=0, the kernel of T is all of u , because vXu=0 with any u, if is always zero. the range would be zero.

when v does not equal 0, the kernel will be a line through the origin spanned by V. ... the range would be a plane that is orthogonal to U..
 
  • #28
Myr73 said:
the range would be a plane that is orthogonal to U..

Orthogonal to V you mean?
Otherwise, that is correct.
 
  • #29
oh, yes, to v. Thank you :)! is that how I would finish it , or is there a way to write it out? R(T)=
 
  • #30
You might write "plane orthogonal to ##\mathbf{v}##" in terms of the dot product, if you have seen that.
 
  • #31
oh ok.. so like ( R(T)=v•u | v does not equal zero) ?
 
  • #32
What is ##\mathbf{u}## here?
 
  • #33
a vector
 
  • #34
is tha right?
 
  • #35
It's not. The notation {x|some statement about x} means "the set of all x such that the statement is true".
 
  • #36
oh ok, so how would I write out when v does not equal zero,that the range of T is the plane that is orthogonal to v
 
  • #37
First of all, there's more than one such plane. Do you know which one of them is the range of T?
 
  • #38
v• (vXu)=0 , v is orthogonal to vXu
 
  • #39
Yes, v is orthogonal to v×u for all u. But that doesn't answer the question. Which one of the infinitely many planes that are orthogonal to v is the range of T? You can answer this one in words.

By the way, if you click the Σ symbol in the green bar at the top of the editor area, you will see a list of symbols that you can use when you type. No need to type X instead of × for example.
 
  • #40
Oh, ok,cool thanks. I will use the symbol bars..
And , I do not know.. I really do not know.
 
  • #41
Are you aware that the range of a linear operator is a subspace? You should verify that for yourself, if you haven't done it recently. It's a nice exercise. Now, of all the planes that are orthogonal to v, which ones are subspaces and which ones are not?
 
  • #42
yes, I am aware of that. and I know that to find a subspace of R3 that is orthogonal to v, we would find the orthogonal projection of R3 on W. where w is a subspace of R3. witch would be the Ax, A being the column space of given vectors of v, and x being the least squares solution of Ax=v.
 
  • #43
So you know that

(1) The range of T is a plane orthogonal to v.
(2) The range of T is a subspace of V.

Now the question is, which of the planes that are orthogonal to v are subspaces? Think about what a subspace is.
 
  • #44
Ok, well i have answered a and b to the best of my knowledge and ability, is there someone that can help me on part c)?
Here is the question again.
1- Question
Let V be a fixed vector in R^3. a)Show that the transformation defined by T( u)= v X u is a linear transformation.
b) Find the range ot the linear transformation
c) If v=i , find the matrix for this linear transformation.

2-I do not know u in this equation, I figured to just take it as u=(u1,u2,u3), I know that v would equal v=(1,0,0). And that T(U)= v×u={v2u3-v3u2,v3u1-v1u3,v1u2-v2u1}, and that
[T]=[T(e1)|T(e2)|T(e3)].
From the first formula, I found it to be ( 0,-u3,u2). However I do not think it is correct.

Thank you
 
  • #45
Myr73 said:
2-I do not know u in this equation,
What equation? If you mean the definition of T, that's not a statement about a vector called u. It's a statement about all vectors.

Myr73 said:
And that T(U)= v×u={v2u3-v3u2,v3u1-v1u3,v1u2-v2u1}
Just some advice about the notation: Don't type U when you mean u. These are two different symbols, so they don't automatically represent the same thing.

Myr73 said:
[T]=[T(e1)|T(e2)|T(e3)].
From the first formula, I found it to be ( 0,-u3,u2). However I do not think it is correct.
It looks like you ignored the formula that tells you what [T] is, and instead calculated T(u) for an arbitrary u.
 
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  • #46
yes, I am sorry , a couple of minor errors there. I meant "u" not "U".

And yes I ignored the formula with [T], because I did not know what to do with it. But I thought they wanted to know if V=i then find T(u), but I guess not. How am I suppose to compute [T] if v=(1,0,0)?
 
  • #47
The formula tells you that ##Te_1##, ##Te_2## and ##Te_3## are the columns of [T]. You should probably take a look at the FAQ post I linked to early in this thread.
 
  • #48
Does [T]= 1 0 0
0 0 0
0 0 0 , make sense? And I will re-read it, I do not really understand much of it, but I will reread it.
 
  • #49
It's not the right answer. For example, we have ##Te_1=e_1\times e_1=0##.

If you ask about the things you find difficult to understand in the FAQ post, I will answer.
 
  • #50
So Fredrik already gave you ##Te_1##. What is ##Te_2## and ##Te_3##?
 
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