Linear Algebra; Transformation of cross product

[Myr73]

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,

 

[eifphysics]

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.

 

[Myr73]

I am not sure what you mean.

 

[eifphysics]

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.

 

[Myr73]

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).

 

[eifphysics]

No, here i is just an example. I couldn't tell what i is from your information. Is this the whole problem?

 

[Myr73]

yes, that is all the information I am given ?

 

[Fredrik]

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.

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?

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?

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/

 

[Myr73]

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?

 

[Myr73]

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)

 

[Fredrik]

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.

 

[Myr73]

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.

 

[Fredrik]

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.

 

[Myr73]

Oh ok, you makes sense, thanks :).. So how would I determine what the kernel and the range is here then ?

 

[Fredrik]

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.

 

[Myr73]

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?

 

[micromass]

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 $t\in \mathbb{R}$.

 

[Myr73]

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) ??

 

[micromass]

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.

 

[Myr73]

Ok- Can someone help me find what the kernel and the range of T is?

 

[micromass]

Your attempt in your previous post was very good, don't give up now! Try to think about the remarks I gave you.

 

[Myr73]

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.

 

[micromass]

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 $\mathbf{v}= \mathbf{0}$)?

 

[Myr73]

ok...ummm.. I don't know.. I'm not even sure where to go from there lol

 

[Myr73]

any chance you can you give me little more guidance? :/

 

[Fredrik]

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).

 

[Myr73]

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..

 

[micromass]

the range would be a plane that is orthogonal to U..

Orthogonal to V you mean?
Otherwise, that is correct.

 

[Myr73]

oh, yes, to v. Thank you :)! is that how I would finish it , or is there a way to write it out? R(T)=

 

[micromass]

You might write "plane orthogonal to $\mathbf{v}$" in terms of the dot product, if you have seen that.

 

[Myr73]

oh ok.. so like ( R(T)=v•u | v does not equal zero) ?

 

[micromass]

What is $\mathbf{u}$ here?

 

[Myr73]

a vector

 

[Myr73]

is tha right?

 

[Fredrik]

It's not. The notation {x|some statement about x} means "the set of all x such that the statement is true".