Question

anyone got any information on kernel and range...

Answer 1

Groups? :)

Answer 2

T: r2 → r2 is the contraction of T(x) = ½ x is that what you mean

Answer 3

Linear Transformation? :/

Answer 4

yes...find the kernel and range of T...which transformations if any are one-to-one? or onto?

Answer 5

Kernel, is the subset of r2 which are sent to 0, in the transformation.

Answer 6

so ker(T) = (0, 0) and ran (T) = R2? is that the answer...

Answer 7

@PeterPan why are you helping me?

Answer 8

I like to help? I don't really know :)
And yes, it seems like you have the answer. I don't really know what R2 is. Maybe you could enlighten me?

Answer 9

\[R\]\[^{2}\]

Answer 10

a 2 dimensional space

Answer 11

Oh. Okay. It seems like you got it right :)
You might have to prove that, though :P

Answer 12

Linear algebra online is tough...

Answer 13

Why is that? :)

Answer 14

T: r3 → r3 is the rotation about the z axis through an angle of Pi/4 .... @peterpan can you answer that one?

Answer 15

@peterpan it is tough because I feel like I need more feedback...so I gotta figure alot out without knowing what I am doing for sure...

Answer 16

You teach Linear Algebra? :/

Answer 17

not yet...I teach regular math...ged prep...basic...

Answer 18

Linear Transformations can be represented with matrices :)

Answer 19

yes...I know...but I need to show the kernel and the range of each linear transformation...that is what I am not sure i understand...I am sure someone could explain it to me...if I was sitting next to them...and I could ask questions ...but sitting here by myself makes it a bit difficult...that is why I came to this site...I am in my last week of this class...working on second to last assignment before final...

Answer 20

I probably already understand it but just don't know it yet...16 weeks crammed into 6 through phoenix...

Answer 21

Well, then, say we're only working in 2d, or just R2, we have what's called the matrix of rotation by an angle θ.

If we have the coordinates (x,y), and we wish to rotate them θ radians with respect to the origin, we do this:

\[\huge \left[\begin{matrix}x' \\ y'\end{matrix}\right]=\left[\begin{matrix}\cos \ \theta & -\sin \ \theta \\ \sin \ \theta & \cos \ \theta \end{matrix}\right]\left[\begin{matrix}x \\ y\end{matrix}\right]\]

Answer 22

right...I got that...I know how to make the matrices...and I think it is cool...but My assigment asks me to express what the kernal is of that transformation and the range also...or the transformation I asked a little earlier in 3d

T: r3 → r3 is the rotation about the z axis through an angle of Pi/4

ker(T) =??? ran(T)=???

I hope my question makes sense?

Answer 23

It does. Now, the rotation is about the z axis, right? So if you rotate it, the z value won't change at all... right?

Answer 24

right...

Answer 25

So, actually, the Linear transformation should be like this :
\[\huge \left[\begin{matrix}x' \\ y' \\ z'\end{matrix}\right]=\left[\begin{matrix}\cos \ \theta & -\sin \ \theta & 0\\ \sin \ \theta & \cos \ \theta & 0 \\ 0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}x \\ y \\ z\end{matrix}\right]\]

Answer 26

yes...so How do I take that...and turn it into

ker(T) = {(x,y,1)-∞ < x,y < ∞} ?

Answer 27

Is that correct?

Answer 28

I don't get it, the kernel of this transformation is just the set containing (0,0,0), unless I'm mistaken, of course...

Answer 29

and is it a 1 to 1 or an onto transformation?

Answer 30

You know determinants? :)

Answer 31

@PeterPan yes... ker(T) = (0, 0, 0) that is what my book says...but I don't understand kernel I guess...sorry...I must sound like an idiot...so...what I think is the kernal(T) is what ever values for (x,y,z) that will make (x,y,z) = (0,0,0)

Is that correct?

I really want to have a grasp of kernel...not just stick in an answer...but I want to know why that answer is best...

and No... T: r3 → r3 is the rotation about the z axis through an angle of Pi/4 is all they gave me...no determinants...

Answer 32

and is the ran(T) = R3?

Answer 33

It really helps if you have a mental picture of what's really going on :)
Let's look at the rotation.


Say we want to rotate this point.

Answer 34

pi/4, right? or 45 degrees.

It would go somewhere around there.

Answer 35

Now, when you really think about it, there is no point that you could rotate in this manner that would lead to the origin, (0,0,0) except for the origin itself, (0,0,0) because it does not really undergo rotation :D