Computing Matrix, finding kernel and image

[specialnlovin]

Let T: R[x]₂\rightarrow R[x]₃ be defined by T(P(x))=xP(x). Compute the matrix of x with respect to bases {1,x,x²} and {1,x,x²,x³}. Find the kernel and image of T.

I know how to do this when given bases without exponents, however I do not know exactly what this is saying and therefore am having a hard time starting it.

 

[micromass]

Let's compute the first column.

You take the first basis element, that is 1. Now you'll have to express T(1)=x in terms of the basis {1,x,x²,x³}. That would be (0,1,0,0). So the first column would consist out of

\left(\begin{array}{ccc}<br /> 0 &amp; ? &amp; ? \\<br /> 1 &amp; ? &amp; ? \\<br /> 0 &amp; ? &amp; ? \\<br /> 0 &amp; ? &amp; ? <br /> \end{array} \right)

Now for the second and third column, you'll have to express T(x) and T(x²) in terms of the basis {1,x,x²,x³}.

 

[specialnlovin]

Okay so matrix T(x)=x² and with respect to the basis {1,x,x²,x³} the second column would be (0,0,1,0), T(x²)=x³ and with respect to the basis {1,x,x²,x³} would be (0,0,0,1).
I would then say that
T=(0 0 0)
(1 0 0)
(0 1 0)
(0 0 1)
with respect to {1,x,x²,x³} right? Not with respect to {1,x,x²} and {1,x,x²,x³}
so then to find the im(T) I just use
(0 0 0)
(1 0 0)
(0 1 0)
(0 0 1)
and solve it with respect to the basis {1,x,x²,x³} right?
Then how would I go about solving for the ker(T)?

 

[micromass]

For the Ker(T). Take your matrix A. Then wonder when Ax=0. This you should be able to solve easily (it's a system with 3 unknowns and 4 equations)

 

[micromass]

So, if (x,y,z) is in the kernel, then you must have

<br /> \left( \begin{array}{ccc}<br /> 0 &amp; 0 &amp; 0\\<br /> 1 &amp; 0 &amp; 0\\<br /> 0 &amp; 1 &amp; 0\\<br /> 0 &amp; 0 &amp; 1<br /> \end{array} \right) \left( \begin{array}{c}<br /> x\\<br /> y\\<br /> z<br /> \end{array} \right) = 0<br />

It's not hard to see that this can only be the case iff x=y=z=0

 

[specialnlovin]

right, that just seemed way too easy I thought I was doing it wrong.
Then the im(T) is the span of e₂, e₃, and e₄

 

[micromass]

Yes, your image is correct to.

And remember, math doesn't have to be difficult