Linear Transformations help?

MoogPoo

So far in my course we've been learning only abstractly. I know formal definitions of kernel, eigenvalues, image and stuff but can you explain in detail how all this relates to my example.

Create a transformation for differentiating a cubic equatiion. We did this in class and we got the matrix but i'm a bit confused still. I know why it works and can use it for any example. But what is the kernel and image in this example? Also,how does integration fit in, I know that when you integrate it back you get the constant as many cubics will have the same differential. Does this have to do with the kernel? Also the basis we used was [1,x,x^2,x^3]. So could we also use a basis say [5, 2+x, -3x^2, x^3- x ] , despite it being harder? how would you use the change of basis matrix? is the dimKernel = 1 because of the constant in the cubic that goes away? And finally, if you want to integrate back i know you can't find the same cubic you started with as the constant would show up. But why is this in terms of kernel matrix?
how does invertibilty relate to all this....

Any help greatly appreciated.
Thanks a lot.

Fremen

The kernel is the set of all vectors which are mapped to 0.

The set of polynomials whose differential is 0 is precicely the constants. Thus, the kernel is the set of all (c,0,0,0). It's easy to see that this has dimension 1.

The image of the transform is the set of all vectors which have some vector mapped to them. You can't have a polynomial of degree 3 whose differential is also degree 3, so the image is the set of all polynomials of degree 2 or less.

Yes, you could have any other basis. The corresponding matrix would look different and everything would be more awkward, but it would work.
Try to repeat the example from clsss with the basis
{1, 2x ,x^2 ,x^3}
or the basis
{1, x -1, x^2, x^3}

In the second example, the polynomial
a + bx + cx^2 + dx^3
is represented as
(a+b, b, c , d)

The matrix for differentiation is not invertible. This is because the kernel has dimension 1. In other words, there's a whole subspace of vectors which are mapped to zero. You "lose information" when you apply the transform (you lose the information about what the constant term was).

Suppose you gave me a vector v in the image, and I wanted to invert the transform (i.e. un-differentiate, or integrate). I could easily find some vector V such that f(V) = v. However, this is not unique.
Let k be in the kernel (the set {c,0,0,0}). Then by linearity of f

f(V+k) = f(V)+f(k) = f(V) = v

so we have a whole family of vectors V being mapped to each vector v in the image. This corresponds to a load of functions which differ by a constant having the same differential.
V+k corresponds to the "integral" of v.

I have a feeling that was a bit incoherent, but hopefully some of it will be helpful.