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For example: $T: P_2 \to P_5$ given by $T(p(x)) = x^3(p(x))$

How would I find the kernel and range for this?

All I know is that $\dim(\ker T)+ \dim(\operatorname{image}T)= \dim(p_2)$.

Therefore, $\dim(\ker T)+ \dim(\operatorname{image}T)=3$?

Also, how do I represent T by a matrix with respect to bases for the domain and codomain?

Any help at all is appreciated, as I really have absolutely no idea what I am doing.

Thanks in advance!

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1 Answer 1

For your specific problem, you can find by inspection that the kernel is just 0 (the only polynomial that gives 0 when multiplied by $x^3$ is just 0) and the image is all the 5th-degree polynomials with no quadratic, linear, or constant terms (as everything has just been multiplied by $x^3$ and there can be no terms with degree less than 2 because of that).

However, in general, we can proceed this way: your linear transformation can be represented as a matrix:

$\left[ \begin{matrix} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} \right]$

To find the kernel, row reduce this matrix as augmented with the identity, and look for all the zero rows: $\left[ \begin{array}{cccccc|ccc} 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right]$

Conveniently, it is already row-reduced. So the kernel has no basis and therefore is trivial. If it did have a basis, the basis vectors would be the entries on the right matrix that corresponded to zero rows on the left matrix (remember that the kernel is expressed in terms of the domain space, not the range space).

I believe that to find a basis of the image, you look for all the nonzero rows after row reduction, but I forgot that part. (It seems to work in this case, however, as $\{x^3, x^4, x^5\}$ is indeed a basis for the image.)

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Ok, I used another problem because I didn't want to cheat myself out of finding out how to do it. . . but the problem in my HW is T:P2- - >P4 given by T(p(x)) = x^2(P(x)) so I guess kerT is still 0 then. . . but I am still confused about how to find the range. . . is the basis of the image analogous to the range? –  So Confused Dec 17 '12 at 4:12
    
Oh, by "image" I mean "range". They're used interchangeably. –  Joe Z. Dec 17 '12 at 4:13
    
Oh okay, thank you! –  So Confused Dec 17 '12 at 4:14
    
You are correct in saying that the kernel of T is still 0 in your case. The range is slightly different - just shift the exponents in the answer above down by 1. –  Joe Z. Dec 17 '12 at 5:43

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