# Matrix of linear transformation (range of transformation)

I'm doing a bit of homework and want to check if I'm doing it right. The question is as follows:

Let $T: P_2(\mathbb{R}) \rightarrow P_3(\mathbb{R})$ be defined by $T(p)(x) = (x-1)p(x)$

We're then asked to compute the matrix of $T$ with respect to the standard basis $\{1, x, x^2\}$. I go through that and get:

$A = \begin{bmatrix}-1&0&0\\ 1&-1&0\\0&1&-1\\0&0&1\end{bmatrix}$

Is this correct? I think it is but I'm not really sure.

However my main question is regarding $range T$. How would one describe the range of $T$ without using the matrix? I tried doing it like

Let $p \in P_3({\mathbb{R})}$ then

$T(p)(x) = (x-1)(a + bx + cx^2)$ for $a,b,c, \in \mathbb{R}$

$T(p)(x) = ax + bx^2 + cx^3 -a -bx -cx^2 = (a-b)x + (b-c)x^2 + cx^3 - a$

So, the range of T would be $\{ (a-b)x + (b-c)x^2 + cx^3 - a : a,b,c \in \mathbb{R}\}$?

-

Yes, your matrix is correct. Remember that $x-1$ is a factor of a polynomial $p(x)$ if and only if $p(1)=0$, so you can describe the range of $T$ as the set of cubic polynomials $p(x)$ such that $p(1)=0$. Now notice that if $p(x)=ax^3+bx^2+cx+d$, then $p(1)=a+b+c+d$, so $p(1)=0$ if and only if $a+b+c+d=0$: the range of $T$ consists precisely of the cubic polynomials whose coefficients sum to $0$.
I was going over this question again and was trying to deduce a basis for rangeT. Would a basis simply by $\{(-1,0,0), (0,-1,0), (0,0,-1)\}$ from $a = -b -c -d$? – user1520427 Sep 26 '12 at 8:43
@user1520427: The range of $T$ is a subspace of $P_3(\Bbb R)$, so the elements of any basis have to be polynomials, not tuples. One basis that works, based on what I think is your underlying idea, is $\{x^3-x^2,x^3-x,x^3-1\}$. Certainly these are all in the range of $T$; if you show that they’re linearly independent, you’ll know that they’re a basis. I’d choose $\{x^3-1,x^2-1,x-1\}$, though: it looks just a little nicer to work with, though there’s really not much difference. – Brian M. Scott Sep 26 '12 at 8:55