Find a basis for the range of linear operator and linear transformation. A)  Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the linear operator given by the formula $T(x,y) = (2x-y, -8x+4y)$. 
Find a basis for the range of the linear operator. 
B)  Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be the linear transformation given by the formula:
$T(x_1, x_2, x_3, x_4) = (4x_1+x_2-2x_3-3x_4, 2x_1+x_2+x_3-4x_4, 6x_1-9x_3+9x_4)$
Find a basis for the range of the linear transformation.
C)  Let $T:P_2 \to P_3$ be the linear transformation defined by $T(p(x)) = xp(x)$.
Find the basis for the range of the linear transformation.
*Update 7/16/13:
Working on part b:
I believe (correct me if I'm wrong) that the basis of the range of a linear transformation is just the column space of the linear transformation. If so, I should set the transformation up in a matrix and reduce to row echelon. Then, I think I'll use the columns in the reduced matrix that have pivots and correspond those columns to the original matrix - thus giving me my basis.
However, I set up the following matrix.
B=\begin{bmatrix}
4 & 1 & -2 & -3 \\
2 & 1 &  1 & -4 \\
6 & 0 & -9 &  9
\end{bmatrix}
When I reduced this, I got
B=\begin{bmatrix}
1 & 0 & -(3/2) & 0 \\
0 & 1 &  4 & 0 \\
0 & 0 & 0 &  1
\end{bmatrix}  
From this, I would think that the column one, two and four contain my pivots. Therefore, (going back to the original matrix) my basis would be the column vectors <4,2,6>, <1,1,0> and <-3,-4,9>. I'm afraid I've gone wrong somewhere. Can someone help me with this?
 A: In cases A and B, you can find the matrix of the linear transformation with respect to the canonical bases; in case A it is
$$
A=\begin{bmatrix}
 2 & -1 \\
-8 &  4
\end{bmatrix}
$$
and in case B it is
$$
B=\begin{bmatrix}
4 & 1 & -2 & -3 \\
2 & 1 &  1 & -4 \\
6 & 0 & -9 &  9
\end{bmatrix}
$$
In general, when you have a linear transformation $T\colon\mathbb{R}^n\to\mathbb{R}^m$ and $\{e_1,e_2,\dots,e_n\}$ is the canonical basis of $\mathbb{R}^n$, you just write down (as columns), the vectors $T(e_1), T(e_2), \dots, T(e_n)$. A basis for the range can easily be computed by Gaussian elimination.
For case C, you don't have a "canonical basis", but you still can compute the matrix associated to the bases $\{1,x,x^2\}$ of $P_2$ (assuming it's the space of polynomials having degree at most 2) and $\{1,x,x^2,x^3\}$ of $P_3$. Since $T(1)=x=0\cdot1+1x+0x^2+0x^3$, $T(x)=x^2$, $T(x^2)=x^3$, the matrix is
$$
C=\begin{bmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
$$
The rank of this matrix is? And what can you conclude from this?
A: Hints for you to sort and apply:
== If we have a linear map $\,T:V\to W\;,\;\;V,W\;$ finite dimensional vector spaces over the same field , then we have the dimension theorem:
$$\dim V=\dim\ker T+\dim\text{Im}\,T$$
== We clearly have $\;(-4)\cdot(2x,-y)=(-8x,4y)\implies\;$ the range of $\,T\,$ is $\,1-$dimensional
== In (B), what is the dimension of the range? Try reducing a coefficients matrix.
== In (C), the minimal degree of any non-zero element in the image is one...
