Linear Algebra - linear transformation- range of T Let $\ T:\Bbb R^3\rightarrow \Bbb R^3$ be the linear tranformation defined by
$\ T(a,b,c)=(2a-b,a+b+c,-a+c)$,
Find a basis for the Range (T).
I already solved the standard matrix $\ A=
$$ \left[    
  \begin{matrix}
    2 & -1 & 0 \\
    1 & 1 & 1 \\
    -1 & 0 & 1 \\
    \end{matrix}
\right] $$
$ and $ R_A=
$$ \left[    
  \begin{matrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    \end{matrix}
\right] $$
$ (if I was solving correctly). can $\ Range(T)=Col(A)=Col (R_A)$ in this case?
 A: You have a small typo in the first column of your $A$ matrix.
The range of $T$ is the span of the columns of $A$. If the columns are linearly independent, then you automatically have a basis for the range.
A: The basis for a set of vectors must be linearly independent. As you've stated we are searching for the basis of $Range(T)$. As you've noted $T$ can be given by $A$, that is to say $T:\overrightarrow{x} \mapsto A\overrightarrow{x}$. Using this notation, it is a bit clearer to see that $Range(T)=A\overrightarrow{x}$. Notice $A\overrightarrow{x} = x_1 \begin{bmatrix} 2 \\
1 \\
-1 \\\end {bmatrix}+x_2\begin{bmatrix} -1 \\
1 \\
0 \\\end {bmatrix}+x_3\begin{bmatrix} 0 \\
1 \\
1 \\\end {bmatrix}$. Setting this vector equation equal to $0$ will allow one to solve for linear independence. Since a set of vectors $\{v_0,v_1, \dots ,v_n \}$ is defined to be linearly independent if and only if the only solution to $c_0v_0+c_1v_1+\dots+c_nv_n=0$ is for all $c_i=0$. So, using an augmented matrix $B,$ $B=\begin{bmatrix} 2 & -1 &0 & 0\\
1 & 1 & 1 & 0\\
-1 & 0 & 1& 0\end{bmatrix}$. Using Gauss-Jordan reduction gives $\begin{bmatrix} 1 & 0 & 0 &0 \\ 
0&1&0&0 \\ 
0&0&1&0\end{bmatrix}$. Since these solutions are unique, the previously mentioned vector equation is in fact linearly independent. As a result it is a suitable basis for $Range(T)$. 
