I am going over a solution of a problem in a Linear Algebra book by Hoffman and Kunze. Here is it:
Now I understand how the basis vectors for the null space were calculated, but I am a little puzzled by the way the basis for the range of $T$ are being calculated here. What I would do is the row-reduce the matrix, say $A$, to get $$R_1=(1,0,-1),R_2=(0,1,1)\text{ and } R_3=(0,0,0)$$ where $R_1,R_2$ and $R_3$ are the rows obtained by reducing $A$ (as done while calculating the null space). Then I would conclude that the column vectors $(1,0,-1)$ and $(2,1,3)$ are the basis vectors of the column space and hence they are the basis vectors for the range of $T.$ Is this correct or am I missing something?
You are basically correct.
I'm not sure why they transpose the matrix, because you can do both computations starting from the given matrix.
Gaussian elimination proceeds as follows: \begin{align} \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ -1 & 3 & 4 \end{bmatrix} &\to \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 5 & 5 \end{bmatrix} && R_3\gets R_3+R_1 \\ &\to \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} && R_3\gets R_3-5R_2 \tag{*} \\ &\to \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} && R_1\gets R_1-2R_2 \tag{**} \end{align} The step marked (*) tells us that the first two columns of every matrix in the display have their column space generated by columns $1$ and $2$, which are linearly independent. So a basis for the column space is $$ \left\{\, \begin{bmatrix}1\\0\\-1\end{bmatrix}, \begin{bmatrix}2\\1\\3\end{bmatrix} \,\right\} $$ The relevant fact here is that elementary row operations don't change the linear relations between the columns.
The step marked (**) tells us that a nonzero vector in the null space is $$ \begin{bmatrix}1\\-1\\1\end{bmatrix} $$ (just solve the system $x_1=x_3$, $x_2=-x_3$), which therefore forms a basis for the null space.