Calculating Null T and Range T for the linear transformation $T(x,y,z)=(x+2y-z,y+z,x+y-2z)$?

Question

I have a doubt regarding the calculation of range of a linear transformation. I will explain my doubt with an example.

Suppose, $T:R^3 \to R^3 \ni$

$T(x,y,z)=(x+2y-z,y+z,x+y-2z)$ is a Linear Transformation.

Null $T$ will be the null space of the coefficient matrix obtained from $T(x,y,z)=0$, i.e,

$x+2y-z=0$

$y+z=0$

$x+y-2z=0$

$$ \begin{bmatrix} 1&2&-1\\0&1&1\\1&1&-2 \end{bmatrix}--(1)$$

that reduces to:

$$ \begin{bmatrix} 1&0&-3\\0&1&1\\0&0&0 \end{bmatrix}--(2)$$

Therefore,

Null $T$ = Span{$(3,-1,1)$}

we know Range $T$ $=$ span$\{T(e1),T(e2),T(e3)\}$ $=$ span$\{(1,0,1),(2,1,1),(-1,1,-2)\}$

Again we get the same matrix (1):

$$ \begin{bmatrix} 1&2&-1\\0&1&1\\1&1&-2 \end{bmatrix}$$

and from $(2)$, the first and second column contain the pivots so the first and second column of $(1)$ are Linearly Independent.

Therefore Range T = span{$(1,0,1),(2,1,1)$}

However, the textbook that I am following has solved for Range T differently, what it has done is, taken the transpose of $(1)$ and found its rows space instead,

$$ \begin{bmatrix} 1&0&1\\2&1&1\\-1&1&-2 \end{bmatrix}--(3)$$

that reduces to

$$ \begin{bmatrix} 1&0&1\\0&1&-1\\0&0&0 \end{bmatrix}--(4)$$

therefore Range T = span{$(1,0,1),(0,1,-1)$}

Now, I know that the row space of $A^{T}$ will be the column space of $A$, but why the extra effort to take the transpose and reduce it when we can simply get the answer from matrix $(1)$. Or is it the case that my way of finding range $T$ is wrong?

Answer 1

Notice that $2(1,0,1)+(0,1,-1)=(2,1,1)$, hence both spans are equal.

Answer 2

Both ways are correct. Transposing the coefficient matrix and row-reducing, which is equivalent to column-reducing the original matrix, gives you a “nicer” basis in the sense that the first $\operatorname{rank}T$ elements of each vector consist of zeros and a single $1$.