Calculating Null T and Range T for the linear transformation $T(x,y,z)=(x+2y-z,y+z,x+y-2z)$?
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?
$\endgroup$2 Answers
$\begingroup$Notice that $2(1,0,1)+(0,1,-1)=(2,1,1)$, hence both spans are equal.
$\endgroup$ 2 $\begingroup$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 $1$.
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