I have a linear transformation $$T(x_1,x_2,x_3) = (x_1-x_2+2x_3,2x_1+x_2,-x_1-2x_2+2x_3) $$
Assuming an ordered basis $\{u_1,u_2,u_3\}$ I get the linear transformation of the basis vectors as below: $$T(u_1) = u_1+2u_2-u_3,\\ T(u_2) = -u_1+u_2-2u_3, \\T(u_3) = 2u_1+u_3\\$$ There are two matrix representations of the above transformation as below: $$[x_1,x_2,x_3]\left[ \begin{matrix} 1 & 2 & -1 \\ -1 & 1 & -2 \\ 2 & 0 & 2 \\ \end{matrix}\right] = [y_1,y_2,y_3] $$ and $$\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 0 \\ -1 & -2 & 2 \\ \end{matrix}\right]\left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ \end{matrix}\right] = \left[\begin{matrix} y_1 \\ y_2 \\ y_3 \\ \end{matrix}\right] $$
Here $y_1,y_2,y_3 $ element of Range(T). I believed both are same, however I am not able to determine certain relations from each. Example, row reducing 1 I get $$\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{matrix}\right]$$ which gives me the basis for R(T) and linear combination of these basis vectors yield $(x_1,x_2, x_1-x_2)$ i.e. I get the relationship between the coordinates and the augmented matrix gives the constraint on $T(u_1),T(u_2),T(u_3)$ as $-2T(u_1)+4T(u_2)+3T(u_3) = 0$, this relationship however I do not get from the second representation. $$\\$$ The second representation gives us relationship between coordinates and also gives us the Null space as $$\{(-a/2,a,(3/4)a):a e R\}$$ and I am not able to get the null space from the first representation.
Are there any differences between the two representations? And few relations that can be derived from one and not from another?
As per my Analysis both of them are same and can be used interchangeably. Putting a bit more effort I got the same results from both the representations. Working shown with non-standard representation.
Row Reducing $$\left[ \begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0\\ -1 & 1 & -2 & 0 & 1 & 0 \\ 2 & 0 & 2 & 0 & 0 & 1 \\ \end{array}\right]$$ we get $$\left[ \begin{array}{ccc|ccc} 1 & 0 & 1 & 1/3 & -2/3 & 0\\ 0 & 1 & -1 & 1/3 & 1/3 & 0 \\ 0 & 0 & 0 & -1/6 & 1/3 & 1/4 \\ \end{array}\right]$$
Above gives us that (1,0,1) and (0,1,-1) are the basis for Range Space and linear combination of these vectors give us the transformation as $T(x_1,x_2,x_3) = (x_1,x_2,x_1-x_2)$ also with $$[y_1,y_2,y_3]\left[ \begin{matrix} 1/3 & -2/3 & 0\\ 1/3 & 1/3 & 0 \\ -1/6 & 1/3 & 1/4 \\ \end{matrix}\right]$$ We get $$x_1 = \frac{y_1+y_2}{3} -y_3/6\\ x_2 = \frac{y_2-2y_1}{3} +y_3/3\\ x_3 = y_3/4$$
The Null space is also given by the third row as: $$\frac{-T(u_1)}{6}+\frac{T(u_2)}{3}+\frac{T(u_3)}{4} = 0\\ \implies T(\frac{-(u_1)}{6}+\frac{(u_2)}{3}+\frac{(u_3)}{4}) = 0\\ $$ Null space is spanned by vector $\frac{-(u_1)}{6}+\frac{(u_2)}{3}+\frac{(u_3)}{4}$ or linear combination of $\frac{-(u_1)}{2}+(u_2) +\frac{(3u_3)}{4}$ $$\therefore Ker(T) = \{(\frac{-\alpha}{2},\alpha,\frac{3\alpha}{4}):\alpha\ e\ R\}$$
Please correct me in case of any issue.