For this question, I am asked to find a matrix in standard coordinates for the linear transformation below. My problem is that I can't seem to set it up properly.
$T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ is a linear transformation with $T(0,0,-1)=(1,-1,1)$ and $Null([T])=\{(x,y,z)|2x+y-z=0\}$.
My understanding was that the matrix would simply be the following:
$$[T]=\left[\begin{array}{rrrr}
2 & 1 & -1 \\
2 & 1 & 1 \\
2 & 1 & -1 \\
\end{array}\right]$$
Because when I multiply $$[T]=\left[\begin{array}{rrrr}
2 & 1 & -1 \\
2 & 1 & 1 \\
2 & 1 & -1 \\
\end{array}\right]$$ and $$\left[\begin{array}{r}
0 \\
0 \\
-1 \\
\end{array}\right]$$
I should get $$\left[\begin{array}{r}
1 \\
-1 \\
1 \\
\end{array}\right]$$
Am I even going about this the right way? If not, can someone suggest where I might be going wrong and help me out? Thanks.
Best Answer
The equation of the null space, $2x+y−z=0$, can be written $z= 2x+ y$ so that all vectors in that null space can be written as $$<x, y, 2x+ y>= <x, 0, 2x>+ <0, y, y>= x<1, 0, 2>+ y<0, 1, 1>$$
Writing the matrix $\begin{bmatrix}a & b & c \\ d & e & f\\ g & h & i\end{bmatrix}$ we must have $\begin{bmatrix}a & b & c \\ d & e & f\\ g & h & i\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix}= \begin{bmatrix}a+ 2c \\ d+ 2f \\ g+ 2i\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$ so $a+ 2c= 0$, $d+ 2f= 0$, and $g+ 2i= 0$.
We must have $\begin{bmatrix}a & b & c \\ d & e & f\\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}b+ c \\ e+ f \\ h+ i\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ so $b+ c= 0$, $e+ f= 0$, and $h+ i= 0$.
Finally, since $T(0, 0, -1)= (1, -1, 1)$ we must have $\begin{bmatrix}a & b & c \\ d & e & f\\ g & h & i\end{bmatrix}\begin{bmatrix}0 \\ 0 \\ -1\end{bmatrix}= \begin{bmatrix}-c \\ -f \\ -i\end{bmatrix}= \begin{bmatrix}1 \\ -1 \\ 1\end{bmatrix}$ so we must have $-c= 1$, $-f= -1$, and $-i= 1$.
So solve the 9 linear equations, $a+ 2c= 0$, $d+ 2f= 0$, $g+ 2i= 0$, $b+ c= 0$, $e+ f= 0$, $h+ i= 0$, $-c= 1$, $-f= -1$, and $-i= 1$ for the $9$ unknown values $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$, and $i$.