$$T(x, y, z) = \left[ \begin{array}{ccc}
x – 2z \\
-3x + y + 3z \\
x – y + z\end{array} \right]$$
I've taken the matrix of coefficients:
$A=\begin{bmatrix} 1&0&-2 \\ -3&1&3 \\ 1&-1&1 \end{bmatrix}$
And reduced it to row-echelon form as follows:
$A=\begin{bmatrix} 1&0&2 \\ 0&1&3 \\ 0&0&0 \end{bmatrix}$
But, I'm not sure where to go from here in proving that this transformation is not one-to-one.
I understand that a one-to-one transformation has a pivot in every column, but I'm not sure what this means and how it applies to this matrix.
Where can I go from here in proving this?
Best Answer
You should, instead, obtain the reduced $A$ as:
$$A' = \begin{bmatrix} 1 &0& -2 \\ 0&1&-3\\0&0&0\end{bmatrix}$$
From the reduced $A$, we notice that a general solution to $A\vec x = \vec 0$ is given by:
$$\vec x = \begin{bmatrix} 2t\\3t \\t\end{bmatrix}$$
We also see that $A \begin{bmatrix} x\\y \\z\end{bmatrix} = T(x,y, z)$.
Hence we obtain $T(2t,3t, t) = \vec0$ for any $t$.
This shows that $T$ is not injective.
In fact, the null space is $\left\langle\begin{bmatrix} 2\\3 \\1\end{bmatrix}\right\rangle$.