$$A =\begin{pmatrix} 1&4& 7& 10\\
2 &5 &8 &11\\
3 &6 &9 &12 \end{pmatrix}$$
is $T(x) = Ax$ onto? Is $T(x)$ one-to-one?
I know it's onto if there's a pivot position in every row, and 1-to-1 if no free variables, but I'm confused when it's saying if $T(x)=Ax$ is onto/1-to-1.
Best Answer
A basis for the column space of matrix $A$ is the first $2$ columns. Since the rank is $2$, it cannot be onto.
Alternatively, notice that the third row is $2$ times the second row minus the first row. Hence, there i no pivot position in every row when you reduce it to reduced row echelon form.
It is also not one to one. As you reduce the problem to reduced row echelon form, there can be at most $3$ pivot columns (there is $2$ in this case), you will have a free variable.