i'm having trouble answering this question.
$T:M_{2x3}(F) \mapsto M_{2x2}(F), T $$ \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ \end{pmatrix} $$ = $$ \begin{pmatrix} 2a_{11}-a_{12} & a_{13} + 2a_{12} \\ 0 & 0 \\ \end{pmatrix} $
for the range, i have $R(T)=$$ \{\begin{pmatrix} a & b\\ 0 & 0 \\ \end{pmatrix}|a,b \in R\} $ and the basis of $R(T)$ is $ \begin{pmatrix} 1 & 0\\ 0 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1\\ 0 & 0 \\ \end{pmatrix} $, so $dim(R(T)) = rank(T)=2$
but i am stuck on calculating nullity. this is my work so far.
i believe the null space $N(T)$ is $ \begin{pmatrix} 0 & 0 & 0 \\ a & b & c \\ \end{pmatrix} $ where $a,b,c \in R$. i believe this implies a basis of three elements, $ \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} $, $ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} $, $ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $, so $rank(T)=dim(N(T))=3$, but the answer in the back of the book says it's 4.
where have i gone wrong here? or may it be the case that there's an error in the book?
thanks in advance for any feedback.
Assuming the $a_2$ in the first entry is supposed to be $a_{12}$, you get an additional independent solution if you solve $2a_{11} - a_{12} = 0$ and $a_{13} + 2a_{12} = 0$, e.g. $a_{11} = 1$, $a_{12} = 2$, $a_{13} = -4$.