I am having troubles with one of my homework exercises.
The problem states:
Prove that T is a linear transformation, and find bases for both N(T) and R(T). Then compute the nullity and rank of T, and verify the dimension theorem. Finally, use the appropriate theorems in this section to determine whether T is one-to-one or onto.
$T:M$$2\times3$$(F) \to M$$2\times2$$(F)$ defined by
$T\begin{bmatrix}a&b&c\\d&e&f\end{bmatrix} =\begin{bmatrix}2a-b&c+2b\\0&0\end{bmatrix}.$
Showing that this is a linear transformation is where I'm getting hung up. Any help would be appreciated.
Thanks.
Let $\phi : M_{2 \times 3}(F) \to F \,\,; \,\, \left[ \begin{matrix} a & b & c \\ d & e & f \end{matrix} \right] \mapsto 2a - b$
Let $\psi : M_{2 \times 3}(F) \to F \,\,; \,\, \left[ \begin{matrix} a & b & c \\ d & e & f \end{matrix} \right] \mapsto c + 2b$
Let $\theta: M_{2 \times 3}(F) \to F \,\,; \,\, \left[ \begin{matrix} a & b & c \\ d & e & f \end{matrix} \right] \mapsto 0$
Can you see why $\phi,\psi,\theta$ are linear transformations? You can use this to show that
$T : M_{2 \times 3}(F) \to M_{2 \times 2}(F) \,\,; \,\, A := \left[ \begin{matrix} a & b & c \\ d & e & f \end{matrix} \right] \mapsto \left[ \begin{matrix} \phi(A) & \psi(A) \\ \theta(A) & \theta(A) \end{matrix} \right]$
is linear as follows: for any scalar $\lambda \in F$ and any matrices $A,B \in M_{2 \times 3}(F)$ we find
$T(\lambda A + B) = \left[ \begin{matrix} \phi(\lambda A + B) & \psi(\lambda A + B) \\ \theta(\lambda A + B) & \theta(\lambda A + B) \end{matrix} \right]$
$= \left[ \begin{matrix} \lambda\phi (A) + \phi(B) & \lambda \psi(A) + \psi(B) \\ \lambda\theta( A) + \theta(B) & \lambda\theta(A) + \theta(B) \end{matrix} \right]$
$ = \lambda \left[ \begin{matrix} \phi (A) & \psi(A) \\ \theta( A) & \theta(A) \end{matrix} \right] = \left[ \begin{matrix} \phi (B) & \psi(B) \\ \theta( B) & \theta(B) \end{matrix} \right] = \lambda T(A) + T(B)$,
using the linearity of $\phi,\psi,\theta$ and the definition of the vector space operations of a space of matrices.