Consider the vector space of 2-by-2 matrices $M_{2\times 2}(\mathbb{R})$ over $\mathbb{R}$. For a given matrix A=$\begin{pmatrix}a_{1,1} &a_{1,2}\\a_{2,1}&a_{2,2}\end{pmatrix}$, consider the map $T:M_{2\times 2}(\mathbb{R}) \to \mathbb{R}$ given by $T(A)=a_{2,1}$
$a)$ Show that T is a linear transformation.
Definition: Let $V$ and $W$ be vector spaces over $F$. We call a function $T : V \to W$ a linear
transformation from $V$ to $W$ if, for all $x, y \in V$ and $c \in F$, we have
$(1) T (x + y) = T (x) + T (y)$
$(2) T (cx) = cT (x)$
Using this definition this is the solution I came up with
$T(c\begin{pmatrix}a_{1,1} &a_{1,2}\\a_{2,1}&a_{2,2}\end{pmatrix}+\begin{pmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{pmatrix}) =c(a_{2,1})+(b_{2,1})=cT(A)+T(B)$
$b)$ Find a basis for $N(T)$
Definition: Null Space $N(T )=\{x \in V : T (x) = 0\}$.
Not sure where to begin with this. The following questions I assume can only be solved if finding a solution to $b)$
$c)$ Calculate the dimension of $N(T)$.
$d)$ Find a basis for $R(T)$
Thanks in advance.
Best Answer
You solution to part (a) seems to be just fine. For part (b):
Consider what is needed for a matrix $A\in\mathbb{R}^{(2,2)}$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_{2,1}=0$. Thus
$$ N(T)=\{A\in\mathbb{R}^{(2,2)}\mid a_{2,1}=0\}=\{\begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix}\in\mathbb{R}^{(2,2)}\} $$
The classical basis for $\mathbb{R}^{(2,2)}$ is the basis associated with the canonical basis for $\mathbb{R}^4$:
$$ \begin{pmatrix}1 &0\\0 &0\end{pmatrix}, \begin{pmatrix}0 &1\\0 &0\end{pmatrix},\begin{pmatrix}0 &0\\1 &0\end{pmatrix}, \begin{pmatrix}0 &0\\0 &1\end{pmatrix} $$
You may check that these are indeed linear independent and span $\mathbb{R}^{(2,2)}$. Coming back to the null space of $T$, you can see that every matrix $A\in N(T)$, i.e. every matrix of the form
$$ \begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix} $$
can be created as a linear combination of
$$ \begin{pmatrix}1 &0\\0 &0\end{pmatrix}, \begin{pmatrix}0 &1\\0 &0\end{pmatrix},\begin{pmatrix}0 &0\\0 &1\end{pmatrix} $$
with
$$ \begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix}=a_{1,1}\begin{pmatrix}1 &0\\0 &0\end{pmatrix} + a_{1,2} \begin{pmatrix}0 &1\\0 &0\end{pmatrix}+ a_{2,2}\begin{pmatrix}0 &0\\0 &1\end{pmatrix} $$
as the defining condition for $N(T)$ is a zero at position $a_{2,1}$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $\mathrm{dim}(N(T))=3$.
Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that
$$ 4=\mathrm{dim}(\mathbb{R}^{(2,2)})=\mathrm{dim}(N(T))+\mathrm{dim}(R(T))=3+\mathrm{dim}(R(T)) $$
Therefore, $\mathrm{dim}(R(T))=1$, i.e. as $T:\mathbb{R}^{(2,2)}\to\mathbb{R}$, we have $R(T)=\mathbb{R}$. A typical basis for $\mathbb{R}$ as a vector space is the number $1$.