If
$\displaystyle T: R^5 \rightarrow R^3 Linear Transformation, (a_1, a_2, a_3, a_4, a_5)$ $\displaystyle \mapsto$ $\displaystyle (a_1+a_5, -a_1+a_2+a_3, 3a_1-a_2-a_3+2a_5)$

So I have so far:

Nullity is:
$\displaystyle
\left(
\begin{array} {cccc}
1 & -1 & 3 & 0 \\
0 & 1 & -1 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 2 & 0 \\
\end{array}
\right)
$
$\displaystyle \downarrow$
$\displaystyle \left(
\begin{array} {cccc}
1 & -1 & 3 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)$

So,
$\displaystyle x_1=-2x_3$
$\displaystyle x_2=x_3$
which gives...
$\displaystyle
x_3
\left(
\begin{array} {c}
-2 \\
1 \\
0 \\
\end{array}
\right)$

So the Kernal or N(T) = {(-2a, a, 0) | a $\displaystyle \epsilon$ R}
What are the values of the Nullity/Dimension Theorem?
N(T)=? R(T)=?
N(T) + R(T) = dim(T)
dim(T)=5 yes?