If
T: R^5 \rightarrow R^3 Linear Transformation, (a_1, a_2, a_3, a_4, a_5) \mapsto (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:
<br />
\left(<br />
 \begin{array} {cccc}<br />
1 & -1 & 3 & 0 \\<br />
0 & 1 & -1 & 0 \\<br />
0 & 1 & -1 & 0 \\<br />
0 & 0 & 0 & 0 \\<br />
1 & 0 & 2 & 0 \\<br />
 \end{array}<br />
\right)<br />
 \downarrow
\left(<br />
 \begin{array} {cccc}<br />
1 & -1 & 3 & 0 \\<br />
0 & 1 & -1 & 0 \\<br />
0 & 0 & 0 & 0 \\<br />
0 & 0 & 0 & 0 \\<br />
0 & 0 & 0 & 0 \\<br />
 \end{array}<br />
\right)

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

So the Kernal or N(T) = {(-2a, a, 0) | a \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?