Answer: (i) The rank of the matrix is \(2\).
(ii) A basis for the range space of \(T\) is \(\left\{\begin{pmatrix}1\\2\\3\\6\end{pmatrix},\begin{pmatrix}3\\8\\13\\24\end{pmatrix}\right\}\).
(iii) A basis for the null space of \(T\) is \(\left\{\begin{pmatrix}-29\\5\\0\\2\end{pmatrix},\begin{pmatrix}-19\\3\\2\\0\end{pmatrix}\right\}\).
Let the columns of \(A\) be \(\mathbf{c}_1,\mathbf{c}_2,\mathbf{c}_3,\mathbf{c}_4\).
We row-reduce \(A\):
\(\begin{pmatrix}1&3&5&7\\2&8&7&9\\3&13&9&11\\6&24&21&27\end{pmatrix}\)
Use the first row to eliminate entries below the first pivot:
\(R_2\to R_2-2R_1\), \(R_3\to R_3-3R_1\), \(R_4\to R_4-6R_1\).
This gives
\(\begin{pmatrix}1&3&5&7\\0&2&-3&-5\\0&4&-6&-10\\0&6&-9&-15\end{pmatrix}\).
Now eliminate below the second pivot:
\(R_3\to R_3-2R_2\), \(R_4\to R_4-3R_2\).
So we obtain
\(\begin{pmatrix}1&3&5&7\\0&2&-3&-5\\0&0&0&0\\0&0&0&0\end{pmatrix}\).
There are 2 non-zero rows, so the rank is \(2\).
For the range space, a basis is given by the pivot columns of the original matrix. The pivot columns are the first and second columns, so a basis for the range space is
\(\left\{\begin{pmatrix}1\\2\\3\\6\end{pmatrix},\begin{pmatrix}3\\8\\13\\24\end{pmatrix}\right\}.\)
For the null space, solve \(A\mathbf{x}=\mathbf{0}\), where \(\mathbf{x}=\begin{pmatrix}x\\y\\z\\t\end{pmatrix}\).
The row-reduced form gives the equations
\(x+3y+5z+7t=0\),
\(2y-3z-5t=0\).
From \(2y-3z-5t=0\),
\(y=\frac{3z+5t}{2}\).
Substitute into the first equation:
\(x+3\left(\frac{3z+5t}{2}\right)+5z+7t=0\),
so
\(2x+9z+15t+10z+14t=0\), hence \(2x+19z+29t=0\).
Thus \(x=-\frac{19z+29t}{2}\), and to avoid fractions choose parameters that give integer basis vectors. Let \(z=2u\) and \(t=2v\). Then
\(y=3u+5v\), \(x=-19u-29v\).
So
\(\begin{pmatrix}x\\y\\z\\t\end{pmatrix}=u\begin{pmatrix}-19\\3\\2\\0\end{pmatrix}+v\begin{pmatrix}-29\\5\\0\\2\end{pmatrix}.\)
Therefore a basis for the null space is
\(\left\{\begin{pmatrix}-29\\5\\0\\2\end{pmatrix},\begin{pmatrix}-19\\3\\2\\0\end{pmatrix}\right\}.\)
