Let \(T:\mathbb{R}^ 3 \to \mathbb{R}^ 4 \) be the linear transformation given by

\[T\left( \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \right) = \left[\begin{array}{c} -x - 4 \, y + 5 \, z \\ -x - 4 \, y + 3 \, z \\ -2 \, x - 8 \, y + 7 \, z \\ x + 4 \, y - 5 \, z \end{array}\right] .\]

  1. Explain how to find the image of \(T\) and the kernel of \(T\).
  2. Explain how to find a basis of the image of \(T\) and a basis of the kernel of \(T\).
  3. Explain how to find the rank and nullity of \(T\), and why the rank-nullity theorem holds for \(T\).

Answer:

\[\operatorname{RREF} \left[\begin{array}{ccc} -1 & -4 & 5 \\ -1 & -4 & 3 \\ -2 & -8 & 7 \\ 1 & 4 & -5 \end{array}\right] = \left[\begin{array}{ccc} 1 & 4 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} -1 \\ -1 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 5 \\ 3 \\ 7 \\ -5 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} -4 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} -1 \\ -1 \\ -2 \\ 1 \end{array}\right] , \left[\begin{array}{c} 5 \\ 3 \\ 7 \\ -5 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} -4 \\ 1 \\ 0 \end{array}\right] \right\} \)
  3. The rank of \(T\) is \( 2 \), the nullity of \(T\) is \( 1 \), and the dimension of the domain of \(T\) is \( 3 \). The rank-nullity theorem asserts that \( 2 + 1 = 3 \), which we see to be true.