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

\[T\left( \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] \right) = \left[\begin{array}{c} 5 \, x_{1} + 4 \, x_{2} - 8 \, x_{3} \\ -2 \, x_{1} - 3 \, x_{2} + 6 \, x_{3} \\ 2 \, x_{1} + x_{2} - 2 \, x_{3} \\ -3 \, x_{1} - 4 \, x_{2} + 8 \, x_{3} \end{array}\right] .\]

  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:

\[\operatorname{RREF} \left[\begin{array}{ccc} 5 & 4 & -8 \\ -2 & -3 & 6 \\ 2 & 1 & -2 \\ -3 & -4 & 8 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]

  1. \[\operatorname{Im}\ T = \operatorname{span}\ \left\{ \left[\begin{array}{c} 5 \\ -2 \\ 2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 1 \\ -4 \end{array}\right] \right\} \]

    \[\operatorname{ker}\ T = \left\{ \left[\begin{array}{c} 0 \\ 2 \, a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \]

  2. A basis of \(\operatorname{Im}\ T\) is \( \left\{ \left[\begin{array}{c} 5 \\ -2 \\ 2 \\ -3 \end{array}\right] , \left[\begin{array}{c} 4 \\ -3 \\ 1 \\ -4 \end{array}\right] \right\} \). A basis of \(\operatorname{ker}\ T\) is \( \left\{ \left[\begin{array}{c} 0 \\ 2 \\ 1 \end{array}\right] \right\} \)
  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.