Exercise D.C31
Given set W below is a subspace of \({ C }^{ 4 }\), find the dimension of W.
\({ W }=\left\{ \left[ \begin{matrix} 2 \\ -3 \\ \begin{matrix} 4 \\ 1 \end{matrix} \end{matrix} \right] ,\quad \left[ \begin{matrix} 3 \\ 0 \\ \begin{matrix} 1 \\ -2 \end{matrix} \end{matrix} \right] ,\quad \left[ \begin{matrix} -4 \\ -3 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix} \right] \right\}\)
We perform some row reduction operations to find the pivot columns, which in turn will give the the dimension of W.
\(\left[ \begin{matrix} 2 \\ -3 \\ \begin{matrix} 4 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ 0 \\ \begin{matrix} 1 \\ -2 \end{matrix} \end{matrix}\begin{matrix} -4 \\ -3 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix} \right] \quad \rightarrow \quad \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ 1 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 5 \\ -2 \\ \begin{matrix} -2 \\ -2 \end{matrix} \end{matrix} \right] \quad \rightarrow \quad \left[ \begin{matrix} 1 \\ 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ -2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right]\)
We can see there is a pivot at the first and second vector, thus the pivor columns are D = {1,2} the dimension of W is 2, and T =
\(T=\left\{ \left[ \begin{matrix} 2 \\ -3 \\ \begin{matrix} 4 \\ 1 \end{matrix} \end{matrix} \right] ,\quad \left[ \begin{matrix} 3 \\ 0 \\ \begin{matrix} 1 \\ -2 \end{matrix} \end{matrix} \right] \right\}\)