Example FS2 Four subsets, no. 2

Now let us return to the matrix $A$ that we used to motivate this section in Example CSANS, \begin{equation*} A= \begin{bmatrix} 10 & 0 & 3 & 8 & 7 \\ -16 & -1 & -4 & -10 & -13 \\ -6 & 1 & -3 & -6 & -6 \\ 0 & 2 & -2 & -3 & -2 \\ 3 & 0 & 1 & 2 & 3 \\ -1 & -1 & 1 & 1 & 0 \end{bmatrix} \end{equation*}

We form the matrix $M$ by adjoining the $6\times 6$ identity matrix $I_6$, \begin{equation*} M= \begin{bmatrix} 10 & 0 & 3 & 8 & 7 & 1 & 0 & 0 & 0 & 0 & 0 \\ -16 & -1 & -4 & -10 & -13 & 0 & 1 & 0 & 0 & 0 & 0 \\ -6 & 1 & -3 & -6 & -6 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 2 & -2 & -3 & -2 & 0 & 0 & 0 & 1 & 0 & 0 \\ 3 & 0 & 1 & 2 & 3 & 0 & 0 & 0 & 0 & 1 & 0 \\ -1 & -1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \end{equation*} and row-reduce to obtain $N$ \begin{equation*} N= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 2 & 0 & 0 & 1 & -1 & 2 & -1 \\ 0 & \leading{1} & 0 & 0 & -3 & 0 & 0 & -2 & 3 & -3 & 3 \\ 0 & 0 & \leading{1} & 0 & 1 & 0 & 0 & 1 & 1 & 3 & 3 \\ 0 & 0 & 0 & \leading{1} & -2 & 0 & 0 & -2 & 1 & -4 & 0 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 0 & 3 & -1 & 3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & \leading{1} & -2 & 1 & 1 & -1 \end{bmatrix} \end{equation*}

To find the four subsets for $A$, we only need identify the $4\times 5$ matrix $C$ and the $2\times 6$ matrix $L$, \begin{align*} C&= \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 2\\ 0 & \leading{1} & 0 & 0 & -3\\ 0 & 0 & \leading{1} & 0 & 1\\ 0 & 0 & 0 & \leading{1} & -2 \end{bmatrix} & L&= \begin{bmatrix} \leading{1} & 0 & 3 & -1 & 3 & 1 \\ 0 & \leading{1} & -2 & 1 & 1 & -1 \end{bmatrix} \end{align*}

Then we apply Theorem FS, \begin{align*} \nsp{A}&=\nsp{C}= \spn{\set{ \colvector{-2\\3\\-1\\2\\1} }} &&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}\\ \rsp{A}&=\rsp{C}= \spn{\set{ \colvector{1 \\ 0 \\ 0 \\ 0 \\ 2},\, \colvector{0 \\ 1 \\ 0 \\ 0 \\ -3},\, \colvector{0 \\ 0 \\ 1 \\ 0 \\ 1},\, \colvector{0 \\ 0 \\ 0 \\ 1 \\ -2} }} &&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}}\\ \csp{A}&=\nsp{L}= \spn{\set{ \colvector{-3\\2\\1\\0\\0\\0},\, \colvector{1\\-1\\0\\1\\0\\0},\, \colvector{-3\\-1\\0\\0\\1\\0},\, \colvector{-1\\1\\0\\0\\0\\1} }} &&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}\\ \lns{A}&=\rsp{L}= \spn{\set{ \colvector{1 \\ 0 \\ 3 \\ -1 \\ 3 \\ 1},\, \colvector{0 \\ 1 \\ -2 \\ 1 \\ 1 \\ -1} }} &&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}} \end{align*}