Example FS1 Four subsets, no. 1

In Example SEEF we found the five relevant submatrices of the extended echelon form for the matrix \begin{equation*} A= \begin{bmatrix} 1 & -1 & -2 & 7 & 1 & 6 \\ -6 & 2 & -4 & -18 & -3 & -26 \\ 4 & -1 & 4 & 10 & 2 & 17 \\ 3 & -1 & 2 & 9 & 1 & 12 \end{bmatrix} \end{equation*}

To apply Theorem FS we only need $C$ and $L$, \begin{align*} C&= \begin{bmatrix} \leading{1} & 0 & 2 & 1 & 0 & 3 \\ 0 & \leading{1} & 4 & -6 & 0 & -1 \\ 0 & 0 & 0 & 0 & \leading{1} & 2 \end{bmatrix} & L&= \begin{bmatrix} \leading{1} & 2 & 2 & 1 \end{bmatrix} \end{align*}

Then we use Theorem FS to obtain \begin{align*} \nsp{A}&=\nsp{C}= \spn{\set{ \colvector{-2\\-4\\1\\0\\0\\0},\, \colvector{-1\\6\\0\\1\\0\\0},\, \colvector{-3\\1\\0\\0\\-2\\1} }} &&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}\\ \rsp{A}&=\rsp{C}= \spn{\set{ \colvector{1\\0\\2\\1\\0\\3 },\, \colvector{0\\1\\4\\-6\\0\\-1},\, \colvector{0\\0\\0\\0\\1\\2} }} &&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}}\\ \csp{A}&=\nsp{L}= \spn{\set{ \colvector{-2\\1\\0\\0},\, \colvector{-2\\0\\1\\0},\, \colvector{-1\\0\\0\\1} }} &&\knowl{./knowls/theorem.BNS.knowl}{\text{Theorem BNS}}\\ \lns{A}&=\rsp{L}= \spn{\set{ \colvector{1\\2\\2\\1} }} &&\knowl{./knowls/theorem.BRS.knowl}{\text{Theorem BRS}} \end{align*}

Boom!