Let
\(W' \) be the corresponding subspace of
\(\R^6 \text{,}\) spanned by the coordinate vectors of the provided spanning vectors for
\(W \text{,}\) relative to the standard basis of
\(\matrixring_{2 \times 3}(\R) \text{.}\) In the solution to
Exercise 28, we determined a simplified basis for
\(W' \text{:}\)
\begin{equation*}
\begin{split}
W = \Span \left\{
\left( 1, 0, 0, 0, -\tfrac{15}{2}, -\tfrac{71}{2} \right),
\left( 0, 1, 0, 0, -\tfrac{ 9}{2}, -\tfrac{97}{6} \right),
\\
\left( 0, 0, 1, 0, \tfrac{ 1}{2}, \tfrac{ 1}{2} \right),
\left( 0, 0, 0, 1, \tfrac{ 3}{2}, \tfrac{35}{6} \right)
\right\}
\end{split}\text{.}
\end{equation*}
So every vector \(\uvec{x} = (x_1,x_2,x_3,x_4,x_5,x_6) \) in \(W' \) is somehow a linear combination of these basis vectors:
\begin{equation*}
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix}
= r \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ -\frac{15}{2} \\ -\frac{71}{2} \end{bmatrix}
+ s \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \\ -\frac{ 9}{2} \\ -\frac{97}{6} \end{bmatrix}
+ t \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \frac{ 1}{2} \\ \frac{ 1}{2} \end{bmatrix}
+ u \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ \frac{ 3}{2} \\ \frac{35}{6} \end{bmatrix}\text{.}
\end{equation*}
Working as if reversing the process of calculating a basis for a null space, calculate the linear combination on the right and compare entries on either side of the equality to create parametric expressions for the components of \(\uvec{x} \text{.}\)
\begin{align*}
x_1 \amp = r \\
x_2 \amp = s \\
x_3 \amp = t \\
x_4 \amp = u \\
x_5 \amp = -\tfrac{15}{2} \, r - \tfrac{ 9}{2} \, s + \tfrac{1}{2} \, t + \tfrac{ 3}{2} \, u \\
x_6 \amp = -\tfrac{71}{2} \, r - \tfrac{97}{6} \, s + \tfrac{1}{2} \, t + \tfrac{35}{6} \, u
\end{align*}
Continuing to reverse the null-space-basis procedure, in the two expressions that are not merely parameter assignments, replace the parameters with the corresponding components and rearrange to become a homogeneous system.
\begin{equation*}
\begin{sysofeqns}{rcrcrcrcrcr}
15 x_1 \amp + \amp 9 x_2 \amp - \amp x_3 \amp - \amp 3 x_4 \amp + \amp 2 x_5 \amp = \amp 0
\\
213 x_1 \amp + \amp 97 x_2 \amp - \amp 3 x_3 \amp - \amp 35 x_4 \amp + \amp 6 x_6 \amp = \amp 0
\end{sysofeqns}
\end{equation*}
(Note that we have cleared fractions by multiplying each equation through by a common denominator.)
To translate this from \(\R^6 \) back to \(\matrixring_{2 \times 3}(\R) \text{,}\) replace the component variables with matrix entry variables, consistent with how matrix entries become coordinates relative to the standard basis of \(\matrixring_{2 \times 3}(\R) \text{.}\)
\begin{equation*}
\begin{sysofeqns}{rcrcrcrcrcr}
15 a_{11} \amp + \amp 9 a_{12} \amp - \amp a_{13} \amp - \amp 3 a_{21} \amp + \amp 2 a_{22} \amp = \amp 0
\\
213 a_{11} \amp + \amp 97 a_{12} \amp - \amp 3 a_{13} \amp - \amp 35 a_{21} \amp + \amp 6 a_{23} \amp = \amp 0
\end{sysofeqns}
\end{equation*}
That is,
\(W \) consists precisely of those
\(2 \times 3 \) matrices
\(A = \begin{bmatrix} a_{ij} \end{bmatrix} \) whose entries represent solutions to the homogeneous system above. You may verify that each of the original spanning vectors for
\(W \) provided in
Exercise 28 represent such solutions.
