3.5: Row Space
The Row Space
As the columns of \(A^{T}\) are simply the rows of \(A\) we call \(Ra(A^{T})\) the row space of \(A^{T}\). More precisely
The row space of the m-by-n matrix A is simply the span of its rows, i.e.,
\[Ra(A^{T}) \equiv \{A^{T} \textbf{y} | \textbf{y} \in \mathbb{R}^{m}\} \nonumber\]
This is a subspace of \(\mathbb{R}^n\)
Let us examine the matrix:
\[A = \begin{pmatrix} {0}&{1}&{0}&{0}\\ {-1}&{0}&{1}&{0}\\ {0}&{0}&{0}&{1} \end{pmatrix} \nonumber\]
The row space of this matrix is:
\[\mathscr{Ra}(A^{T}) = \left \{ y_{1} \begin{pmatrix} {0}\\{1}\\{0}\\{0} \end{pmatrix}+y_{2} \begin{pmatrix} {-1}\\{0}\\{1}\\{0} \end{pmatrix}+y_{3} \begin{pmatrix} {0}\\{0}\\{0}\\{1} \end{pmatrix} | y \in \mathbb{R}^{3} \right \} \nonumber\]
As these three rows are linearly independent we may go no further. We "recognize" then \(\mathcal{Ra}(A^{T})\) as a three dimensional subspace of \(\mathbb{R}^{4}\)
Method for Finding the Basis of the Row Space
Regarding a basis for \(\mathscr{Ra}(A^T)\) we recall that the rows of \(A_{red}\), the row reduced form of the matrix \(A\), are merely linear \(A\) combinations of the rows of \(A\) and hence
\[\mathscr{Ra}(A^T) = \mathscr{Ra}(A_{red}) \nonumber\]
This leads immediately to:
Suppose \(A\) is m-by-n. The pivot rows of \(A_{red}\) constitute a basis for \(\mathscr{Ra}(A^{T})\).
With respect to our example,
\[\left \{ \begin{pmatrix} {0}\\{1}\\{0}\\{0} \end{pmatrix}, \begin{pmatrix} {-1}\\{0}\\{1}\\{0} \end{pmatrix}, \begin{pmatrix} {0}\\{0}\\{0}\\{1} \end{pmatrix} \right \} \nonumber\]
comprises a basis for \(\mathscr{Ra}(A^{T})\).