3.4: Left Null Space
If one understands the concept of a null space, the left null space is extremely easy to understand.
The Left Null Space of a matrix is the null space of its transpose, i.e.,
\[\mathcal{N}(A^T) = \{ \textbf{y} \in \mathbb{R}^{m} | A^{T} \textbf{y} = 0\} \nonumber\]
The word "left" in this context stems from the fact that \(A^{T} \textbf{y} = 0\) is equivalent to \(\textbf{y}^{T} A = 0\) where \(\textbf{y}\) "acts" on A from the left.
Example
As \(A_{red}\) was the key to identifying the null space of A, we shall see that \(A^{T}_{red}\) is the key to the null space of \(A^T\). If
\[A = \begin{pmatrix} {1}&{1}\\ {1}&{2}\\ {1}&{3} \end{pmatrix} \nonumber\]
then
\[A^{T} = \begin{pmatrix} {1}&{1}&{1}\\ {1}&{2}&{3} \end{pmatrix} \nonumber\]
and so
\[A^{T}_{red} = \begin{pmatrix} {1}&{1}&{1}\\ {0}&{1}&{2} \end{pmatrix} \nonumber\]
We solve \(A^{T}_{red} = 0\) by recognizing that \(y_{1}\) and \(y_{2}\) are pivot variables while \(y_{3}\) is free. Solving \(A^{T}_{red} \textbf{y} = 0\) for the pivot in terms of the free we find \(y_{2} = -(2y_{3})\) and \(y_{1} = y_{3}\) hence
\[\mathcal{N}(A^{T}) = \begin{equation} \left \{ y_{3} \begin{pmatrix} {1}\\ {-2}\\ {1} \end{pmatrix} | y_{3} \in \mathbb{R} \right \} \end{equation} \nonumber\]
Finding a Basis for the Left Null Space
The procedure is no different than that used to compute the null space of A itself. In fact
Suppose that \(A^{T}\) is n-by-m with pivot indices \(\{c_{j} | j = \{1, \cdots, r\}\}\) and free indices \(\{c_{j} | j = \{r+1, \cdots, n\}\}\). A basis for \(\mathcal{N}(A^T)\) may be constructed of \(m-r\) vectors \(\{z^{1}, z^{2}, \cdots, z^{m-r}\}\) where \(z^{k}\) and only \(z^k\), possesses a nonzero in its \(c_{r+k}\) component.