3.4: Left Null Space

If one understands the concept of a null space, the left null space is extremely easy to understand.

Definition: Left Null Space

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

Definition: A Basis for the Left Null Space

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.