Why Transpose the Matrix?
A matrix equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) asks whether the right-hand side can be made from the columns of \(\mathbf{A}\). A related question concerns the rows: are there coefficients that combine them to give the zero row? The left null space records exactly those coefficient choices. It is found by solving a homogeneous system, but the matrix in that system is the transpose of the original one.
Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\). Its transpose \(\mathbf{A}^{\top}\) has \(n\) rows and \(m\) columns. Thus the equation \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) has an unknown \(\mathbf{y}\in\mathbb{R}^{m}\): its number of entries is the number of rows of \(\mathbf{A}\), not the number of columns. This dimension check is essential, especially when \(\mathbf{A}\) is rectangular.
The word “left” reflects the equivalent equation \(\mathbf{y}^{\top}\mathbf{A}=\mathbf{0}^{\top}\). The vector \(\mathbf{y}\) is written as a column in the defining equation, while \(\mathbf{y}^{\top}\) is a row in this equivalent form. In particular, a left-null vector gives coefficients for combining the rows of \(\mathbf{A}\).
Proof. By the definition of matrix multiplication, the \(j\)th entry of \(\mathbf{A}^{\top}\mathbf{y}\) is the dot product of row \(j\) of \(\mathbf{A}^{\top}\) with \(\mathbf{y}\). That row is \(\mathbf{a}_j^{\top}\), so the entry is \(\mathbf{a}_j^{\top}\mathbf{y}=\mathbf{y}\cdot\mathbf{a}_j\). A vector in \(\mathbb{R}^{n}\) is zero exactly when each of its \(n\) entries is zero. Therefore \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) holds exactly when all \(n\) of these dot products are zero. \(\square\)
Solving the Transposed System
To compute the left null space, form \(\mathbf{A}^{\top}\) and solve \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\). Row-reduce \(\mathbf{A}^{\top}\), keeping its columns as the coefficient columns for the entries of \(\mathbf{y}\). Row operations are valid because swapping equations, multiplying an equation by a nonzero scalar, and adding a multiple of one equation to another each preserve the set of solutions. After elimination, assign parameters to any free variables and express the pivot variables in terms of them.
Worked Example: A Rectangular Matrix with Only the Zero Left-Null Vector
Find the left null space of \(\mathbf{A}=\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}\). Here \(\mathbf{A}\) has two rows, so a left-null vector has the form \(\mathbf{y}=(y_1,y_2)^{\top}\in\mathbb{R}^{2}\).
Transpose and row-reduce the coefficient matrix of the system:
The first step is \(R_3\to R_3-R_1\); the second is \(R_3\to R_3-R_2\). The equations from the final matrix are \(y_1=0\) and \(y_2=0\). Hence the left null space consists only of \(\mathbf{0}\).
Check: Substituting \(\mathbf{y}=(0,0)^{\top}\) into the original transposed system gives \(\mathbf{A}^{\top}\mathbf{y}=(0,0,0)^{\top}=\mathbf{0}\). The solution is valid, and the reduced system shows there are no other solutions.
This example has three equations but only two unknowns. The extra equation does not create another variable; it must be checked for compatibility with the others. In general, the transposed system has \(n\) equations and \(m\) unknowns when \(\mathbf{A}\) is \(m\times n\).
Worked Example: A Nonzero Left Null Space
Find all left-null vectors of \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\). Since \(\mathbf{A}\) has three rows, write \(\mathbf{y}=(y_1,y_2,y_3)^{\top}\).
The transposed system is
The two equations are \(y_1+y_3=0\) and \(y_2+y_3=0\). Let the free variable \(y_3=t\), where \(t\in\mathbb{R}\). Then \(y_1=-t\) and \(y_2=-t\), so every solution is \(\mathbf{y}=(-t,-t,t)^{\top}\).
Check: For every \(t\in\mathbb{R}\),
Thus the parameterized vectors all solve the system. Conversely, the equations force \(y_1=-y_3\) and \(y_2=-y_3\), so every solution has this form.
Left-Null Vectors as Relations Among Rows
Proof. The \(j\)th entry of the row vector \(\mathbf{y}^{\top}\mathbf{A}\) is \(\sum_{i=1}^{m}y_i a_{ij}\). The \(j\)th entry of \(y_1\mathbf{r}_1+\cdots+y_m\mathbf{r}_m\) is the same sum, so these row vectors are equal. Also, \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) if and only if its transpose is zero, which is equivalent to \(\mathbf{y}^{\top}\mathbf{A}=\mathbf{0}^{\top}\). Combining these equalities proves both directions. \(\square\)
This gives a useful interpretation: the entries of a left-null vector are coefficients in a linear relation among the rows. Zero coefficients mean that the corresponding rows do not contribute to that particular relation. The left null space collects all such coefficient vectors, not the row combinations themselves.
Worked Example: Finding Row Relations by Elimination
Let \(\mathbf{A}=\begin{bmatrix}1&2&0\\0&1&1\\1&3&1\\2&5&1\end{bmatrix}\). Find all vectors in its left null space by solving \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\).
The transposed coefficient matrix is
Apply \(R_2\to R_2-2R_1\), then \(R_3\to R_3-R_2\), using the updated second row in the second operation:
The equations are \(y_1+y_3+2y_4=0\) and \(y_2+y_3+y_4=0\). Set \(y_3=s\) and \(y_4=t\). Then \(y_1=-s-2t\), \(y_2=-s-t\), and the full solution set is \(\mathbf{y}=(-s-2t,-s-t,s,t)^{\top}\), with \(s,t\in\mathbb{R}\).
Check: The rows of \(\mathbf{A}\) are \(\mathbf{r}_1=(1,2,0)\), \(\mathbf{r}_2=(0,1,1)\), \(\mathbf{r}_3=(1,3,1)\), and \(\mathbf{r}_4=(2,5,1)\). The coefficients above give \[ (-s-2t)\mathbf{r}_1+(-s-t)\mathbf{r}_2+s\mathbf{r}_3+t\mathbf{r}_4=\mathbf{0}^{\top}. \] Coordinate by coordinate, the sums are \(-s-2t+s+2t=0\), \(-2s-4t-s-t+3s+5t=0\), and \(-s-t+s+t=0\). This verifies the row relation and the computed solution.
A Necessary Condition for Consistency
Left-null vectors can test whether a proposed right-hand side is compatible with \(\mathbf{A}\mathbf{x}=\mathbf{b}\). The following implication is useful even without a full test for consistency.
Proof. Consistency means that some \(\mathbf{x}\in\mathbb{R}^{n}\) satisfies \(\mathbf{b}=\mathbf{A}\mathbf{x}\). For a left-null vector \(\mathbf{y}\), \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\). Therefore \(\mathbf{y}\cdot\mathbf{b}=\mathbf{y}^{\top}\mathbf{b}=\mathbf{y}^{\top}\mathbf{A}\mathbf{x}=(\mathbf{A}^{\top}\mathbf{y})^{\top}\mathbf{x}=\mathbf{0}^{\top}\mathbf{x}=0\). Thus the condition is necessary. \(\square\)
For the matrix in the preceding example, choose \(s=1,t=0\), giving \(\mathbf{y}=(-1,-1,1,0)^{\top}\). Any consistent right-hand side must therefore satisfy \(-b_1-b_2+b_3=0\). This condition comes from a row relation: the third row of \(\mathbf{A}\) is the sum of its first two rows. The proposition asserts necessity; this argument alone does not claim that one selected left-null vector supplies every condition needed for consistency.
Common Mistake: Solving the Wrong Homogeneous System
Do not confuse the left null space with \(\operatorname{Nul}\mathbf{A}\). For the matrix \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\), a vector in the left null space has three entries, because it belongs to \(\mathbb{R}^{3}\). Solving \(\mathbf{A}\mathbf{z}=\mathbf{0}\) instead gives a different question: \(\mathbf{z}\) has two entries and belongs to \(\mathbb{R}^{2}\).
Indeed, the incorrect substitution \(\mathbf{A}\mathbf{z}=\mathbf{0}\) gives \(z_1=0\), \(z_2=0\), and \(z_1+z_2=0\), hence only \(\mathbf{z}=\mathbf{0}\). Concluding from this that the left null space is trivial is wrong. The correct calculation uses \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) and gives the nonzero left-null vectors \((-t,-t,t)^{\top}\). The reliable check is to count dimensions before solving: for an \(m\times n\) matrix, left-null vectors have \(m\) entries, and the coefficient matrix is \(n\times m\).
Keep the two questions distinct. The null space of \(\mathbf{A}\) describes solutions of \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and has vectors in \(\mathbb{R}^{n}\). The left null space describes solutions of \(\mathbf{A}^{\top}\mathbf{y}=\mathbf{0}\) and has vectors in \(\mathbb{R}^{m}\). Transposing first, then solving for all variables, preserves that distinction.
Check Your Understanding
Use the dimensions of each matrix to decide what the unknown vector must look like before solving.
- For \(\mathbf{A}=\begin{bmatrix}1&2\\3&6\\0&1\end{bmatrix}\), write the dimensions of \(\mathbf{A}^{\top}\) and the number of entries in a left-null vector.
- Find the left null space of \(\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\) by solving the transposed system.
- Explain why a left-null vector gives a linear relation among the rows of the original matrix.
- If \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent, what equation must every left-null vector \(\mathbf{y}\) satisfy with \(\mathbf{b}\)?
- For an \(m\times n\) matrix, state the dimensions of the unknown and coefficient matrix in the system used to find its left null space.
