From a Homogeneous System to Column Relations
A matrix equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\) asks which coefficient vectors produce the zero output. The answer is not just a list of solutions to a system: each solution records a specific linear combination of the columns of \(\mathbf{A}\) that equals zero. Nonzero solutions therefore reveal exactly how the columns can depend on one another.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{m\times n}\), with columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\). The input \(\mathbf{x}\) has \(n\) entries, one coefficient for each column, while the output \(\mathbf{A}\mathbf{x}\) has \(m\) entries. By matrix multiplication, if \(\mathbf{x}=(x_1,\dots,x_n)^\top\), then \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\).
The null space is a set of coefficient vectors in \(\mathbb{R}^n\), not a set of outputs in \(\mathbb{R}^m\). It always contains \(\mathbf{0}\), since \(\mathbf{A}\mathbf{0}=\mathbf{0}\). The important question is whether it contains any nonzero vector. If it does, the entries of that vector give coefficients, not all zero, whose combination of columns vanishes.
Proof. To verify the subspace conditions, first note that \(\mathbf{0}\in\operatorname{Nul}\mathbf{A}\), because \(\mathbf{A}\mathbf{0}=\mathbf{0}\). Now let \(\mathbf{x},\mathbf{y}\in\operatorname{Nul}\mathbf{A}\) and let \(c\in\mathbb{R}\). By definition, \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and \(\mathbf{A}\mathbf{y}=\mathbf{0}\). Linearity of matrix multiplication gives \(\mathbf{A}(\mathbf{x}+\mathbf{y})=\mathbf{A}\mathbf{x}+\mathbf{A}\mathbf{y}=\mathbf{0}\) and \(\mathbf{A}(c\mathbf{x})=c\mathbf{A}\mathbf{x}=\mathbf{0}\). Thus \(\mathbf{x}+\mathbf{y}\) and \(c\mathbf{x}\) are both in \(\operatorname{Nul}\mathbf{A}\). The set contains zero and is closed under addition and scalar multiplication, so it is a subspace of \(\mathbb{R}^n\). \(\square\)
Each Null Vector Is a Dependence Relation
Proof. By the definition of matrix multiplication, \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). Therefore, \(\mathbf{x}\in\operatorname{Nul}\mathbf{A}\) means exactly that this linear combination equals \(\mathbf{0}\), proving the first statement in both directions. By definition, the columns are linearly dependent exactly when there are coefficients \(x_1,\dots,x_n\), not all zero, for which \(x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n=\mathbf{0}\). The first statement says those coefficients form a nonzero vector in \(\operatorname{Nul}\mathbf{A}\). Conversely, any nonzero vector in that null space supplies coefficients not all zero for such a relation. \(\square\)
This equivalence makes the zero vector an important boundary case. Its presence in every null space says nothing about dependence: it represents the trivial equation \(0\mathbf{a}_1+\cdots+0\mathbf{a}_n=\mathbf{0}\). It is a nonzero null vector that certifies a nontrivial dependence relation. If the only null vector is \(\mathbf{0}\), then the columns are independent.
Worked Example: An Independent Pair of Columns
Find \(\operatorname{Nul}\mathbf{A}\) for \(\mathbf{A}=\begin{bmatrix}1&2\\3&7\end{bmatrix}\), and determine whether its columns are dependent.
Write the homogeneous system and row-reduce its coefficient matrix. The operation \(R_2\to R_2-3R_1\) gives
The second row says \(x_2=0\). The first row then says \(x_1+2x_2=0\), so \(x_1=0\). Thus \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\), and the columns are independent by the theorem.
Check: The only candidate is \(\mathbf{x}=(0,0)^\top\), and \(\mathbf{A}\mathbf{x}=(0,0)^\top\). The original columns are \((1,3)^\top\) and \((2,7)^\top\); the row reduction shows that no nontrivial coefficients combine them to give zero.
Worked Example: Reading a Relation from the Null Space
Find the null space of \(\mathbf{A}=\begin{bmatrix}1&2&3\\0&1&1\\1&3&4\end{bmatrix}\), and express the relation it gives among the columns.
Apply \(R_3\to R_3-R_1\), then \(R_3\to R_3-R_2\):
Next apply \(R_1\to R_1-2R_2\). The resulting equations are \(x_1+x_3=0\) and \(x_2+x_3=0\), so \(x_1=-x_3\) and \(x_2=-x_3\). Writing \(t=x_3\), every solution has the form \(\mathbf{x}=t(-1,-1,1)^\top\), with \(t\in\mathbb{R}\). Hence \(\operatorname{Nul}\mathbf{A}=\{t(-1,-1,1)^\top:t\in\mathbb{R}\}\).
Choosing \(t=1\) gives a nonzero null vector, so the columns are dependent. Its entries state that \(-\mathbf{a}_1-\mathbf{a}_2+\mathbf{a}_3=\mathbf{0}\), or \(\mathbf{a}_3=\mathbf{a}_1+\mathbf{a}_2\).
Check: The columns are \(\mathbf{a}_1=(1,0,1)^\top\), \(\mathbf{a}_2=(2,1,3)^\top\), and \(\mathbf{a}_3=(3,1,4)^\top\). Adding the first two gives \((1+2,0+1,1+3)^\top=(3,1,4)^\top=\mathbf{a}_3\), as claimed. Direct multiplication also gives \(\mathbf{A}(-1,-1,1)^\top=(-1-2+3,-1+1,-1-3+4)^\top=\mathbf{0}\).
Why Row Reduction Preserves the Null Space
Row reduction is useful for finding solutions, but its role must be understood precisely. An elementary row operation replaces one system of equations with an equivalent system: swapping equations does not change which vectors satisfy them; multiplying an equation by a nonzero number does not change its solutions; and adding a multiple of one equation to another preserves the solutions because the operation can be reversed. Therefore, if \(\mathbf{B}\) is obtained from \(\mathbf{A}\) by row operations, then \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and \(\mathbf{B}\mathbf{x}=\mathbf{0}\) have the same solutions. Row reduction can reveal the null space, though it does not change the original columns whose dependence is being interpreted.
In particular, a dependence relation should be read from the solution coefficients, not from a claim that the rows of a reduced matrix are the original columns. The vector \(\mathbf{x}\) always has one entry for each original column. Its multiplication by the original matrix gives the corresponding combination of those original columns.
Worked Example: A Relation Among Three Columns
For \(\mathbf{A}=\begin{bmatrix}1&0&2\\0&1&3\end{bmatrix}\), verify a dependence relation among its columns and state the associated null vector.
The columns are \(\mathbf{a}_1=(1,0)^\top\), \(\mathbf{a}_2=(0,1)^\top\), and \(\mathbf{a}_3=(2,3)^\top\). The third column is \(2\mathbf{a}_1+3\mathbf{a}_2\), so \(-2\mathbf{a}_1-3\mathbf{a}_2+\mathbf{a}_3=\mathbf{0}\). The coefficient vector is \(\mathbf{x}=(-2,-3,1)^\top\), which has one entry for each of the three columns.
Check: Multiplying by the matrix gives
Thus \(\mathbf{x}\in\operatorname{Nul}\mathbf{A}\), and the nonzero vector confirms that the columns are dependent.
Null Vectors and Non-Unique Coefficients
A null vector also explains when two different coefficient vectors produce the same output. This matters whenever a vector is expressed as a linear combination of matrix columns: dependence means that one representation can be changed without changing the resulting vector.
Proof. If \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{y}\), subtract the right side from the left and use linearity to obtain \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\mathbf{A}\mathbf{x}-\mathbf{A}\mathbf{y}=\mathbf{0}\). By the definition of the null space, \(\mathbf{x}-\mathbf{y}\in\operatorname{Nul}\mathbf{A}\). Conversely, if \(\mathbf{x}-\mathbf{y}\in\operatorname{Nul}\mathbf{A}\), then \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\mathbf{0}\), so \(\mathbf{A}\mathbf{x}-\mathbf{A}\mathbf{y}=\mathbf{0}\) and therefore \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{y}\). \(\square\)
For a fixed right-hand side \(\mathbf{b}\), suppose \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has a solution \(\mathbf{x}\). Any other solution \(\mathbf{y}\) satisfies \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{y}\), so their difference is in \(\operatorname{Nul}\mathbf{A}\). Conversely, if \(\mathbf{z}\in\operatorname{Nul}\mathbf{A}\), then \(\mathbf{A}(\mathbf{x}+\mathbf{z})=\mathbf{A}\mathbf{x}+\mathbf{A}\mathbf{z}=\mathbf{b}+\mathbf{0}=\mathbf{b}\). Thus null vectors describe exactly the changes to a coefficient vector that leave its output unchanged.
Common Mistakes
Confusing the number of columns with the number of rows. For \(\mathbf{A}\in\mathbb{R}^{m\times n}\), a null vector has \(n\) entries because it supplies one coefficient for each column. The output \(\mathbf{A}\mathbf{x}\) has \(m\) entries. Trying to place an \(m\)-entry vector in the null space when \(m\ne n\) uses the wrong space.
Counting the zero solution as evidence of dependence. Every homogeneous system has the zero solution. For example, the independent columns in the first worked example still have \(\mathbf{0}\in\operatorname{Nul}\mathbf{A}\). Dependence requires a nonzero null vector.
Using coefficients in the wrong order or with the wrong signs. For the third worked example, the relation is \(-2\mathbf{a}_1-3\mathbf{a}_2+\mathbf{a}_3=\mathbf{0}\). A tempting but incorrect coefficient vector is \((-2,3,1)^\top\). Multiplication checks the claim: \(\mathbf{A}(-2,3,1)^\top=(-2+0+2,\;0+3+3)^\top=(0,6)^\top\ne\mathbf{0}\). The correction is to use \(-3\) in the second entry, giving \((-2,-3,1)^\top\), which multiplies to \(\mathbf{0}\).
The practical test is to keep the coefficient vector aligned with the original column order and multiply it by the original matrix. A correct nonzero null vector immediately verifies a dependence relation; a solution set containing only zero certifies independence.
Check Your Understanding
For each question, distinguish the coefficient vector from the output and use matrix multiplication to check your conclusion.
- For \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\), find a nonzero vector in \(\operatorname{Nul}\mathbf{A}\) and state the corresponding column relation.
- If \(\mathbf{A}\in\mathbb{R}^{3\times 5}\), how many entries does each vector in \(\operatorname{Nul}\mathbf{A}\) have? Explain using the columns of \(\mathbf{A}\).
- Prove directly from the definition that \(\mathbf{0}\in\operatorname{Nul}\mathbf{A}\) for every matrix \(\mathbf{A}\).
- Suppose \(\mathbf{A}\mathbf{x}=\mathbf{b}\) and \(\mathbf{z}\in\operatorname{Nul}\mathbf{A}\). Show that \(\mathbf{x}+\mathbf{z}\) also solves the equation.
- For \(\mathbf{A}=\begin{bmatrix}1&0&1\\0&1&2\end{bmatrix}\), verify whether \(( -1,-2,1)^\top\) belongs to \(\operatorname{Nul}\mathbf{A}\), then express the resulting column relation.
