Why a Nonzero Right-Hand Side Changes the Set
A homogeneous linear equation asks for vectors sent to zero. Its solution set contains the zero vector and is closed under linear combinations, so it is a subspace of its ambient vector space. A nonhomogeneous equation instead asks for vectors sent to a specified vector that may not be zero. Its solutions, when they exist, are shifted away from the origin. This difference is easy to overlook: a line or plane described by linear equations need not be a subspace.
We work over \(\mathbb{R}\). For a matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\), the equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has unknown \(\mathbf{x}\in\mathbb{R}^n\) and right-hand side \(\mathbf{b}\in\mathbb{R}^m\). The solution set is the set of all such \(\mathbf{x}\) satisfying the equation. Recall that a subspace of \(\mathbb{R}^n\) must contain \(\mathbf{0}\) and be closed under addition and scalar multiplication. In particular, failing the zero-vector test is enough to show that a set is not a subspace.
A translate can have the same directions and shape as a subspace without being a subspace itself. The key geometric distinction is whether it contains the origin. The examples below make this distinction concrete, then a general result will show why translates arise from nonhomogeneous systems.
A Line That Misses the Origin
Worked Example: One Nonhomogeneous Equation in Two Variables
Describe the solutions of \(x+y=3\), and decide whether they form a subspace of \(\mathbb{R}^2\).
Choose \(y=t\), where \(t\in\mathbb{R}\). The equation then gives \(x=3-t\), so every solution, and no other vector, has the form
The vector \(\begin{bmatrix}3\\0\end{bmatrix}\) is one particular solution. The direction vector \(\begin{bmatrix}-1\\1\end{bmatrix}\) satisfies the associated homogeneous equation, since \(-1+1=0\). The solution set is therefore a translate of the line through the origin in that direction.
Check: For every \(t\), the coordinates add to \((3-t)+t=3\), so every displayed vector is a solution. Conversely, any solution has \(y=t\) for some real \(t\), and then \(x=3-t\), so it appears in the displayed form. The zero vector is not a solution because \(0+0\ne3\). Thus this nonempty solution set is not a subspace.
Closure also fails directly. The vectors \(\begin{bmatrix}3\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\3\end{bmatrix}\) both solve the equation, but their sum \(\begin{bmatrix}3\\3\end{bmatrix}\) does not: its coordinates add to \(6\), not \(3\). A subspace must be closed under addition.
The General Form of a Consistent Solution Set
For a matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\), define its null space by \(\operatorname{Nul}\mathbf{A}=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\mathbf{0}\}\). This is a subspace: it contains \(\mathbf{0}\), and if \(\mathbf{u},\mathbf{v}\) lie in it and \(a,c\in\mathbb{R}\), then \(\mathbf{A}(a\mathbf{u}+c\mathbf{v})=a\mathbf{A}\mathbf{u}+c\mathbf{A}\mathbf{v}=\mathbf{0}\). Here we use the distributive and scalar properties of matrix multiplication.
Proof. Assume first that \(\mathbf{p}\) is a solution, so \(\mathbf{A}\mathbf{p}=\mathbf{b}\). If \(\mathbf{x}\) is any other solution, then \(\mathbf{A}\mathbf{x}=\mathbf{b}\), and therefore \(\mathbf{A}(\mathbf{x}-\mathbf{p})=\mathbf{A}\mathbf{x}-\mathbf{A}\mathbf{p}=\mathbf{b}-\mathbf{b}=\mathbf{0}\). Hence \(\mathbf{x}-\mathbf{p}\in\operatorname{Nul}\mathbf{A}\), so \(\mathbf{x}=\mathbf{p}+\mathbf{w}\) for some \(\mathbf{w}\in\operatorname{Nul}\mathbf{A}\). Conversely, if \(\mathbf{w}\in\operatorname{Nul}\mathbf{A}\), then \(\mathbf{A}(\mathbf{p}+\mathbf{w})=\mathbf{A}\mathbf{p}+\mathbf{A}\mathbf{w}=\mathbf{b}+\mathbf{0}=\mathbf{b}\); thus every vector in \(\mathbf{p}+\operatorname{Nul}\mathbf{A}\) is a solution. If no solution exists, by definition there are no vectors in the solution set, so it is empty. \(\square\)
This result gives a useful interpretation: once one solution is known, every other solution differs from it by a solution of the homogeneous system. Conversely, adding any homogeneous solution to the particular solution produces another solution. The result requires consistency to choose \(\mathbf{p}\); it does not claim that an inconsistent system has a translated null space as its solution set.
Worked Example: Two Equations in Three Variables
Find all solutions of \(x+y+z=1\) and \(x-y=2\), and identify the translated subspace.
Set \(y=t\). The second equation gives \(x=t+2\). Substituting these expressions into the first equation gives \((t+2)+t+z=1\), hence \(z=-1-2t\). Thus
The first vector is a particular solution. The direction vector solves both homogeneous equations: \(1+1-2=0\) and \(1-1=0\). Therefore the solution set is a translate of the subspace consisting of multiples of \(\begin{bmatrix}1\\1\\-2\end{bmatrix}\).
Check: Substituting the parameterized coordinates into the first equation gives \((2+t)+t+(-1-2t)=1\), and into the second gives \((2+t)-t=2\). Every real \(t\) therefore gives a solution. Conversely, the second equation forces \(x=y+2\), and the first then forces \(z=-1-2y\), so taking \(t=y\) recovers every solution. The zero vector fails both equations, so the set is not a subspace.
Exactly When Is the Solution Set a Subspace?
The zero-vector test gives a complete answer, including the possibility that the system is inconsistent.
Proof. Suppose first that the solution set is a subspace. Every subspace contains \(\mathbf{0}\), so \(\mathbf{0}\) must solve the equation. But \(\mathbf{A}\mathbf{0}=\mathbf{0}\), and therefore \(\mathbf{A}\mathbf{0}=\mathbf{b}\) implies \(\mathbf{b}=\mathbf{0}\). Conversely, suppose \(\mathbf{b}=\mathbf{0}\). The solution set is then \(\operatorname{Nul}\mathbf{A}\). It contains \(\mathbf{0}\), and for any \(\mathbf{u},\mathbf{v}\in\operatorname{Nul}\mathbf{A}\) and \(a,c\in\mathbb{R}\), \(\mathbf{A}(a\mathbf{u}+c\mathbf{v})=a\mathbf{A}\mathbf{u}+c\mathbf{A}\mathbf{v}=\mathbf{0}\). It is therefore closed under linear combinations and is a subspace. \(\square\)
This theorem does not require the equations to be independent, or the solution to be unique. When \(\mathbf{b}=\mathbf{0}\), the zero vector is always a solution, even if other solutions exist. When \(\mathbf{b}\ne\mathbf{0}\), the zero vector cannot be a solution. The system may be consistent, in which case its solutions form a nonempty translate, or inconsistent, in which case there are no solutions; neither possibility gives a subspace.
Worked Example: An Inconsistent System
Consider the equations \(x+y=1\) and \(x+y=2\). Determine the solution set and explain its subspace status.
If a vector \(\begin{bmatrix}x\\y\end{bmatrix}\) solved both equations, the same number \(x+y\) would have to equal both \(1\) and \(2\). Subtracting the first equation from the second gives \(0=1\), a contradiction. There are no solutions, so the solution set is empty.
Check: The contradiction proves there is no vector satisfying both equations; hence the empty set is indeed the solution set. It is not a subspace of \(\mathbb{R}^2\), because it does not contain \(\mathbf{0}\). In matrix form, the right-hand side is \(\begin{bmatrix}1\\2\end{bmatrix}\ne\mathbf{0}\), in agreement with the theorem.
Common Mistake: Treating Every Linear Equation Set as a Subspace
A common error is to reason that the equation is linear, so its solutions must form a subspace. The missing question is whether the equation is homogeneous. For example, the claim that the solutions of \(x+y=3\) form a subspace is wrong: substituting the zero vector gives \(0+0=0\), which is not \(3\). The correction is to apply the zero-vector test first. Since the right-hand side is nonzero, the solution set cannot be a subspace.
Another tempting but invalid check is to verify only that two selected solutions are present and then assume all subspace properties follow. In the equation \(x+y=3\), both \(\begin{bmatrix}3\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\3\end{bmatrix}\) are solutions, but their sum is \(\begin{bmatrix}3\\3\end{bmatrix}\), whose coordinates add to \(6\). The correction is to distinguish the shifted set from its direction subspace: differences of solutions satisfy the homogeneous equation, while the solutions themselves satisfy the original nonhomogeneous equation.
Thus the useful procedure is brief but exact. First check whether the system is consistent. If it is, find one particular solution and describe the homogeneous solutions; together they give the full translate. To decide whether the solution set is a subspace, test whether the right-hand side is zero, equivalently whether the zero vector belongs to the set.
Check Your Understanding
For each question, justify the conclusion using the equations or the general result.
- Write all solutions of \(x-2y=4\) in the form of a particular solution plus a direction vector.
- Can the solution set of \(\mathbf{A}\mathbf{x}=\mathbf{b}\) be a subspace when \(\mathbf{b}\ne\mathbf{0}\)? Explain using the zero-vector test.
- For the system \(x+y+z=0\), \(x-y=0\), explain why its solution set is a subspace.
- Suppose \(\mathbf{p}\) solves \(\mathbf{A}\mathbf{x}=\mathbf{b}\), and \(\mathbf{w}\in\operatorname{Nul}\mathbf{A}\). Verify directly that \(\mathbf{p}+\mathbf{w}\) is a solution.
- Explain why an inconsistent system’s empty solution set is not a subspace.
