From One Solution to the Whole Solution Set
A particular solution to \(\mathbf{A}\mathbf{x}=\mathbf{b}\) answers one question: can the system be solved, and if so, what is one vector that solves it? To describe every solution, we need to account for how any other solution can differ from that first one. The key is to compare two solutions by subtracting their equations. The right-hand sides then cancel, leaving a homogeneous system.
Throughout, let \(\mathbf{A}\in\mathbb{R}^{m\times n}\) and \(\mathbf{b}\in\mathbb{R}^m\). A solution of \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is a vector \(\mathbf{x}\in\mathbb{R}^n\) satisfying that equation. The system is consistent if it has at least one solution. Its homogeneous system is \(\mathbf{A}\mathbf{z}=\mathbf{0}\), and its solution set is the null space \(\operatorname{Nul}\mathbf{A}=\{\mathbf{z}\in\mathbb{R}^n:\mathbf{A}\mathbf{z}=\mathbf{0}\}\).
The proposed description is an equality of sets. It says that each vector in \(\mathbf{x}_p+\operatorname{Nul}\mathbf{A}\) solves the original system, and that no solution of the original system is missing from this translated set. Both directions matter: showing that the constructed vectors work alone would not establish that they are all the solutions.
The Translate Structure Theorem
Proof. Fix a particular solution \(\mathbf{x}_p\), so \(\mathbf{A}\mathbf{x}_p=\mathbf{b}\). First take any vector in \(\mathbf{x}_p+\operatorname{Nul}\mathbf{A}\). By the definition of a translate, it has the form \(\mathbf{x}_p+\mathbf{z}\), where \(\mathbf{A}\mathbf{z}=\mathbf{0}\). By matrix multiplication and the distributive law,
Thus every vector in the translated set is a solution of the original system. For the reverse inclusion, take any solution \(\mathbf{x}\), so \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Define \(\mathbf{z}=\mathbf{x}-\mathbf{x}_p\). Then
Therefore \(\mathbf{z}\in\operatorname{Nul}\mathbf{A}\), and \(\mathbf{x}=\mathbf{x}_p+\mathbf{z}\) belongs to \(\mathbf{x}_p+\operatorname{Nul}\mathbf{A}\). We have proved both set inclusions, so the sets are equal. \(\square\)
The proof relies on consistency exactly where a particular solution is chosen. It does not require that \(\mathbf{x}_p\) be unique, nor that \(\mathbf{A}\) be square. The same argument works for any real \(m\times n\) matrix: the vector dimensions match because both \(\mathbf{x}_p\) and each null-space vector lie in \(\mathbb{R}^n\).
Worked Example: A System with One Free Parameter
Describe every solution of \(\begin{bmatrix}1&2\\2&4\end{bmatrix}\mathbf{x}=\begin{bmatrix}3\\6\end{bmatrix}\) using one particular solution and the null space.
Write \(\mathbf{x}=(x_1,x_2)^\top\). The system gives the two equations \(x_1+2x_2=3\) and \(2x_1+4x_2=6\). The second equation is twice the first, so it adds no new restriction. Choose \(x_2=0\); then \(x_1=3\), giving the particular solution \(\mathbf{x}_p=(3,0)^\top\). Its check is
For the homogeneous system, the equations are \(z_1+2z_2=0\) and \(2z_1+4z_2=0\). Set \(z_2=t\), where \(t\in\mathbb{R}\). The first equation gives \(z_1=-2t\), and the second is then satisfied because \(2(-2t)+4t=0\). Thus every homogeneous solution is \(t(-2,1)^\top\). The theorem gives
Check: multiplying this general vector by the matrix gives \(\begin{bmatrix}(3-2t)+2t\\2(3-2t)+4t\end{bmatrix} =\begin{bmatrix}3\\6\end{bmatrix}\) for every real \(t\). Conversely, the first equation of the original system forces \(x_1=3-2x_2\), so every solution has the displayed form with \(t=x_2\).
Why the Description Includes Every Solution
The subtraction in the theorem is not just a way to generate some new solutions. Given any solution \(\mathbf{x}\), there is exactly one difference vector \(\mathbf{z}=\mathbf{x}-\mathbf{x}_p\) that places it in the translate. This gives a precise correspondence between the homogeneous solutions and the solutions of the nonhomogeneous system.
Proof. If \(\mathbf{z}\in\operatorname{Nul}\mathbf{A}\), then \(\mathbf{A}(\mathbf{x}_p+\mathbf{z})=\mathbf{b}\), as shown in the theorem, so the rule produces a solution. If two null-space vectors \(\mathbf{z}_1,\mathbf{z}_2\) produce the same solution, then \(\mathbf{x}_p+\mathbf{z}_1=\mathbf{x}_p+\mathbf{z}_2\). Subtracting \(\mathbf{x}_p\) from both sides gives \(\mathbf{z}_1=\mathbf{z}_2\), so no two different null-space vectors produce the same solution. Finally, for any solution \(\mathbf{x}\), the vector \(\mathbf{z}=\mathbf{x}-\mathbf{x}_p\) belongs to \(\operatorname{Nul}\mathbf{A}\) and satisfies \(\mathbf{x}=\mathbf{x}_p+\mathbf{z}\). Thus every solution is produced, and it is produced by exactly one null-space vector. \(\square\)
This correspondence explains why finding the null space is enough to complete a solution description after one particular solution is known. It also separates the tasks: the particular solution locates the translated set, while the homogeneous solutions specify all the directions in which one can move without changing the right-hand side.
Worked Example: A Two-Equation System
Find the full solution set of \(\begin{bmatrix}1&1&0\\0&1&1\end{bmatrix}\mathbf{x}=\begin{bmatrix}2\\1\end{bmatrix}\) in translate form.
The equations are \(x_1+x_2=2\) and \(x_2+x_3=1\). Taking \(x_3=0\) gives \(x_2=1\), and then \(x_1=1\). Thus \(\mathbf{x}_p=(1,1,0)^\top\). Direct multiplication confirms the choice:
For a homogeneous solution \(\mathbf{z}=(z_1,z_2,z_3)^\top\), the equations are \(z_1+z_2=0\) and \(z_2+z_3=0\). Let \(z_3=t\). The second equation gives \(z_2=-t\), and the first gives \(z_1=t\). Hence \(\mathbf{z}=t(1,-1,1)^\top\). The null-space vector checks directly:
Therefore the full solution set is \(\begin{bmatrix}1\\1\\0\end{bmatrix}+t\begin{bmatrix}1\\-1\\1\end{bmatrix}\), with \(t\in\mathbb{R}\). For a final check, multiplying the resulting vector \((1+t,1-t,t)^\top\) by the original matrix gives \(((1+t)+(1-t), (1-t)+t)^\top=(2,1)^\top\). Every solution is included because the two original equations force \(x_2=1-x_3\) and \(x_1=1+x_3\), exactly this parametrization with \(t=x_3\).
Changing the Particular Solution
The word any in the theorem is important. Choosing a different particular solution changes the point used to start the translate, but it does not change the full solution set. In fact, the difference between any two particular solutions must itself be a homogeneous solution: subtract their equations and the two copies of \(\mathbf{b}\) cancel.
Worked Example: Two Particular Solutions, One Solution Set
Consider \(\begin{bmatrix}1&1\\2&2\end{bmatrix}\mathbf{x}=\begin{bmatrix}4\\8\end{bmatrix}\). The equations reduce to \(x_1+x_2=4\). Both \(\mathbf{x}_{p,1}=(4,0)^\top\) and \(\mathbf{x}_{p,2}=(0,4)^\top\) are particular solutions, since
The homogeneous equation is \(z_1+z_2=0\), so its solutions are \(t(1,-1)^\top\). Using the first particular solution gives vectors \((4+t,-t)^\top\). Using the second gives \((s,4-s)^\top\). These are the same set: choosing \(s=4+t\) in the second expression gives \((4+t,-t)^\top\), and every real \(t\) gives a real \(s\). Conversely, given any \(s\), taking \(t=s-4\) recovers \((s,4-s)^\top\) from the first expression.
Check: for every \(t\), the sum of the coordinates of \((4+t,-t)^\top\) is \(4+t-t=4\), so the matrix product is \((4,8)^\top\). The two parametrizations therefore describe the same full solution set.
More generally, if \(\mathbf{x}_{p,1}\) and \(\mathbf{x}_{p,2}\) are particular solutions, then \(\mathbf{A}(\mathbf{x}_{p,1}-\mathbf{x}_{p,2})=\mathbf{b}-\mathbf{b}=\mathbf{0}\). So changing the starting point shifts it by a null-space vector, which leaves the translated set unchanged. This is a direct consequence of the equations and does not depend on a special matrix shape.
Common Mistake: Using a Homogeneous Solution Directly
A vector in \(\operatorname{Nul}\mathbf{A}\) solves \(\mathbf{A}\mathbf{z}=\mathbf{0}\), not necessarily \(\mathbf{A}\mathbf{x}=\mathbf{b}\). For instance, in the system from the previous example, \(\mathbf{z}=(1,-1)^\top\) is homogeneous because
A mistaken calculation might conclude that \((1,-1)^\top\) solves the original system. It does not: its product is \((0,0)^\top\), not \((4,8)^\top\). The correction is to add the homogeneous vector to a particular solution. For example, \((4,0)^\top+(1,-1)^\top=(5,-1)^\top\), and
The distinction is simple but essential: homogeneous solutions describe the allowable changes to a particular solution; they are not, by themselves, generally solutions for a nonzero right-hand side. If the system has no particular solution, there is no \(\mathbf{x}_p\) to translate by, so the theorem's consistency hypothesis cannot be omitted.
Check Your Understanding
Use the theorem by identifying a particular solution and describing the corresponding homogeneous solutions.
- For \(\begin{bmatrix}1&-1\\2&-2\end{bmatrix}\mathbf{x}=\begin{bmatrix}3\\6\end{bmatrix}\), find one particular solution and write the full solution set in translate form.
- For the system in Question 1, verify directly that every vector in your parametrization gives the stated right-hand side.
- If \(\mathbf{x}_{p,1}\) and \(\mathbf{x}_{p,2}\) both solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\), what equation does their difference satisfy? Derive it by subtraction.
- Explain why a vector in \(\operatorname{Nul}\mathbf{A}\) need not solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\) when \(\mathbf{b}\ne\mathbf{0}\).
- State why proving only that \(\mathbf{x}_p+\mathbf{z}\) solves the original system is not enough to prove the translate theorem.
