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# 16a Proof of the structural theorem of solutions to nonhomogeneous systems.
Let us prove this theorem:
> **Theorem. Structural theorem of solutions to nonhomogeneous linear systems.**
> Suppose $A\vec x=\vec b$ is consistent, so there is a **particular solution** $\vec x_{p}$ to this nonhomogeneous system. Then each solution $\vec x$ to this nonhomogeneous system is **precisely** a vector of the form $$
\vec x= \vec x_{h} + \vec x_{p}
$$where $\vec x_{h}$ is **some homogeneous solution** to the equation $A\vec x =\vec 0$. And all vectors of this form is a solution to this nonhomogeneous system.
To prove this we need to show two things: (1) That every solution to $A\vec x = \vec b$ has that form $\vec x_{h}+\vec x_{p}$, and conversely (2) every vector of the form $\vec x_{h}+\vec x_{p}$ is a solution of $A\vec x=\vec b$. The subtle usage of the word **precisely** meaning each of these two conditions implies the other.
Notice this amounts to showing two sets are the same. On one hand the set of all solutions to $A\vec x = \vec b$, which let us denote it as $$
S_{1} = \{\vec x: A\vec x = \vec b\}.
$$And on the other hand the set of vectors of the form $\vec x_{h}+\vec x_{p}$, for $\vec x_{h}$ some homogeneous solution, which we denote it as $$
S_{2} = \{\vec x_{h} + \vec x_{p} : \vec x_{h} \text{ is any homogeneous solution}\},
$$for a fixed nonhomogeneous particular solution $\vec x_{p}$ to $A\vec x = \vec b$. So this theorem is equivalent to showing $S_{1}=S_{2}$. To show equality of two sets, it is equivalent to showing **mutual containment**, that each set contains each other.
$\blacktriangleright$ Proof.
$(S_{1}\subset S_{2})$
Take $\vec x \in S_{1}$, so $\vec x$ is a solution to $A\vec x = \vec b$. Let us consider the difference $\vec x - \vec x_{p}$. Note that $$
A(\vec x - \vec x_{p})=A\vec x- A\vec x_{p}=\vec b-\vec b=\vec 0.
$$This shows $\vec x-\vec x_{p}$ solves the homogeneous equation, so let write this as $\vec x_{h}=\vec x - \vec x_{p}$, where $\vec x_{h}$ is a homogeneous solution. Hence $\vec x = \vec x_{h}+\vec x_{p}$, that is, $\vec x \in S_{2}$. So we showed $S_{1}\subset S_{2}$.
$(S_{1}\supset S_{2})$
Take $\vec x \in S_{2}$, then we can write $\vec x = \vec x_{h} + \vec x_{p}$ where $\vec x_{h}$ is some homogeneous solution, and $\vec x_{p}$ is this given fixed particular solution. But note that $$
A\vec x= A(\vec x_{h}+\vec x_{p})=A\vec x_{h}+A\vec x_{p}=\vec 0+\vec b=\vec b.
$$This shows $\vec x$ is a solution to $A\vec x=\vec b$, that is $\vec x \in S_{1}$. So we showed $S_{2}\subset S_{1}$.
Whence we have $S_{1} = S_{2}$ as desired. $\blacksquare$
Note well that the set $S_{2}=\{\vec x_{h}+\vec x_{p}: \vec x_{h} \text{ is any homogeneous solution}\}$ is just a translation of the set of all homogeneous solutions by $\vec x_{p}$, we have $S_{2} = NS(A) + \vec x_{p}$. And hence what we showed is $$
S_{1}=S_{2}=NS(A)+\vec x_{p},
$$ that the solution set to $A\vec x = \vec b$ is the translation of the solution set of $A\vec x = \vec 0$ by a fixed particular solution $\vec x_{p}$. Again, this means each solution to $A\vec x = \vec b$ is just some translation of some homogeneous solution by $\vec x_{p}$.