Why is Ax = 0 to be even considered when finding the complete solution to Ax = B?

Question

In this write-up, "Homogeneous and non-homogeneous equations", it says:

If the matrix equation

$\displaystyle AX = B$

has one particular solution $ X_p$, and the associated homogeneous equation

$\displaystyle AX = 0$

has the complete solution $ X_h$, then the complete solution to the original non-homogeneous equation is

$\displaystyle X = X_p + X_h.$

Question: What I do not understand is the motivation or intuition behind even attempting to look for the nullspace via $\displaystyle AX = 0$ when you already have not just one solution but an infinite number of solutions as the complete solution (via the parameter t) of a system of equations where there are more unknowns than equations! (If needed, please see the Example that immediately follows in the write-up.)

Answer 1

I read the relevant excerpt of the referenced article more carefully this time.

It seems $Ax = B$ won't require venturing into nullspace solutions via $Ax = 0$ provided you express the solution as parameterized (when the system of equations has free variables).

The complete solution even in Intro to Linear Algebra, 4ed book by Strang, page 158, is something we can obtain just by solving $Ax = B$ and remembering to use parameters. In other words, the parameterized solution is the complete solution that includes in it the particular solution automatically.

The confusion, I think, arose because I failed to notice how some authors were arriving at their particular solution. They were setting free variables to 1, one at a time. But if you set them to parameters, $s$, $t$ etc from the get go, there would be no need to even mention nullspace and conceptually complicate things for newbies (like myself).

Answer 2

Suppose $X_1$ and $X_2$ are solutions to $AX=B$, that is, $$ AX_1=B \qquad\text{and}\qquad AX_2=B $$ In particular, $AX_2=AX_1$, which implies $A(X_2-X_1)=0$, that is, $X_0=X_2-X_1$ is a solution to $AX=0$. Note that $$ X_2=X_1+X_0 $$ so $X_2$ has the form “solution of $AX=B$ plus solution of $AX=0$”.

Conversely, if $X_p$ is a solution of $AX=B$ and $X_h$ is a solution of $AX=0$, then also $X_p+X_h$ is a solution of $AX=B$, because $$ A(X_p+X_h)=AX_p+AX_h=B+0=B $$

It's not difficult to complete the proof that every solution of $AX=B$ can be obtained as $X_p+X_h$ with $X_p$ a fixed particular solution of $AX=B$ and $X_h$ any solution of $AX=0$.

Another way to look at the subject is by considering linear maps. Suppose $A$ is $m\times n$; then we can define a linear map $f\colon \mathbb{R}^n\to\mathbb{R}^m$.

If $B\in\mathbb{R}^m$, then the set of solutions to $AX=B$ is precisely $$ f^{-1}(\{B\})=\{X\in\mathbb{R}^n:f(X)=B\} $$ If a solution $X_p$ exists, then $$ f^{-1}(\{B\})=X_p+\ker f=\{X_p+X_h:X_h\in\ker f\} $$