Show that if $\exists \vec{x} : \vec{x} = \vec{x}_0+\vec{x}_1 \iff \exists \vec{x}_0: A\vec{x} = \vec{b}$ and $\exists \vec{x}_1 : A\vec{x}=\vec{0}$.

Question

Matrix $A\vec{x}=\vec{b}$ represents a system of $m \times n$ unknowns with a specific solution $\vec{x}_0$.

Show that any solution of this system $\vec{x}$ can be written in the form $\vec{x} = \vec{x}_0+\vec{x}_1$ where $\vec{x}_1$ is a solution of $A\vec{x}=\vec{0}$.

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I don't quite understand how this proof would progress, but I also don't (very well) visualise the claim conceptually.

Would I need to use an augmented matrix and the inversion algorithm on an abstractly constructed $m \times n$ matrix?

I'm failing to find an appropriate approach, but I had some disorganised thought along the line of thought that considering the solution of a homogeneous system includes the 0 vectors for it's $m\times n $ column of unknowns.

Answer 1

Suppose $Ax=b$. Since $Ax_0=b$ we can say that $$Ax-Ax_0=b-b$$ $$A(x-x_0)=0$$ so define $x_1=x-x_0$ and then $Ax_1=0$ with $x=x_0+x_1$

Answer 2

Hint: Use linearity properties $A(\vec x_0 + \vec x_1) = A \vec x_0 + A \vec x_1$, then suppose $\vec x_1$ is a solution to the homogeneous system. Now, what can you say about the vector $\vec x_0 + \vec x_1$?