why isn't my proof valid? system of linear equations

The problem was: given the matrix $$A \in M_{m\times n}$$ and let $$b \in F^{m}_{col}$$. $$y \in F^{m}_{col}$$ is a solution of the system of equations $$AX=b$$. prove: every solution of the system of equations $$AX=b$$ can be represented as $$y+x$$, where $$x \in F^{m}_{col}$$ is the solution of the homogeneous system $$AX=0$$.

so my proof was like that: because $$y \in F^{m}_{col}$$ is a solution of the system of equations $$AX=b$$ we can conclude: $$(1) A \cdot y=b$$ and because $$x \in F^{m}_{col}$$ is a solution of the system of equations $$AX=0$$ we can conclude: $$(2) A \cdot x=0$$ from (1) and (2) we get: $$A \cdot y+A \cdot x=0+b$$ according to the rules of matrix multiplication we get: $$(3) A \cdot ( y+ x)=0+b=b$$ and therefore according to (3) the solution which is represented by $$z=x+y$$ is also a solution of the system of equations $$AX=b$$

my instructor gave me only 3 points for that telling that "it isn't a valid proof". Why is that? what's the problem with it? and what should I say in order to appeal his decision. thank in advance.

• You are only verifying that $y+x$ is a solution, not every solution is of the form $y+x$. Commented Feb 27, 2021 at 15:08

1 Answer

This is the part that you missed: let $$\;z\;$$ be a solution, any solution, to $$\;Ax=b\;$$ , then

$$A(z-y)=Az-Ay=b-b=0\implies \;z-y\;\text{is a solution to the hom. system}\;\;Ax=0$$

and thus there exists $$\;t\in\;$$ space of all solutions to the hom. sytem, such that

$$z-y=t\implies z=t+y$$

with $$\;t\;$$ a solution to the hom. system, and $$\;y\;$$ a particular solution to the non-hom. system