Prove that a consistent linear system $Ax = b$ has infinitely many solutions if and only if $Ax = 0$ has a nontrivial solution. (Note: A is a homogenous matrix)
In order to prove this, we have to prove two separate statements, namely:
1) If $Ax = b$ has infinitely many solutions, $Ax = 0$ has a nontrivial solution.
2) If $Ax = 0$ has a nontrivial solution, $Ax = b$ has infinitely many solutions.
Proof of statement 1:
Since $Ax = b$ has infinitely many solutions, there are at least two distinct solutions, say $y_1$ and $y_2$, where $y_1 ≠ y_2$.
So $Ay_1 = b$ and $Ay_2 = b.$
Then, $A(y_1 – y_2) = Ay_1 – Ay_2 = b – b = 0.$
So $y_1 – y_2 ≠ 0$ is a nontrivial solution to $Ax = 0.$
Proof of statement 2:
If $Ax = 0$ has a nontrivial solution $y$, then $Ay = 0$, where $y≠0$.
If $Ax = b$, then
$A(y+x) = Ay + Ax = 0 + b = b$
According to the uniqueness theorem, $A(αy) = 0$ for any scalar $α$ if $A(y) = 0.$
Therefore, $A(αy + x) = A(αy) + Ax = 0 + Ax = b$ for any scalar $α$
So if $Ax = 0$ has a nontrivial solution, $Ax = b$ has infinitely many solutions.
Proof by contradiction:
If $Ax = 0$ has a trivial solution $y$, then $Ay = 0.$
Also, if $Ax = b$, then $A(y + x) = Ay + Ax = 0 + b = b.$
According to the uniqueness theorem, $A(αy) = 0$ for any scalar $α$ if $A(y) = 0.$
However, if $y = 0$, then $αy = 0$ for any scalar $α$, which implies that there is only one solution, not infinitely many.
Could someone verify my proof and see if I have worded anything incorrectly or missed anything?