$Ax=0$ has trivial solution $\implies$ $Ax=b$ has a unique solution.

Question

I'm trying to prove that these three conditions are equivalent.

1. $Ax=0$ has only trivial solution.
2. operator $A$ is invertible.
3. $Ax=B$ has a unique solution.

I proved $1 \implies 2$ and $2 \implies 3$ but I need help writing the proof of $3 \implies 1$. Can you help me with this?

IMO I think this is trivial, but I need help writing it clearly.

Answer 1

Starting from the assumption that $Ax = b$ has a unique solution, suppose that $Ax = 0$ had a nontrivial solution, i.e. $A\bar{x} = 0$ for some $\bar{x} \neq 0$. Then, assuming $x^*$ is the solution of $Ax = b$:

$$ A(x^* + \bar{x}) = Ax^* + A(\bar{x}) = b + 0 = b $$ which would show that $Ax = b$ has at least two solutions, a contradiction.

Answer 2

If $A x = b$ has two solutions $x_1,x_2$ or

$$ A x_1 = b\\ A x_2 = b $$

then by linearity

$$ A(x_1-x_2) = b-b = 0 $$

but the solution for $A(x_1-x_2) = 0$ is $x_1-x_2 = 0\Rightarrow x_1 = x_2$