What does Ax=0 has only the trivial solution imply?

Question

There is a true or false question that asks if A is a n x n matrix and if Ax=0 has only the trivial solution, then the system Ax = b has a unique solution for every b that is real.

I believe that this is true. But in order to show it I want to verify the statement "if Ax=0 has only the trivial solution" implies that there are no free variables. Hence Ax = b will have unique solutions since rank(A) = the number of variables in the system.

Answer 1

As for proof of the equivalence of the two specific conditions you cite without referring to other conditions... consider doing so by contrapositive. Suppose that there exists some $b$ such that $Ax=b$ has more than one solution, in other words there exist $x_1\neq x_2$ such that $Ax_1=Ax_2=b$. Then subtracting, we have $A(x_1-x_2)=Ax_1-Ax_2=b-b=0$ and so $Ax=0$ has a solution different than the trivial solution, namely $x_1-x_2$.

In the other direction, supposing that $Ax=0$ has a solution different than the trivial solution, namely some $x_1$ for which $Ax_1=0$, then here we have both $A0=0$ and $Ax_1=0$ and so letting $b=0$ we have an example of a $b$ for which there is more than one solution.

Answer 2

It is true, let $v_1$ and $v_2$ be two solutions for the system $Ax=b$. If we calculate $A(v_1-v_2)$ we get: $$A(v_1-v_2) = Av_1-Av_2=b-b=0$$ But we know that $Ax=0$ iff $x=0$, so it follows that $v_1-v_2=0$ and hence $v_1=v_2$.

Now let's show that the solution always exists. Let $\{e_1,...,e_n\}$ be a base for our vector space $V$, we will show that $\{Ae_1,...,Ae_n\}$ is a base for the image of the function.

Let $Av$ be an element of the image, we can write $v$ as $v = \sum_{k=1}^{n}a_ke_k$, then applying $A$ we get $Av = A(\sum_{k=1}^{n}a_ke_k) = \sum_{k=1}^{n}a_kAe_k$, so the set $\{Ae_1,...,Ae_n\}$ generates $\text{Im}(A)$.

We now only need to show that $\{Ae_1,...,Ae_n\}$ are linearly independent, in fact $\sum_{k=1}^{n}a_kAe_k = 0$ iff $A(\sum_{k=1}^{n}a_ke_k) = 0$ and we know by our hypotesis that this is true iff $\sum_{k=1}^{n}a_ke_k=0$ and hence (since $\{e_1,...,e_n\}$ is a base) iff $a_k = 0$ for every $1\le k\le n$.

So know we constructed a base of $n$ vectors for $\text{Im}(A)$ that it's contained in an $n-$dimensional vector space, hence $\text{Im}(A)$ is the whole arrival vector space (i.e. $A$ is surjective).

This is a corollary of a more general formula, that is, given a linear map $f:V\rightarrow W$, with $\dim V = n$ it is true that $$\dim V = \dim \text{Ker}f + \dim \text{Im}f$$ (try to show this by yourself, the idea is the same as the one I used in the last proof)