We are of course assuming $A$ is an $n\times n$ matrix.
I know there's a proof of it going the other way (invertibility implies a unique solution), but I'm trying to work out a proof going this way.
I'm looking for a proof that only uses the most basic concepts of linear algebra.
I know that:
1) Every matrix can be reduced to rref.
2) Elementary row operations do not change the solution set of $Ax = b$.
3) I also know that $Ix = y$, where $y$ is the particular solution.
I'm not sure how to prove that $[A|b]$ can be reduced to $[I|y]$, however, and that's where the problem lies.
Any help with this would be greatly appreciated.
$\endgroup$ 24 Answers
$\begingroup$Suppose $A$ were not invertible. Then, there would be a non-zero vector in its null space. Call this vector $z$. Compute $y$ such that $z = x-y$. Since $z$ is non-zero, $x\neq y$.
Then, $$Az = 0 \implies A(x-y) = 0 \implies Ax-Ay = 0 \implies Ax = Ay.$$
But since $Ax = b$ then $Ay = b$. But $x$ was the unique solution to $Ax=b$, so $x=y$ necessarily, contradicting $x \neq y$.
$\endgroup$ 1 $\begingroup$if there's no solutions then you know $A$ is not invertible. on the other hand if there is at least one solution and if $A$ is not invertible then the set of vectors $x$ such that $Ax = 0$ is not trivial (contains some nonzero vector). do you see how to show that your solution to $Ax = b$ is not unique in this case?
$\endgroup$ $\begingroup$If $A\in M_n(F)$ then let $f:F^n\to F^n$ be the canonically associated endomorphism to the matrix $A$. The equation $Ax=b$ can be written $f(x)=b$ and saying that this equation has a unique solution is only saying that $b$ has a unique preimage by $f$. So in order to say that $f$ is bijective we need to have that for each $b$ the equation $f(x)=b$ has a unique solution $x$. In matrix language, you need that the equation $Ax=b$ has a unique solution for each $b$, not only for a given $b$, to say that $A$ is invertible.
$\endgroup$ 1 $\begingroup$Since $A$ is a $n$-by-$n$ matrix, the linear transformation $T: x \mapsto Ax$ is one-to-one $\implies$ linear transformation $T: x \mapsto Ax$ is invertible $\implies$ $A$ is invertible.
Suppose that $Ax=b$ and $Av = Au = 0, v\neq u$. Then $A(x+v) = A(x+u) = b$, whereby $x+v \neq x+u$.
$\therefore Ax=b $ for some unique $x \implies$ either there is no $v$ such that $Av=0$ or there is a unique $v$ such that $Av=0$.
Since $A0=0$, we conclude that there is a unique $v$ such that $Av=0$, and thus $T$ is one-to-one.
$\therefore A$ is invertible. (Line 1)
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