# Are these statements equivalent?(invertible matrix theorem)

Suddenly I got a question about the invertible matrix theorem.

Among lots of equivalent statements suggested in my lin-alg text, I'm confused whether the statement that 'The equation Ax=b has "at least" one solution for each b in R^n.' is equivalent or not to the statement 'A is a invertible matrix.' (A is n-by-n matrix)

The phrase "at least" in the above statment implies, of course you guys know, it's ok when there are more than one solution of the equation.

However, I think 'at least' should be corrected as 'only' so that THE statement is equivalent to the different but having same mean with 'the linear transformation x to Ax is one-to-one.'

If I have wrong concept, let me know what i miss is.

• How can we say if that statement is true or not if we don't know anything about the problem? What are the other statements? Commented Aug 13, 2015 at 15:09
• For the invertible matrix theorem, I believe the statement should be $Ax=b$ has a unique solution for any $b$ in $\mathbb{R}^n$. By the way, here is a tutorial on how to type math formulas on this site: meta.math.stackexchange.com/questions/5020/… Commented Aug 13, 2015 at 15:11
• @bartgol Oops! I missed one. I added what you said.
– KIM
Commented Aug 13, 2015 at 15:18

The statement is that $Ax=b$ has to have solutions for each $b$, so that means $Ax=e_i$ is solvable for each of the standard basis elements $e_i$. Do you see how this implies that $A$ is invertible? In fact, this implies that there is a unique solution for each $b$.
It does sound false, but it is actually correct. If $A$ is not invertible, then you can find a right hand side such that there is no solution to the linear system. In other words, it is impossible to build a matrix $A$ such that there are infinitely many solution for all $\mathbf{b}\in\mathbb{R}^n$. Indeed, if there is a right hand side $\mathbf{b}_0$ that admits infinitely many solution, then for every possible right hand side $\mathbf{b}$ you either have infinitely many or zero solutions, and for at least one (in fact, a nontrivial dimensional subspace of $\mathbb{R}^n$) the system does not have a solution.