# How to prove my statement is correct about invertible matrices?

1. SO far i showed that if A matrix is left invertible (L) then in Ax = b, x has at most 1 solution. I got that LAx = x = Lb, so x = Lb

2. for right inverse (R) of A, in Ax = b, x has at least one solution. I got that x = Rb.

In the book it says that x has 1 solution for case 1, and at least one solution for case 2. Can someone explain my WHY? HOW?

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Yes you are right. In 1., you SOLVED the equation, and thus you PROVED that $x=Lb$.

In 2., you can see easily that $x=Rb$ works, but this only proves that this is one solution. It could be the only one, or there could be others....

PS I don't know how you got the $x=Rb$, it is easy to guess or, to observe that $b=ARb$. Note that once you see that $Ax=ARb$, you cannot cancel the $A$ unless $A$ has a left inverse...

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ARb=Ib=b => x=Rb. that is how i found the x ;) – ASROMA Oct 19 '12 at 5:40
@everybody The thing is that this claims are known, so i need to prove them, not to disprove. Since the book says that for one case it has only one solution and for the other case more than 1, this means that somehow x=Lb implies to a single solution – ASROMA Oct 19 '12 at 5:48
I am just thinking if we can find more then 1 left inverses for A then we can have many Lbs, so thus many solutions... – ASROMA Oct 19 '12 at 5:49
@VaheMusinyan The second one doesn't say that you can find more than one. It sais that you can find at least one... And this is exactly how you prove it, by finding ONE.... Without knowing more about $A$ you cannot conclude anything more... – N. S. Oct 19 '12 at 14:52
@VaheMusinyan If the problem wanted you to prove that you can find more than one, it would had asked you to prove that in the second case there are infinitely many solutions ;) – N. S. Oct 19 '12 at 14:53

What you did for the left inverse is fine, but what you did for the right inverse doesn’t work: you know that $AR=I$, but you don’t know that $RA=I$, so you can’t say that $x=Rb$. (Actually, you may have seen that $ARb=Ib=b$ and realized on that basis that $Rb$ is a solution; if so, you’re right, but that doesn’t say anything one way or the other about whether it’s the only solution. In any case, you should justify it.)

Suppose that $A$ is an $m\times n$ matrix with a right inverse $R$. Then $R$ is $n\times m$, and $AR$ is $m\times m$. This implies that the $m$ rows of $A$ are linearly independent (why?), i.e., that the rank of $A$ is $m$, and hence that $m\le n$. From here there are several ways to argue that $Ax=b$ has at least one solution, depending on what you’ve already proved. One very straightforward way, however, is to consider what happens to the augmented matrix $\begin{bmatrix}A&b\end{bmatrix}$ when you row-reduce it.

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Actually I think he saw that $x=Rb$ works. It is easy to guess that this is a solution. – N. S. Oct 19 '12 at 5:32
@N.S.: You may be right; I’m not sure. But I’ll modify the wording to account for that possibility. – Brian M. Scott Oct 19 '12 at 5:34