1
$\begingroup$
  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?

2 Answers 2

0

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...

  • 0
    Well, if $L_1, L_2$ are left inverses of$A$then $L_1b=L_2b$ because $b \in Col(A)$ ;)2012-10-21
1

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.

  • 0
    @N.S.: You may be right; I’m not sure. But I’ll modify the wording to account for that possibility.2012-10-19