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Consider a general system $AX=B$ of m linear equations in n unkons, where m and n are not necessarily equal. The coefficient matrix $A$ may have a left inverse $L$, a matrix such that $LA=I_n$. If so we may try to solve the system as follows: $$AX=B, LAX=LB, X=LB$$

But when we try to check our world by running the solution backward, we run into trouble: If $X=LB$ then $AX=ALB$. We seem to want L to be a right inverse, which isn't what was given.

Exactly what does the sequence of steps made above show? What would the existence of a right inverse show?

Amr
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amir
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3 Answers3

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What you've shown is that if $AX=B$ and $L$ is a left inverse, then $X=LB$. However there's no guarantee that $X=LB$ is indeed a solution, since the implication is not necessarily reversible. If $L$ also happens to be a right inverse, then you avoid the trouble you mention.

It's worth noting that the implication sign ($\Rightarrow$) is useful in this context to clarify the logical connections between your statements. Your argument above then goes something like this. Suppose that $LA=I_n$. Then

\begin{eqnarray*}AX=B&\Rightarrow& LAX=LB\\ &\Rightarrow& I_nX=LB\\ &\Rightarrow& X=LB.\end{eqnarray*}

Daniel King
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Shane O Rourke
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The reasoning above establishes that $X=LB$ is a necessary condition for having a solution. After all one has obtained this equation by supposing $X$ was a solution, and applying valid operations to that. Since there is only one $X$ that satisfies the equation, either $LB$ is the unique solution, or there is no solution at all. As your check shows, we cannot be sure that $LB$ is a solution (and indeed it does not have to be is $A$ is a rectangular matrix): the given condition may not be a sufficient one.

The existence of a right inverse$~R$ to$~A$ would establish the $X=RB$ is a sufficient condition for having a solution; stated differently, $RB$ is certainly a solution, but there might well be other solutions.

(Talking about being necessary and/or sufficient condition for an equation that completely specifies$~X$ mays seem a bit odd, but it is perfectly valid.)

Marc van Leeuwen
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Since $LA=I$ the matrix $L$ is regular and hence if $LZ=0$ for some matrix $Z$ we have $Z=0$.

Now, note that, if $LA=I$ then $L(AL-I)=LAL-L=0$, so we have $AL-I=0$, that is $AL=I$.

zacarias
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