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The proofs are already presented at length here. And I found that most of the solutions used the fact that if the linear map $B$ is one-one then it must be onto.

But this fact appears trivial because of $AB = I$. Consider the equation $Bx = y$ and if there are two solutions $x, x'$ then $Bx = Bx' = y$ by multiplying by $A$ on left we get $x = x'$. This is what is referred to as $B$ is one-one. But the same argument also shows that $x = x' = Ay$ so that $B$ is onto (for each $y$ we have found $x = Ay$). I wonder why do the proofs in the linked question used some linear-algebra to show that $B$ is onto.

It appears that most proofs use the left multiplication by $A$ to ensure that $x = x'$ but somehow this does not seem to guarantee $x = Ay$. Why is that the case? Or am I missing something?

Update: Thanks to all those who replied. I got hold of my mistake. The argument proves that if $Bx = y$ has a solution it must be $x = Ay$. It does not show that there is a solution. Sorry to bother you all for such trivial matter (and if all agree I may delete this silly question). Existence of a left inverse of $B$ does not guarantee the solution of $Bx = y$. Rather it is the existence of right inverse of $B$ which is needed here to get a solution.

Ben Grossmann
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Paramanand Singh
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  • Injectivity of $B$ follows as you pointed out. But how did you conclude that for any $y$ we can find $x$ for which $y = Ax$? It seems to me that your starting point was the equation $y = Bx$, which already assumes that $y \in \operatorname{Ran}(B)$. – Sangchul Lee Apr 14 '17 at 05:28
  • @SangchulLee: If I multiply the equation by $A$ on left I get $x = Ay$. – Paramanand Singh Apr 14 '17 at 05:29
  • @ParamanandSingh For a fixed $y$, it is clear that $Bx = y$ implies that $x = Ay$. It is possible, however, that $Bx = y$ is false for all $x$. Note that the implication does not apply in reverse, which is to say that the converse is untrue. – Ben Grossmann Apr 14 '17 at 05:32
  • @ParamanandSingh: Regarding part of your update, there is no need to apologize! For deletion, I don't know a reason to do so but I don't have a strong opinion about it. For the mechanics of it, you would have to unaccept, and all answerers with positive answer score would have to delete their answers first. So if you decide it is worth trying, asking Omnonnomnom and I to delete our answers (after you unaccept) would come first. – Jonas Meyer Apr 14 '17 at 05:44
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    @JonasMeyer: Ok I will keep the question. It will serve as a guidepost to other people like me. – Paramanand Singh Apr 14 '17 at 05:46
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    @ParamanandSingh: More generally, it is a special property for a ring to satisfy $ab=1\implies ba=1$, given the name "Dedekind finite" by analogy with Dedekind infinite sets. Martin Brandenburg's answer at the linked question points out that matrix rings over commutative rings are examples. Finite rings are also examples: https://math.stackexchange.com/questions/138541/left-inverse-implies-right-inverse-in-a-finite-ring – Jonas Meyer Apr 14 '17 at 05:54
  • @JonasMeyer: Thanks for that link on finite rings. Of late I have started to get hold of linear-algebra and abstract-algebra topics (and that's reason for such questions) and MSE is a big help here (apart from a few textbooks which I posses). – Paramanand Singh Apr 14 '17 at 05:59

2 Answers2

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But the same argument also shows that $x = x' = Ay$ so that $B$ is onto (for each $y$ we have found $x = Ay$)

It's not totally clear to me what you mean by that sentence. Note, however, that in the beginning of that first argument, we say if there are two solutions $Bx = Bx' = y$. We cannot assume (without more information) that there any solutions to $Bx = y$.

Note that there are certainly matrices that have a left inverse, but fail to have a right inverse. For instance, with $$ B = \pmatrix{1&0\\0&1\\0&0} $$ Then $A = B^T$ is a left inverse, so that $B$ is one to one. It is clear, however, that the transformation $x \mapsto Bx$ fails to be onto. Clearly, your argument must go wrong somewhere.

It is also notable that if $V$ is infinite dimensional, then a linear $T:V \to V$ may be one to one but not onto. As a classic example, consider the map $f(x) \mapsto \int_0^x f(t)dt$.

(I like Jonas's answer a lot. Looking at arbitrary functions allows for a useful analogy).

Ben Grossmann
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(There are some parallels with Omnomnomnom's answer, and we even chose the same quote to start our answers. This note helps visually distinguish them.)

But the same argument also shows that $x = x' = Ay$ so that $B$ is onto (for each $y$ we have found $x = Ay$).

To show $B$ is onto, you need to show that for each $y$ there exists $x$ with $Bx=y$. Your claim is that $x=Ay$ will do, so let's see if this works: $Bx=BAy=\cdots$? We'd like to say that's $y$, but it assumes the result that $BA=I$.

You have to use linear algebra because for functions $f,g:X\to X$, $f\circ g(x)=x$ for all $x\in X$ implies that $g$ is one-to-one and $f$ is onto, but does not imply that $g$ is onto or that $f$ is one-to-one. For example, $X=\mathbb R$, $g(x)=e^x$, $f(x)=\ln(x)$ when $x>0$, $f(x)=0$ when $x\leq 0$.

There are linear counterexamples on infinite dimensional spaces. Let $V$ be the space of real sequences $(a_1,a_2,\ldots)$, and let $A$ and $B$ be the left and right shifts, $A(a_1,a_2,a_3,\ldots)=(a_2,a_3,a_4,\ldots)$, $B(a_1,a_2,a_3,\ldots)=(0,a_1,a_2,\ldots)$. Then $AB=I$, but $BA\neq I$.

Jonas Meyer
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  • Got it!. See my update in question. +1 to you – Paramanand Singh Apr 14 '17 at 05:37
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    Going off the notion of functions $f:X \to X$ - just as for finite $X$, $f:X \to X$ is injective iff it is surjective, so does the same hold for linear $T:V \to V$ when $V$ is finite dimensional. – Ben Grossmann Apr 14 '17 at 05:40
  • @Omnomnomnom: I like this analogy! The proofs given in the linked question show in a variety of way how injective leads to surjective for finite dimensional spaces. – Paramanand Singh Apr 14 '17 at 05:42
  • @ParamanandSingh in fact, Bill's "proof by the pigeonhole principle" makes direct use of this analogy – Ben Grossmann Apr 14 '17 at 05:45
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    @Omnomnomnom: Talking of Bill's proof (or proofs), I liked it very much especially the one where he shows that there will be an ifninite chain of subspaces if the map $B$ is not onto. – Paramanand Singh Apr 14 '17 at 05:48