I am interested in the case that AA is a matrix over a commutative ring, not necessarily a field. Is it still true that if Ax=bAx = b has a solution for every bb, then AA is invertible? I know that in the general setting, AA having the trivial nullspace does not imply that it is invertible. However, I cannot seem to find a counterexample to the fact in the title of the question, so I am starting to believe it is true. Any ideas how to prove it?

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3 Answers

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For each standard basis vector e1,…,ene_1, \dots, e_n, we have some b1,…,bnb_1, \dots, b_n such that Abi=eiA b_i = e_i, respectively. Then simply ‘squishing’ the b1,…,bnb_1, \dots, b_n together into a n×nn \times n matrix directly gives A−1A^{-1}.

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You built a matrix BB that satisfies AB=IAB=I. However, does this imply that also BA=IBA=I, over an arbitrary commutative ring?

– Marc van Leeuwen

yesterday

It doesn’t directly, but since the determinant of AA is a unit in RR, then AA will have a two-sided inverse (given by (det A)−1⋅adj A(\textrm{det } A)^{-1} \cdot \textrm{adj } A, which is then necessarily equal to BB.

– D_S

yesterday

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@MarcvanLeeuwen The result that, for square matrices AA and BB, AB=IAB=I implies BA=IBA=I holds over any commutative ring (and one doesn’t need determinants for it, but with determinants it’s easy).

– egreg

yesterday

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@egreg That is what my answer says (except for the “one doesn’t need determinants” part; I guess that is so, but one does need commutativity, and finite rank, so one might as well use determinants).

– Marc van Leeuwen

yesterday

The result is wrong as stated, since by taking for AA a rectangular matrix (more columns than rows) one easily gets counterexamples. I will therefore suppose you implicitly assumed AA to be square (a necessary condition for being invertible).

This is then a complement to the answer by basket, using a simple argument found in this answer. By the hypothesis you can find a matrix BB such that AB=IAB=I, since this amounts for every column of BB to an equation of the form Ax=cAx=c, where cc is the corresponding column of II. Now taking determinants we get det(A)det(B)=1\det(A)\det(B)=1, so the determinant of AA is invertible in your commutative ring. Then AA as well is invertible in the matrix ring, namely \det(B)\det(B) times the cofactor matrix of AA gives A^{-1}A^{-1}.

That was all that was asked for, but multiplying AB=IAB=I to the left by A^{-1}A^{-1} shows that in fact B=A^{-1}B=A^{-1} and hence BA=IBA=I. Note that commutativity of the base ring (which allowed taking determinants) is essential; the result does not hold over non-commutative rings (even for 1\times11\times1 matrices, since for a scalar having a right inverse now does not imply having a left inverse).

As user basket wrote there are solution vectors x_ix_i for

A x_i = e_i

A x_i = e_i

where e_ie_i is the ii-th canonical basis vector.

Then X = (x_1 x_2 \dotsb x_n)X = (x_1 x_2 \dotsb x_n) fulfills A X = IA X = I.

If there exists a YY with Y A = IY A = I, then we would have

Y = Y I = Y (A X) = (Y A) X = I X = X

Y = Y I = Y (A X) = (Y A) X = I X = X

I still lack a simple argument, why there should exist a left-inverse for AA as well, or rather why would XX work as left-inverse too.

Marc’s argument looks nice but would have me to review determinant theory over commutative rings instead of the usual fields. 🙂

Let us assume AA is invertible and

X A =: B \ne I

X A =: B \ne I

Then

A B = A (X A) = (A X) A = I A = A

A B = A (X A) = (A X) A = I A = A

and

A (B – I) = 0

A (B – I) = 0

If B- I \ne 0B- I \ne 0 then at least one of its row vectors is non-zero, let us call it rr and we got

\DeclareMathOperator{ker}{ker}r \in \ker A\DeclareMathOperator{ker}{ker}r \in \ker A.

For regular linear algebra (over a field) this would prevent AA from being invertible. No idea what happens for just commutative rings.

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We have commutivity of the ring that AA and XX are over, but that does not mean by itself that AA commutes with XX.

– Paul Sinclair

yesterday

Thanks, you are right. My mistake.

– mvw

yesterday