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I have read this question page (If $AB = I$ then $BA = I$) by Dilawar and saw that most of proofs are using the fact that the algebra of matrices and linear operators are isomorphic.

But from a simple view, matrix is only an structured set of scalars, and the fact that the dot product of the i-th row of $A$ and the j-th column of $B$ equals to the Kronecker delta is just a componentwise algebraic informarion.

Then I started to wonder if there's any "brutal" proof that does not visit the "higher" domain of algebraic structures and just uses the simple componentwise algebraic operations to prove that the dot product of the i-th row of $B$ and the j-th column of $A$ equals to the Kronecker delta from the given condition. (Proof that even a "machine" can do)

Should we think a matrix as more than a mere 'number box' to show $BA=I$?

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    $\begingroup$ I think you can do this. Just write out the matrix multiplication explicitly in terms of the entries. But why would you ever want to do this when there are easier ways available? $\endgroup$ – Potato Nov 1 '13 at 14:32
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    $\begingroup$ Matrix products are designed to be representative of linear operators. Kind of a shame not to use the extra structure. If you are familiar with the proof of invertibility using elementary row operations to reduce the matrix, that possibly fits the bill. $\endgroup$ – EuYu Nov 1 '13 at 14:35
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    $\begingroup$ Of course the essence of studying linear algebra is to understand that matrices and linear mappings are two faces of one mathematical obeject, but I always felt some kind of reductionistic desire to deal with a matrix as if it is nothing more than an oredered scalar set. It's something like... if we can go around in easy way, then there should be some foolish straight way, too! $\endgroup$ – generic properties Nov 1 '13 at 14:40
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    $\begingroup$ I can somewhat understand your drive to deal with matrices in and of themselves; I myself have felt the opposite drive, to deal with linear operators in and of themselves and avoid matrices as much as possible. Usually we're studying the maps, not the matrices, anyway. $\endgroup$ – Muphrid Nov 1 '13 at 15:09
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    $\begingroup$ Here is the reason why I think the proof will be difficult or at least tedious. You want to prove $\mathbf{b}_{i\rightarrow}\cdot\mathbf{a}_{j\downarrow}=\delta_{ij}$ given the fact that $\mathbf{a}_{i\rightarrow}\cdot\mathbf{b}_{j\downarrow}=\delta_{ij}$. To me, these are local properties and their equivalence is most easily seen as a special case of the global property that $AB = BA = I$ (which you are trying to avoid if I understand correctly). Trying to go from one to the other is like trying to go from point $a$ to point $b$ without stepping through the intermediate space. $\endgroup$ – EuYu Nov 1 '13 at 16:13
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Here is a sketch of a "brutal proof" of the sort you imagine. (On rereading the other answers, you've already seen this. But the notes at the end may be interesting.) A more detailed version of the following can be found in "Bijective Matrix Algebra", by Loehr and Mendes.

Remember that the adjugate matrix, $\mathrm{Ad}(A)$, is defined to have $(i,j)$ entry equal to $(-1)^{i+j}$ times the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$-th row and $j$-th column from $A$.

Write down a brute force proof of the identity: $$\mathrm{Ad}(A) \cdot A = A \cdot \mathrm{Ad}(A) = \det A \cdot \mathrm{Id}_n$$ by grouping like terms on both sides and rearranging.

Likewise, write down a brute force proof of the identity $$\det(AB) = \det(A) \det(B).$$ So, if $AB=\mathrm{Id}$, you know that $\det(A) \det(B)=1$.

Now compute $\mathrm{Ad}(A) ABA$ in two ways: $$(\mathrm{Ad}(A) A)BA = \det(A) BA$$ and $$\mathrm{Ad}(A) (AB) A = \mathrm{Ad}(A) A = \det(A) \mathrm{Id}.$$ Since $\det(A) \det(B) =1$, you also have $\det(B) \det(A)=1$, and get to deduce that $BA = \mathrm{Id}$.

There is some interesting math here. Let $R$ be the polynomial ring in $2n^2$ variables $x_{ij}$ and $y_{ij}$ and let $X$ and $Y$ be the $n \times n$ matrices with these entries. Let $C_{ij}$ be the entries of the matrix $XY-\mathrm{Id}$ and let $D_{ij}$ be the entires of the matrix $YX-\mathrm{Id}$. Tracing through the above proof (if your subproofs are brutal enough) should give you identities of the form $D_{ij} = \sum_{k,\ell} P_{k \ell}^{ij} C_{ij}$. It's an interesting question how simple, either in terms of degree or circuit length, the polynomials $P_{k \ell}^{ij}$, can be made.

I blogged about this question and learned about some relevant papers (1 2), which I must admit I don't fully understand.

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  • $\begingroup$ I like that this answer reframes the question from "Can you prove this 'by computation alone'" to asking exactly how painful would it be to do so, and why. $\endgroup$ – Semiclassical Aug 6 '14 at 21:47
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I started to wonder if there's any "brutal" proof that does not visit the "higher" domain of algebraic structures and just uses the simple componentwise algebraic operations to prove that the dot product of the i-th row of B and the j-th column of A equals to the Kronecker delta from the given condition.

Should we think a matix as more than a mere 'number box' to show BA=I?

It sounds a little bit like you're suggesting that merely having $A$ and $B$ and the formal multiplications of elements in the matrix equation $AB=I$ is enough to show that $BA=I$.

But the truth of this claim depends on properties of the ring that the matrix entries comes from!

So no, the matrices and their multiplication alone are not sufficient to prove the statement $AB=I\implies BA=I$.


In several places on math.SE, there is an example of a ring $R$ with two elements $a,b$ such that $ab=1$ and $ba\neq 1$, and that gives an example of a matrix with $n=1$ where the statement isn't true.

It takes about as long to give the example as to find an appropriate link, so for completeness I'll include it. Take $R$ to be the ring of linear transformations of the vector space of polynomials $\Bbb R[x]$, and take $b$ to be the derivative operator on $\Bbb R[x]$, and $a$ to be the antiderivative with constant set to $0$. It's easy to see that $ab=I$, but $ba\neq I$ because $ba(2)=b(0)=0$.

And if you're really hankering for an example with $n>1$, just use $M_2(R)$ over this ring. Obviously $\begin{bmatrix}a&0\\0&a\end{bmatrix}$ and $\begin{bmatrix}b&0\\0&b\end{bmatrix}$ do the trick.


The proposition is true for large classes of rings, though. A ring $R$ is called stably finite if, for every $n\in \Bbb Z^+$ and $A,B\in M_n(R)$, $AB=I\implies BA=I$. Perhaps the two broadest classes of rings with this property are right Noetherian rings and commutative rings.

If all you are interested in is commutative rings, then I can think of no better solution than the $A\mathrm{Adj}(A)=\mathrm{Adj}(A)A=\det A\cdot I$ proof given already. It is essentially brute force computation with lots of determinants, all phrased in terms of the entries of the matrices. Determinants are not available (or are at least extremely complicated) in noncommutative rings, so this is why commutativity saves the day.

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  • $\begingroup$ I suppose the OP's question can be restated as: "How does one prove that a matrix over a commutative ring is a stably finite ring with algebra alone?" $\endgroup$ – Semiclassical Aug 5 '14 at 13:27
  • $\begingroup$ @Semiclassical : That's true. I've now added a little to weigh in on the commutative case in particular. $\endgroup$ – rschwieb Aug 5 '14 at 13:37
  • $\begingroup$ Nice. It seems to all boil down to the fact that proving that the adjugate formula in a general way isn't tractable 'by algebra alone.' $\endgroup$ – Semiclassical Aug 5 '14 at 13:42
  • $\begingroup$ @Semiclassical If commutativity is part of your rules of algebra, it is, isn't it? $\endgroup$ – rschwieb Aug 5 '14 at 13:59
  • $\begingroup$ I meant in the sense of the OP, i.e. by brute force alone. Obviously that works easily for the 2-by-2 case; proving for arbitrary $n$ by brute force seems a different matter. $\endgroup$ – Semiclassical Aug 5 '14 at 14:04
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Here's an answer I wrote to a question which perhaps gets a bit closer to yours than those in the link provided (only in the sense that it appeals to an explicit formula).

Right invertible and left zero divisor in matrix rings over a commutative ring

The point is: the Laplace rule for determinants yields universal polynomial identities which hold for any commutative ring:

$$\newcommand\adj{\operatorname{adj}}\adj(A) A=A \adj(A)=\det(A)I.$$

From this point of view, the proof is algorithmic, though it relies on the fact that $\det(AB)=\det(A)\det(B)$, which is an algebraic result (the fanciest way to phrase it is through the use of top exterior powers, which might be overkill).

I doubt you might get your question solved, for the following reason: we have one set of linear equations on the coefficients of $B$ (the superindex will denote row no. and the subindex will refer to column no.):

$$A^i B_j= \delta_i^j.$$ Here the variables belonging to $B_j$ appear exactly in $n$ equations, and do not appear explicitly in any other.

The desired result is: $$A_i B^j=\delta_i^j.$$ same thing here, but with $B^j$. In order to perform the linear combinations necessary one needs to use algebra, for instance:

$A^i B_j=\delta_j^i$ ($B_i$) implies that in our ground ring the coefficients $B_j$ generate the ideal $(1)$.

The most elegant way, and perhaps the only possible, is to place the numbers in matrix form and thus to isolate $B$ in terms of the adjoint matrix of $A$, or vice versa.

The solution is a certain quotient of polynomial expressions in the coefficients, which is not an easy formula to come up with without the proper algebraic framework. Thus, I surmise that a solution such as you ask would be very hard to come by.

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