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If $A$ and $B$ are square matrices such that $AB = I$ where $I$ is identity matrix. Show that $BA = I$. I do not understand anything more than the following.

  1. Elementary row operations.
  2. Linear dependence.
  3. Row reduced forms and their relations with the original matrix.

If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.

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Dilawar: By "transform" of a matrix, did you mean transpose? [And if so, can you correct the question?] –  Srivatsan Jan 3 '12 at 5:56
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See also Dedekind finite rings. –  Bill Dubuque Apr 29 '12 at 20:41
    
Suppose $U$ and $V$ are sets and $T:V \to W$ is a function. Then $T$ has a left inverse $\iff$ $T$ is one-to-one, and $T$ has a right inverse $\iff$ $T$ is onto. Suppose now that $V$ is a finite dimensional vector space, and $T:V \to V$ is a linear transformation. Then $T$ has a left inverse iff $T$ is one-to-one iff $T$ is onto iff $T$ has a right inverse. From here, it's easy to show that any left inverse must also be a right inverse. (If $LT = I$ and $TR = I$, then $LTR = L \implies R = L$.) –  littleO Jun 19 at 23:24
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17 Answers

up vote 58 down vote accepted

Dilawar says in 2. that he knows linear dependence! So I will give a proof, similar to that of TheMachineCharmer, which uses linear independence.

Suppose each matrix is $n$ by $n$. We consider our matrices to all be acting on some $n$-dimensional vector space with a chosen basis (hence isomorphism between linear transformations and $n$ by $n$ matrices).

Then $AB$ has range equal to the full space, since $AB=I$. Thus the range of $B$ must also have dimension $n$. For if it did not, then a set of $n-1$ vectors would span the range of $B$, so the range of $AB$, which is the image under $A$ of the range of $B$, would also be spanned by a set of $n-1$ vectors, hence would have dimension less than $n$.

Now note that $B=BI=B(AB)=(BA)B$. By the distributive law, $(I-BA)B=0$. Thus, since $B$ has full range, the matrix $I-BA$ gives $0$ on all vectors. But this means that it must be the $0$ matrix, so $I=BA$.

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@MB: I think this argument is okay. It is very standard in intro linear algebra classes to prove that in a finite-dimensional vector space every spanning set contains a basis and every LI set can be enlarged to a basis. Moreover the proofs are algorithmic and the algorithm is always the same: row reduction. –  Pete L. Clark Sep 2 '10 at 8:38
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@Martin: I'm assuming $n$ is finite - maybe I should have been more explicit. I was assuming that the fact about $n$ dimensional subspaces coinciding with the original space was elementary enough to fit under his requirements. –  Davidac897 Sep 2 '10 at 8:45
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Besides, Martin's proof also uses this fact (how else can you show that the chain is stationary in a finite-dimensional vector space?). –  Davidac897 Sep 2 '10 at 9:16
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Yes I also use this or related statements and we have to use something like that; I just wanted to make precise for the readers what you are using. –  Martin Brandenburg Sep 2 '10 at 11:14
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Thank you David. This is what exactly I've been looking for. @MB : I really appreciate the way you have formulate the solutions. –  Dilawar Sep 2 '10 at 11:56
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So you want to find a proof of this well-known fact, which avoids the usual "indirect" proofs? I've also pondered over this some time ago.

We have the following general assertion:

Let $M$ be a finite-dimensional $K$-algebra, and $a,b \in M$ such that $ab=1$, then $ba=1$. [For example, $M$ could be the algebra of $n \times n$ matrices]

Proof: The sequence of subspaces $... \subseteq b^{k+1} M \subseteq b^k M \subseteq ... \subseteq M$ must be stationary, since $M$ is finite-dimensional. Thus there is some $k$ and some $c \in M$ such that $b^k = b^{k+1} c$. Now multiply with $a^k$ on the left to get $1=bc$. Then $ba=ba1 = babc=b1c=bc=1$. QED

No commutativity condition is needed. The proof shows more general that the claim holds in every left- or right-artinian ring $M$.

Remark that we needed, in a essential way, some finiteness condition. There is no purely algebraic manipulation with $a,b$, which shows $ab = 1 \Rightarrow ba=1$ (and shift operators provide a concrete counterexample). Every argument uses some argument of the type above. For example when you want to argue with linear maps, you have to know that every subspace of a finite-dimensional(!) vector space of the same dimension actually is the whole vector space, for which there is also no "direct" proof. I doubt that there is one.


PS. See here for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.

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Thanks Martin. Though I have to get accustomed to some of the words you used but certainly, this improved my understanding and I have one more approach in my list. –  Dilawar Sep 2 '10 at 7:41
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@MB: This is a very nice answer. A couple of tips: (i) it took me several readings to gather that "quadratic matrix" = "square matrix". This is included in the question, so it seems best to just assume it. (ii) I think you have some $A$'s which should be $a$'s (or vice versa). (iii) Since the OP asked for an elementary answer, please consider rewriting it to first treat exactly the case s/he asked for with no "fancy language" (ideals, Artinian algebras, etc.). Afterwards you can comment on how general your argument can be made. –  Pete L. Clark Sep 2 '10 at 8:23
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"The claim is also true for non-square matrices". No it is not. –  Laurent Lessard Sep 2 '10 at 8:41
    
Thanks for the suggestions. I've edited my answer. –  Martin Brandenburg Sep 2 '10 at 12:51
    
Compare to my proof here. –  Bill Dubuque Sep 2 '10 at 14:11
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If $\rm\:B\:$ is a linear map on a finite dimensional vector space $\rm\: V\:$ over field $\rm\:K\:,\:$ then trivially by finite dimensionality (cf. Note below) there is a nonzero polynomial $\rm\:p(x)\in K[x]\;$ such that $\rm\:p(B) = 0\:$. We may assume $\rm\:p(0) \ne 0\;$ by canceling any factors of $\rm\:B\;$ from $\rm\;p(B)\;$ by left-multiplying by $\rm\:A\;$ and using $\rm\;AB = 1\;$.

But $\rm\;\;\; AB=1 \ \ \Rightarrow\ \ (BA-1)\;\: B^n \;=\; 0\;$ for $\:\rm\;n>0\;$

thus we have $\rm\;\; 0 \:=\: (BA-1)\ p(B)\ =\ (BA-1)\ p(0) \;\Rightarrow\; BA=1 \quad\quad$ QED

This is essentially a special case of computing inverses by the Euclidean algorithm - see my Apr 13 1999 sci.math post.

Note: The existence of $\rm\;p(x)\;$ follows simply from the fact that $\rm\:V\;$ finite-dimensional implies the same for the vector space $\rm\:L(V)\:$ of linear maps on $\rm\:V\:$ (indeed if $\rm\:V\;$ has dimension $\rm n$ then a linear map is uniquely determined by its matrix of $\:\rm n^2\:$ coefficients). So $\rm\: 1,\: B,\: B^2,\: B^3,\:\cdots\;$ are $\rm\:K$-linearly dependent in $\rm\: L(V)$ which yields the sought nonzero polynomial annihilating $\rm\:B\;$.

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Nice proof. 1 + –  Martin Brandenburg Sep 2 '10 at 22:14
    
Nice proof, but why do you write a proof of BA=I⇒AB=I when the question asks a proof of AB=I⇒BA=I? (Of course this difference is inessential.) –  Tsuyoshi Ito Sep 3 '10 at 10:21
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I think it's the best answer so far. It only uses the fact that a system of linear equations with more unknowns than equations has a nontrivial solution. Alternative wording: There is a polynomial q such that q(B) = 1, q(0) = 0. Then BA-1 = (BA-1) q(B) = 0. –  Pierre-Yves Gaillard Sep 3 '10 at 14:18
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Thanks for the comments. I swapped A,B to match the OP's notation. My posts here are excerpts of my linked sci.math posts (which had A,B swapped). They're part of a handful of posts that I composed to provide different perspectives on this FAQ. –  Bill Dubuque Sep 3 '10 at 14:56
    
The Google Groups link no longer works, it seems; here's a replacement‌​. –  Steven Taschuk Feb 17 at 17:58
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Since there seem to be some lingering beliefs that an approach which does not make explicit use of the finite-dimensionality could be valid, here is a familiar counterexample in the infinite dimensional case.

Let $V = \mathbb{R}[t]$ be the vector space of real polynomial functions. Let $B: V \rightarrow V$ be differentiation: $p \mapsto p'$, and let $A: V \rightarrow V$ be anti-differentation with constant term zero: $\sum_{n=0}^{\infty} a_n t^n \mapsto \sum_{n=0}^{\infty} \frac{a_n}{n+1} t^{n+1}$.

These are both $\mathbb{R}$-linear maps and $B \circ A$ is the identity, but $A \circ B$ is not (the constant term is lost).

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1+. Then I also add the following example, because I think it is easier (but very similar): Consider the vector space of infinite sequences (of rational numbers, say), and the linear maps $A : (a_0,a_1,...) \to (0,a_0,a_1,...), B : (a_0,a_1,...) \to (a_1,a_2,...)$. Then $BA=1$, but not $AB=1$ since $AB$ maps $(a,0,...) \mapsto 0$. You may also restrict to finite sequences (without fixed lenght) and then cook up infinite matrices satisfying the properties. –  Martin Brandenburg Sep 2 '10 at 8:19
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@MB: Right, it's similar -- and indeed, also has a polynomial interpretation, as I'm sure you know. –  Pete L. Clark Sep 2 '10 at 8:24
    
see my post here which makes obvious the crucial role of finite dimensionality. See esp. the linked sci.math post. –  Bill Dubuque Sep 2 '10 at 14:24
    
@MB: Polynomials give a nice concrete model: Q[x] where the right shift is R = x = multiplication by x, and left shift L = (f(x)-f(0)/x. Then LR = I but RL f(x) = f(x)-f(0), so RL != I. For more see my old post bit.ly/Shift1-1notOnto –  Bill Dubuque Sep 2 '10 at 19:24
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Let $x_1, x_2, \dots, x_n$ be a basis of the space. At first we prove that $Bx_1, Bx_2, \dots, Bx_n$ is also a basis. To do it we need to prove that those vectors are linearly independent. Suppose it's not true. Then there exist numbers $c_1, c_2, \dots, c_n$ not all equal to zero such that $$c_1 Bx_1 + c_2 Bx_2 + \cdots + c_n B x_n = 0.$$ Multiplying it by $A$ from the left, we get $$c_1 ABx_1 + c_2 ABx_2 + \cdots + c_n ABx_n = 0,$$ hence $$c_1 x_1 + c_2 x_2 + \cdots + c_n x_n = 0$$ and so the vectors $x_1, x_2, \dots, x_n$ are also linearly dependent. Here we get contradiction with assumption that the vectors $x_i$ form a basis.

Since $Bx_1, Bx_2, \dots, Bx_n$ is a basis, every vector $y$ can be represented as a linear combination of those vectors. This means that for any vector $y$ there exists some vector $x$ such that $Bx = y$.

Now we want to prove that $BA = I$. It is the same as to prove that for any vector $y$ we have $BAy = y$. Now given any vector $y$ we can find $x$ such that $Bx = y$. Hence $$BAy = BABx = Bx = y$$ by associativity of matrix multiplication.

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Very nice!!! I just suggest that you replace "from the leftwe get" with "from the left we get", and "and so vectors" with "and so the vectors". –  Pierre-Yves Gaillard Sep 2 '10 at 16:37
    
@Pierre-Yves Gaillard: Corrected, thank you. –  falagar Sep 2 '10 at 16:39
    
Thanks falagra ! I almost got it in the first read itself. :-) I think you missed something to point out in the first paragraph after last line. I think you need to add one more line stating 'A contradiction.'. –  Dilawar Sep 2 '10 at 17:10
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That's essentially the proof that I gave - translated into the language of coordinates (bases). However, adding such extraneous information as bases only serves to obfuscate the simple essence of the manner, namely: injective maps cannot decrease heights (here = dimension = length of max subspace chain). As I stress in my post, the proof is a very intuitive one-line pigeonhole squeeze when viewed this way. –  Bill Dubuque Sep 2 '10 at 18:23
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@Bill, the thing is, «the essence of the matter» is not always, and is not for everybody, the best way to see something; obfuscation is in the eye of the beholder. I enjoyed reading through each of your proofs, but I find it quite natural that they be classified in the may-be-by-the-end-of-this-semester-I'll-be-able-to-understand-it category! –  Mariano Suárez-Alvarez Sep 8 '10 at 15:03
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It follows by the pigeonhole principle. Here's an excerpt from my Dec 11 2007 sci.math post:

Recall (proof below) $\rm\; AB \:\:=\:\:\: I \:\;\Rightarrow\; BA \:\:=\:\: I\;\;\:$ easily reduces to:

THEOREM $\;$ $\rm\;\;B\;$ injective $\rm\;\Rightarrow\:\: B\;$ surjective, $\:$ for linear $\rm\:B\:$ on a finite dim vector space $\rm\:V$

Proof $\rm\ \ \ B\;$ injective $\rm\;\Rightarrow\ B\;$ preserves injections: $\rm\;R < S \;\Rightarrow\; BR < BS\;$
Hence for $\rm\;\;\; R \;\: < \;\; S < \cdots < \; V\;\;$ a chain of maximum length (= dim $\rm V\:$)
its image $\rm\;BR < BS < \cdots < BV \le V\;\;\:\;$ would have length greater
if $\rm\ BV < V\:,\: $ hence, instead $\rm\:\:\:\ \ BV = V\:,\;\:$ i.e. $\rm\; B \;$ is surjective. $\;$ QED

Notice how this form of proof dramatically emphasizes the essence of the matter, namely that
injective maps cannot decrease heights (here = dimension = length of max subspace chain).

Below is said standard reduction to xxxjective form. See said sci.math post for much more,
including references to folklore generalizations, e.g. work of Vasconcelos in the seventies.

First, notice that $\rm\;\;\ AB = I \;\Rightarrow\: B\:$ injective, since $\rm\;A\;$ times $\rm\;B\:x = B\:y\;$ yields $\rm\;x = y\:,\:$ and

$\rm\ B\ $ surjective $\rm\ \Rightarrow\ BA = I \;\;$ since for all $\rm\;x\;$ exists $\rm\; y : \;\; x = B\:y = B(AB)\:y = BA \: x$

Combining them: $\rm\: AB = I \;\Rightarrow\: B\:$ injective $\rm\;\Rightarrow\; B\:$ surjective $\rm\;\Rightarrow\: BA = I$

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May be by the end of this semester, I'll be able to understand it. I have added it into 'things to understand' list. Thanks! –  Dilawar Sep 2 '10 at 14:49
    
@Dilawar: I've reformulated this proof to make it clearer. Please see my other post here which shows that it essentially reduces to Hilbert's infinite hotel on subspaces, viz. math.stackexchange.com/questions/4252 –  Bill Dubuque Sep 8 '10 at 15:35
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I prefer to think in terms of linear operators rather than matrices. A function has a right inverse iff it is surjective and it has a left inverse iff it is injective. For a linear operator, this means that having a right inverse is equivalent to having range equal to the entire space and having a left inverse is equivalent to having trivial kernel. For a linear operator on a finite-dimensional space, the dimension of its kernel + the dimension of its range = the dimension of the entire space, so this does the job.

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Motivation $\ $ If vector space $\rm V\:$ has a $1$-$1$ map $\rm\,B\,$ that's not onto, i.e. $\rm V > BV\:,\:$ then this yields an $\infty$ descending chain of subspaces by $\rm V > \: BV > \;\cdots\: > B^i V$ by repeatedly applying $\rm B\:$.

Theorem $\rm\:\ AB = 1 \;\Rightarrow\; BA=1\:$ for linear maps $\rm\:A,B\:$ on a finite dimensional vector space $\rm\: V $

Proof $\rm\;\; V > BV, \;$ times $\rm\; B^i\:\Rightarrow\: B^i V > B^{i+1} V \;$ (else $\rm\; B^i V = B^{i+1} V, \;$ times $\rm\; A^i \Rightarrow V = BV)$

$\rm\ \ \ \ \Rightarrow\rm\;\;\; V > BV > B^2 V > \cdots \:$ is an $\infty$ descending chain $\rm\; \Rightarrow\; dim\: V = \infty\,\:$ contra hypothesis.

Hence $\rm\ \ \ V = BV \;\Rightarrow\; (BA\!-\!1)V = (BA\!-\!1)BV = B(AB\!-\!1)V = 0 \quad$ QED

Remark $\;\;$ Hence vector space $\rm\:V\;$ has infinite dimension $\rm\;\iff V\:$ is Dedekind infinite, i.e. $\rm\:V\:$ has an isomorphic proper subspace $\rm BV,\:$ viz. the theorem proves $(\Leftarrow)$ and the converse follows by taking $\rm B\:$ to be a basis shift map $\rm\; (v_1,v_2,v_3,\cdots\:)\:\to\: (0,v_1,v_2,v_3,\cdots\:)\:,\;$ i.e. said simply, a vector space is infinite iff it has a subspace chain model of Hilbert's infinite hotel. $\:$ That is the simple essence of the matter.

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@Dilawar / readers: This is a reformulation of my prior pigeonhole proof intended to make clearer the essence of the matter. Please feel free to ask questions if anything is not clear. –  Bill Dubuque Sep 8 '10 at 15:39
    
Very clear exposition! Thanks! –  user1119 Nov 14 '10 at 10:13
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For a more elementary treatment ...

Fact. If the rows of $A$ are linearly dependent, then the rows of $A B$ are linearly dependent.

Proof of fact. Consider the 3x3 case, where the linearly-dependent rows of $A$ are $\mathbf{a}_1$, $\mathbf{a}_2$, $\mathbf{a}_3 = h \mathbf{a}_1 + k \mathbf{a}_2$ (for some scalars $h$ and $k$):

$$A = \begin{bmatrix}\mathbf{a}_1 \\ \mathbf{a}_2 \\ h\mathbf{a}_1 + k\mathbf{a}_2\end{bmatrix} = \begin{bmatrix}p & q & r \\ s & t & u \\ hp + ks & hq + kt & hr + ku \end{bmatrix}$$

Writing $\mathbf{b}_1$, $\mathbf{b}_2$, and $\mathbf{b}_3$ for the rows of $B$, we have

$$A B = \begin{bmatrix}p & q & r \\ s & t & u \\ hp + ks & hq + kt & hr + ku \end{bmatrix} \begin{bmatrix}\mathbf{b}_1 \\ \mathbf{b}_2 \\ \mathbf{b}_3\end{bmatrix} = \begin{bmatrix}p \mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ (hp + ks)\mathbf{b}_1 + (hq + kt)\mathbf{b}_2 + (hr + ku)\mathbf{b}_3 \end{bmatrix}$$

$$= \begin{bmatrix}p \mathbf{b}_1+ q\mathbf{b}_2 + r\mathbf{b}_3 \\ s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3 \\ h(p\mathbf{b_1}+q\mathbf{b}_2+r\mathbf{b}_3) + k(s\mathbf{b}_1 + t\mathbf{b}_2 + u\mathbf{b}_3) \end{bmatrix}$$

Generally, the linear dependence of the rows of $A$ carries over to the rows of the product, proving our Fact. (This reasoning actually shows the more-precise Fact that $rank(AB)\le rank(A)$.)

We can restate the Fact this way:

Re-Fact. If the rows of $AB$ are linearly independent, then the rows of $A$ are linearly independent.

To your question: If $A B = I$, then (by the Re-Fact) the rows of $A$ must be linearly independent. This implies that $A$ can be row-reduced to a diagonal matrix with no zero entries on that diagonal: the row-reduced form of $A$ must be the Identity matrix.

Note that row-reduction is actually an application of matrix multiplication. (You can see this in the equations above, where (left-)multiplying $B$ by $A$ combined the rows of $B$ according to the entries in the rows of $A$.) This means that, if $R$ is the result of some row combinations of $A$, then there exists a matrix $C$ that "performed" the combinations:

$$C A = R$$

If (as in the case of your problem) we have determined that $A$ can be row-reduced all the way down to the Identity matrix, then the corresponding $C$ matrix must be a (the) left-inverse of $A$:

$$C A = I$$

It's then straightforward to show that left and right inverses of $A$ must match. This has been shown in other answers, but for completeness ...

$$A B = I \;\; \to \;\; C (A B) = C \;\; \to \;\; (C A) B = C \;\; \to \;\; I B = C \;\; \to \;\; B = C$$

Once you start thinking (ahem) "outside the box (of numbers)" to interpret matrices as linear transformations of vectors and such, you can interpret this result in terms of mapping kernels and injectivity-vs-surjectivity and all the kinds of sophisticated things other answers are suggesting. Nevertheless, it's worth noting that this problem is solvable within the realm of matrix multiplication, plain and simple.

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+1, and to add: Row reduction/Gaussian elimination/LU decomposition is just a left multiplication of a matrix by a sequence of so-called "Gauss transforms", which are low-rank corrections to the identity matrix. Nothing sledgehammer-y about it! –  J. M. Sep 3 '10 at 2:12
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You might have a look at the following note "Right inverse implies left inverse and vice versa": http://www.lehigh.edu/~gi02/m242/08m242lrinv.pdf

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That paper invokes LU-factorization, which is quite a sledgehammer for a result that is really nothing but a pigeonhole squeezing argument (see my post). –  Bill Dubuque Sep 2 '10 at 14:55
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Bill: LU decomposition is merely Gaussian elimination, rearranged, and I would suppose equivalent to row reduction, which the OP says he can use. –  J. M. Sep 2 '10 at 17:11
    
@J.M. That doesn't stop it from being a sledgehammer compared to some other proofs given here. Not only are they simpler but more conceptual. I think the OP could easily understand at least one of those proofs. –  Bill Dubuque Sep 3 '10 at 18:13
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Coincidentally, a totally unrelated MathSciNet search turned up the following article, which gives a result along the lines of (but slightly stronger than) the one in Martin Brandenburg's answer.

http://www.math.uga.edu/~pete/Hill67.pdf

In particular:

Theorem (Hill, 1967): Let $R$ be a ring satisfying the ascending chain condition on right ideals. Then:
a) For any $n \in \mathbb{Z}^+$, the matrix ring $M_n(R)$ also satisfies ACC on right ideals.
b) If $R$ satisfies ACC on right ideals and $a,b \in R$ are such that $ab = 1$, then also $ba = 1$.

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here is a try. we will use AB = I to show that the columns of $B$ are linearly independent and then use that to show $BA$ is identity on the range of $B$ which is all of the space due to linear independence of the columns of $B$. This implies that $BA$ is identity. linear independence of the columns of follows if we can show Bx = 0 implies x = 0. assume $Bx = 0, x = Ix = ABx = A0 = 0$. now to show that $BA$ is an identity on range of $B$ we have $(BA)Bx = B(AB)x = Bx $ and and we are done.

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First some notation, given a $n\times n$ matrix $M$ write $\mathbf m_j$ for the $j$-th column of $M$. In what follows $E$ is the identity $n\times n$ matrix, so it's clear what the $\mathbf e_1,\ldots,\mathbf e_n$ are.

Assume $AB = E$. Then for each matrix $n\times 1$, $\mathbf{y}$ the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

Indeed, if $\mathbf c$ is solution then \begin{gather*} \mathbf y = B\mathbf c\\ A\mathbf y = (AB)\mathbf c = \mathbf c \end{gather*}

Particularly, for each $j\in \{1,\ldots,n\}$ the system $$B\mathbf x = \mathbf e_j$$ has exactly one solution, namely $$\mathbf x = A\mathbf e_j = \mathbf a_j.$$ Thus $$B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$.

Write $C = BA$. Since $$\mathbf c_j = B\mathbf a_j = \mathbf e_j$$ for each $j\in \{1,\ldots,n\}$, $C$ and the identity matrix both have the same columns, so $$C = E.$$

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Disclaimer: The following proof was written by me, but the main idea was given to me by a friend in some discussion, and he probably proved it too.

The main idea is that this claim is true when $A$ is an elementary row matrix, and we can induct on the number of row operations needed to be done on $A$ in order to reduce it.

  • Some background:

    An elementary row-operation matrix $E$ corresponds to one of 3 operations: row addition, row multiplication and row switching. Every $E$ has a matrix $E'$ which is another row-operation matrix of the same type, such that $EE'=E'E=I$ (you don't need to term "inverse of a matrix" to believe this - this "inverse" $E'$ is described here and you can verify it yourself).

    By performing elementary row operations on a matrix $A$, one gets the following equality: $E_{k} E_{k-1} \cdots E_1 A = C$, where $C$ is in row reduced form and $E_i$ are elementary row matrices.

Claim 1: $A$ can be written as $E_{k} \cdots E_{1}C $, where $E_i$ are some elementary row matrices (not neccesarily the same matrices from before). Proof: Multiply by $E'_1 E'_2 \cdots E'_k$ from the left.

Claim 2: if $AB=I$, then this $C$ is the identity. Proof: By using determinant this is fairly easy. The condition $AB=I$ ensures $\det(C) \neq 0$ by multiplicativity of determinant, and since $C$ is upper triangular with every leading coefficient being 1 (the only nonzero entry in its column), the determinant is non-zero iff $C=I$. This can also be proved without determinants, but it is not the main focus of the proof (but it is an important claim nonetheless).

So we showed that $A=E_{k} E_{k-1} \cdots E_{1}$.

Claim 3: If $k=0$ or $1$ ($k$ is the number of row operations required to reduce $A$), then $AB=I \implies BA=I$. Proof: If $k=0$, $A=I \implies B=I \implies BA=I$. If $k=1$, $A$ is an elementary row matrix: $A=E$, so $AB=EB=I \implies B = E'EB = E' \implies BA=E'E=I$.

Claim 4: $AB = I \implies BA = I$. I will induct on $k$. Let's assume this is true while $k \le n$, $n \ge 1$. I am going to prove this for $k=n+1$. Let $A=E_{n+1}E_{n} \cdots E_{1}$.

$$ AB = E_{n+1} \cdots E_{1} B = I \implies E_{n} \cdots E_{1} B = E'_{n+1} \implies $$ $$ (E_{n} \cdots E_{1}) (B E_{n+1})= I \implies (B E_{n+1})(E_{n} \cdots E_{1}) = I \implies $$ $$ BA= I$$

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The crucial fact is to show: $T\text{ injective} \iff T\text{ surjective}$
(That is hidden by saying: $A B=\mathbb{1}\iff B A=\mathbb{1}$)

But it holds in general: $\dim\mathcal{D}(T)=\dim\mathcal{N}(T)+\dim\mathcal{R}(T)$
So the above assertion holds for finite dimensional spaces.

To get a grasp why the assertions are actually the same think of the following:
As you might know: $f\circ h=id\Rightarrow f\text{ surjective}(h\text{ injective})$
And similarly: $h\circ f=id\Rightarrow f\text{ injective}(h\text{ surjective})$
Conversely: $f\text{ injective}\Rightarrow h_0\circ f=id$ ...for some $h_0$
And similarly: $f\text{ surjective}\Rightarrow f\circ h_0=id$ ...for some $h_0$
Hoewever, the constructions will be more tedious...

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An alternative.

Let $\textbf{A}$ be a $n \times n$ matrix, and let $\lambda_k$ be the eigenvalues, then we can write

$$ \prod_{k=1}^n \Big( \textbf{A} - \lambda_k \textbf{I} \Big) = 0 $$

So we obtain

$$ \textbf{A}^n = a_0 \textbf{I} + \sum_{k=1}^{n-1} a_k \textbf{A}^k $$

Let us write

$$ \textbf{B} = b_0 \textbf{I} + \sum_{k=1}^{n-1} b_k \textbf{A}^k$$

It is clear that

$$ \textbf{A} \textbf{B} = \textbf{B} \textbf{A} $$

We also find that

$$ \begin{eqnarray} \textbf{A} \textbf{B} &=& b_0 \textbf{A} + \sum_{k=1}^{n-1} b_k \textbf{A}^{k+1}\\ &=& b_0 \textbf{A} + \sum_{k=1}^{n-2} b_k \textbf{A}^{k+1} + b_{n-1} \textbf{A}^{n}\\ &=& b_0 \textbf{A} + \sum_{k=1}^{n-2} b_k \textbf{A}^{k+1} + b_{n-1} a_0 \textbf{I} + b_{n-1} \sum_{k=1}^{n-1} a_k \textbf{A}^k\\ &=& b_{n-1} a_0 \textbf{I} + \Big( b_0 + b_{n-1} a_1 \Big) \textbf{A} + \sum_{k=2}^{n-1} \Big( b_{k-1} + b_{n-1} a_k \Big) \textbf{A}^k\\ \end{eqnarray} $$

When we set (for $a_0 \ne 0$)

$$ \begin{eqnarray} b_{n-1} a_0 &=& 1\\ b_{k-1} + b_{n-1} a_k &=& 0 \end{eqnarray} $$

we obtain

$$ \textbf{A} \textbf{B} = \textbf{I} $$

So we can write

$$ \textbf{B} = -a_0^{-1} a_1 \textbf{I} - \sum_{k=1}^{n-2} a_0^{-1} a_{k+1} \textbf{A}^k + a_0^{-1} \textbf{A}^{n-1} $$

or

$$ \textbf{A}^{-1} = -a_0^{-1} \left( a_1 \textbf{I} + \sum_{k=1}^{n-2} a_{k+1} \textbf{A}^k - \textbf{A}^{n-1} \right) $$

the inverse of $\textbf{A}$ can be expressed as a linear sum of $\textbf{A}^k$.

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Suppose $AB=I.$

$(BA)B=B(AB)=BI=B$

So $(BA)B=B \implies BA=I$

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The last implication $(BA)B = B \implies BA = I$ is not clear at all. –  azimut Feb 17 at 17:20
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