If $AB = I$ then $BA = I$

Question

If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the 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.

Answer 1

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 $I=BA$.

Answer 2

We have the following general assertion:

Lemma. Let $A$ be a finite-dimensional $K$-algebra, and $a,b \in A$. If $ab=1$, then $ba=1$.

For example, $A$ could be the algebra of $n \times n$ matrices over $K$.

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

The proof also works in every left- or right-Artinian ring $A$. In particular, the statement is true in every finite ring.

Remark that we need in an essential way some finiteness condition. There is no purely algebraic manipulation with $a,b$ that shows $ab = 1 \Rightarrow ba=1$.

In fact, there is a $K$-algebra with two elements $a,b$ such that $ab=1$, but $ba \neq 1$. Consider the left shift $a : K^{\mathbb{N}} \to K^{\mathbb{N}}$, $a(x_0,x_1,\dotsc) := (x_1,x_2,\dotsc)$ and the right shift $b(x_0,x_1,\dotsc) := (0,x_0,x_1,\dotsc)$. Then $a \circ b = \mathrm{id} \neq b \circ a$ holds in the $K$-algebra $\mathrm{End}_K(K^{\mathbb{N}})$.

See SE/298791 for a proof of $AB=1 \Rightarrow BA=1$ for square matrices over a commutative ring.

Answer 3

If $\rm\,B\,$ is a linear map on a finite dimensional vector space $\rm\, V\,$ over field $\rm\,K,\,$ then easily by finite dimensionality (cf. Note below) there is a polynomial $\rm\,0\ne p(x)\in K[x]\;$ with $\rm\,p(B) = 0.\,$ W.l.o.g.$\rm\,p(0) \ne 0\,$ by canceling any factors of $\rm\,B\;$ from $\rm\;p(B)\;$ by left-multiplying by $\rm A,\,$ using $\rm\, AB = 1.$

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

so by linearity $\rm\, 0 \,=\, (BA-1)\ p(B)\, =\, (BA-1)\ p(0) \;\Rightarrow\; BA=1 \quad\quad$ QED

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

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 $\rm\,B.$

Answer 4

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.

Answer 5

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).

Answer 6

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 Armendariz and 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 $\rm\;\Rightarrow\: BA = I$

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

Coincidentally, a totally unrelated MathSciNet search turned up this article, which gives a result along the lines of (but slightly stronger than) the one in Martin Brandenburg's answer.

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 $a,b \in R$ are such that $ab = 1$, then also $ba = 1$.

Answer 11

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 $A=E_{n+1}E_{n} \cdots E_{1}$.

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

Answer 12

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

Answer 13

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.

Answer 14

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 $AB = E$. Consider the map $\mathbf x\mapsto B\mathbf x$. If $\mathbf x$ and $\mathbf x'$ are $n\times 1$ matrices, then $B\mathbf x = B\mathbf x'$ implies \begin{align*} AB \mathbf x &= AB \mathbf x' \\ E \mathbf x &= E \mathbf x' \\ \mathbf x &= \mathbf x', \end{align*} that is, left multiplication by $B$ is an injective map. By the rank-nullity theorem, this map is also onto, then for each matrix $n\times 1$ $\mathbf{y}$, the system $$B\mathbf x = \mathbf y$$ has exactly one solution, namely $$\mathbf x = A\mathbf y.$$

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 $$C = E.$$

Answer 15

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 $\textbf{A}^k$.

Answer 16

Given a matrix $A$, apply row operations to transform it to reduced form $A'$. Each row operation is left multiplication by a (two-sided) invertible matrix (as noted in Ofir's answer), so there is an invertible matrix $C$ such that $CA=A'$.

Since $A$ is a square matrix, either $A'=I$, or the last row of $A'$ is completely zero. In the first case, $CA=I$, and since $C$ is invertible, $A$ is also. In the second case, $A'$ is not right invertible, since the last row of $A'D$ is completely zero for any $D$. Therefore $A$ cannot be right invertible, since if $AB=I$, then $A'(BC^{-1})=CABC^{-1}=CC^{-1}=I$.

Conclusion: if $A$ has a right inverse $B$, we must be in the first case, so $A$ is (two-sided) invertible.

P.S. I wouldn't be thinking about the proof of this fact if I am not teaching Finite Math...

Answer 17

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: $f\circ g=id\Rightarrow f\text{ surjective}(g\text{ injective})$
And similarly: $g\circ f=id\Rightarrow f\text{ injective}(g\text{ surjective})$
Conversely: $f\text{ injective}\Rightarrow g_0\circ f=id$ ...for some $g_0$
And similarly: $f\text{ surjective}\Rightarrow f\circ g_0=id$ ...for some $g_0$
Hoewever, the constructions will be more tedious...

Answer 18

Suppose $AB=I$.

$\left\{A^j\right\}_{j=0}^{n^2}$ can not be linearly independent as vectors in $\mathbb{R}^{n\times n}$ (or $\mathbb{C}^{n\times n}$), so there is a non-trivial linear combination of the $A^j$ that vanishes. That is, $$ \sum_{j=k}^mc_jA^j=0\tag1 $$ where $0\le k\le m\le n^2$ and $c_k\ne0$. Right multiply $(1)$ by $B^{k+1}$ to get $$ \sum_{j=k+1}^mc_jA^{j-k-1}+c_kB=0\tag2 $$ Left multiply $(2)$ by $A$ to get $$ \sum_{j=k+1}^mc_jA^{j-k}+c_kI=0\tag3 $$ Right multiply $(2)$ by $A$ to get $$ \sum_{j=k+1}^mc_jA^{j-k}+c_kBA=0\tag4 $$ Comparing $(3)$ and $(4)$, we get $$ BA=I\tag5 $$

Answer 19

The following proof appears in the article A Short and Elementary Proof of the Two-sidedness of the Matrix-Inverse [Paparella, P., College Math. J. 48 (2017), no. 5, 366-367; MR3723713] and only uses the criteria stipulated by the OP:

Answer 20

A simple solution: $Rank (AB) \leq Rank(B).$ Since the each row of $AB$ can be written as the linear combination of rows $B.$ since $AB=I_n$ this implies $Rank(B)=n$ thus $B$ is invertible. Now by multiplying $B^{-1}$ to the both sides of $AB=I_n$ (From right) you will get $A=B^{-1}$ this implies $BA=I_{n}$

Answer 21

$AB=I$ => dim of row space of $B =n.$ Now B can be written as $B=BI=B(AB).$ i.e $(I-BA)B=O=> I-BA =O $....(1).

In (1) I am using the fact that all the rows of $B$ are linearly independent and the rows of the matrix $(I-BA)B$ are linear combinations of the rows of $B$ and the coefficients comes from the rows of $(I-BA).$ So as $(I-BA)B=O$ so the coefficients must be all $0$ i.e all rows of $(I-BA)$ are $0$.

Answer 22

$AB=I$, hence $A$ is full rank and we can perform elementary column transformation to transform $A$ into $I$.

Therefore, we can perform elementary row transformation to transform $A$ into $I$. Hence there exists a full rank matrix $C$ s.t. $CA=I$.

$C=CI=C(AB)=(CA)B=IB=B$.

Similar to @Blue 's answer. The argument for $C$'s existence is simplified.

Answer 23

It's easy to see that for square matrices $A$ and $B$, $AB = I$ is equivalent to $ABs_n = s_n$ where $\{s_n\}$ is a basis for $R^n$.

Define $g_n = Bs_n$. We observe the following:

1. $\{g_n\}$ is a basis for $R^n$, since any $s_n$ is a linear combination of $\{g_n\}$ (i.e., $s_n = ABs_n = Ag_n$)
2. $BAg_n = BABs_n = Bs_n=g_n$

Conclusion: Since $\{g_n\}$ is a basis, $BAg_n = g_n$ is equivalent to $BA = I$.

Answer 24

$\newcommand{\mat}[1]{\left(\begin{matrix}#1\end{matrix}\right)}$ Here is a proof using only calculations and induction over the size $n$ of the matrices. Observe that no commutativity is needed; we can work in any division ring.
If $n=1$ then two scalars $a,b$ with $ab=1$ are given. Then $b=1/a$ and we also have $ba=1$.
Now suppose that the statement is true for all matrices of size $n-1$, $n\geq2$. Given two $n$ by $n$ matrices with $AB=I$, we can assume without loss of generality that the upper left element of $A$ is nonzero. Otherwise, since the first row of $A$ cannot be all zero, we can achieve this by permuting two columns of $A$ and the correponding rows of $B$. We can also assume that this upper left element $\alpha$ equals 1, otherwise we multiply the first row of $A$ by the inverse of this element from the left and the first column of $B$ by $\alpha$ from the right.
Now we write $A,B$ in block matrix form: $$A=\mat{1&a_2\\a_3&a_4}, B=\mat{b_1&b_2\\b_3&b_4},$$ where $b_1$ is a scalar, $a_2,b_2$ are matrices of size 1 by $n-1$, $a_3,b_3$ have size $n-1$ by 1 and $a_4,b_4$ are $n-1$ by $n-1$. $AB=I$ means that $$b_1+a_2b_3=1,\ b_2+a_2b_4=0,\ a_3b_1+a_4 b_3=0\mbox{ and }a_3b_2+a_4b_4=I.$$ Here $I$ has size $n-1$ only and "0" abbreviates any matrix of zeros).
First, we calculate $(a_4-a_3a_2)b_4=a_4b_4+a_3b_2=I$. Since both matrices have size $n-1$ we can conclude that $b_4(a_4-a_3a_2)=I$.
Next, we calculate $$(a_4-a_3a_2)(b_3+b_4 a_3)=a_4 b_3-a_3a_2b_3+a_3= (a_4 b_3+a_3b_1)+a_3(-b_1-a_2b_3+1)=0$$ and, multiplying by $b_4$ from the left, we obtain $b_3+b_4 a_3=0$.
Then we obtain that $a_2b_3=-a_2b_4a_3=b_2a_3$ and therefore also $b_1+b_2a_3=1$.
Finally we have $a_2+b_2(a_4-a_3a_2)=(a_2b_4+b_2)(a_4-a_3a_2)=0$ and thus $b_2a_4=(b_2a_3-1)a_2=-b_1a_2$. Altogether we obtain $BA=\mat{b_1&b_2\\b_3&b_4}\mat{1&a_2\\a_3&a_4}=I$ which completes the proof.

Answer 25

It is possible to give a very elementary short proof by combining the arguments of Davidac897, Bill Dubuque and Mohammed, but without using linear functions, images, injective maps or relatively advanced basis notions.

Take the canonical base $\{e_1,e_2,\dots,e_n\}$ where $e_i=[0~0~\dots 1~0~0\dots 0]^\top$ with the value one appearing at position $i$, for all $i\in[1..n]$. Assume $Be_1,~Be_2,\dots Be_n$ are linearly dependent, i.e., there exists a non-zero vector $\alpha\in R^n$ such that $\sum_{i=1}^n \alpha_i Be_i=0$. Multiplying this by $A$, we obtain $A\left(\sum_{i=1}^n \alpha_i Be_i\right)=0$, which is equivalent to $\sum_{i=1}^n\alpha_i ABe_i=0$ or $\sum_{i=1}^n \alpha_ie_i=0$, which is a contradiction for non-zero $\alpha$. The assumption that $Be_1,~Be_2,\dots Be_n$ are linearly dependent is false. These vectors need to be linearly independent, and so, their linear combinations cover a space of dimension $n$, i.e., $R^n$. This means that for any $x\in R^n$ there exist a linear combination $x=\sum_{i=1}^n \beta_i Be_i= B(\sum_{i=1}^n \beta_ie_i)=By$.

From $x=By$ we derive $x=BABy=BAx$. Since this holds for any $x\in R^n$, we can extend it to a matrix version: $X=BAX$ for all $X\in R^{n\times n}$, and so, $BA$ needs to be the identity matrix $I_n$.

Answer 26

Here is an alternative approach. Assume $AB = I$ and let $x \in Im(B)$. Then there exists $y$ such that $By = x$, so it holds that $BAx = B(ABy) = By = x$. If the image of $B$ coincides with the domain of $A$, then the previous reasoning implies that $B$ is a left-inverse of $A$.

The advantage of this approach is that it directly shows what can go wrong in infinite dimensional spaces. If $A$ and $B$ are the left and right shift operators on $\ell^2$ (where the first element is set to 0 for the right shift), for example, then clearly $AB = I$ but $BA x \neq x$ for any $x$ that is not in $Im(B)$ i.e. for any $x$ whose first element is nonzero.

Answer 27

Theorem:

If $A$ and $B$ are two square matrices such that $A B=I$, then $B A=I$.

Proof:

By applying the properties of determinants of square matrices, we get that

$\det A\cdot\det B=\det(A B)=\det I=1\ne0$.

So it results that $\;\det A\ne0\;$ and $\;\det B\ne0\;$.

Now we consider the matrix

$C=\frac{1}{\det A}\cdot\text{adj}(A)$

where $\;\text{adj}(A)\;$ is the adjugate matrix of $A$ which is the transpose of the cofactor matrix of $A$.

$\text{adj}(A)= \begin{pmatrix} A_{1,1} & A_{2,1} & \cdots & A_{n,1} \\ A_{1,2} & A_{2,2} & \cdots & A_{n,2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1,n} & A_{2,n} & \cdots & A_{n,n} \end{pmatrix}$

where $\;A_{i,j}\;$ is the cofactor of the element $a_{i,j}$ of the matrix $A$.

So $\;A_{i,j}=(-1)^{i+j}\det M_{i,j}$

where $\;M_{i,j}\;$ is the submatrix of $A$ formed by deleting the $i^{th}$ row and the $j^{th}$ column.

We are going to use the Laplace expansions which are the following equalities:

$a_{i,1}A_{j,1}+a_{i,2}A_{j,2}+\ldots+a_{i,n}A_{j,n}= \begin{cases} \det A\;,\quad\text{ if } i=j\\ 0\;,\quad\quad\;\,\text{ if } i\ne j \end{cases}$

$A_{1,i}a_{1,j}+A_{2,i}a_{2,j} +\ldots+A_{n,i}a_{n,j}= \begin{cases} \det A\;,\quad\text{ if } i=j\\ 0\;,\quad\quad\;\,\text{ if } i\ne j \end{cases}$

By applying Laplace expansions, we get that

$A C = C A = I$.

Since for hypothesis $\;A B =I\;,\;$ then $\;C(A B)= C I\;,\;$ hence $\;(C A)B=C,\;$ therefore $\;I B = C\;$ and so we get that $\;B=C$.

Consequently,

$B A = C A = I\;,$

so we have proved that

$B A = I$.

Answer 28

Let $EA$ be the RREF of $A,$ where $E$ is a suitable product of elementary matrices. If $A$ is not invertible, then $EA$ has a zero row. Then $EAB$ also has a zero row. However, $EAB=E$ does not have a zero row. Thus $A$ is invertible. Consequently, $B=A^{-1}$ is also invertible, and $BA=I.$

Answer 29

Let us give a more general result:

Result: Let $T:V\rightarrow V$ be a linear operator on a finite dimensional vector space $V$ such that there is another linear operator $U$ on $V$ satisfying $TU=I$. Then show that, $UT=I$.

Proof: Since $TU=I$, so $T$ is onto. Now $V$ is finite dimensional, so from Rank-Nullity theorem $\operatorname{Nullity}(T)=0$, which implies $\operatorname{Ker}(T)=\{0\}$.

Now for any $v\in V$ we have, $T(UT-I)v=(TU-I)Tv=0$. So $(UT-I)v\in \operatorname{Ker}(T)$ for all $v\in V$, which gives $(UT-I)v=0$ for all $v\in V$. Thus, $UT-I=0$.

Note: Finiteness of $\operatorname{dim}(V)$ is necessary. Otherwise, "$T$ is onto $\Rightarrow \operatorname{Ker}(T)=\{0\}$" does not hold in an infinite dimensional vector space.

Now a square matrix $A$ of size $n$ over a field $F$ can be treated as a linear operator on $F^n$ by following linear map $L_A: F^n\rightarrow F^n$, defined by $L_A(v)=Av$ for all column vector $v\in F^n$.

For two $n\times n$ matrix $A$, $B$, we clearly have $L_AL_B=L_{AB}$. Now put $T=L_A$ and $U=L_A$ then the required result follows.

Answer 30

Since matrices represent linear maps between finite-dimensional vector spaces, the question can be recast as:

Let $K$ be a field. Prove that if $f:K^n\to K^n$ and $g:K^n\to K^n$ are linear maps, and $f\circ g=\DeclareMathOperator{\id}{id}\id$, then $g\circ f=\id$.

Since $f\circ g=\id$, the function $g$ has a left-inverse, and so it is injective. Now, the rank-nullity theorem asserts that $$ \dim(\operatorname{im} g)+\dim(\operatorname{ker} g)=n \, . $$ As $g$ is injective, its kernel is $\{\mathbf{0}\}$, which has dimension $0$. Hence, $\dim(\operatorname{im} g)=n$. If a linear subspace $U$ of $V$ has the same dimension as $V$, then $U=V$. Since the image of $g$ is a linear subspace of $K^n$, it follows that the image of $g$ equals $K^n$, and so $g$ is surjective. Hence, $g$ is bijective, with $f$ being its unique two-sided inverse.