2

The complete definition of inverse of a matrix is- if $AB=BA=I$, then $B$ is inverse of $A$.

But l want to know that from $AB=I$, can we comment $B$ is inverse of $A$? Is it possible that $AB=I$ but $BA$ is not equal to $I$?

[ I personally think it is impossible since the inverse of a matrix is unique. Please clarify if I am wrong.]

Bill O'Haran
  • 3,034
DREAM ISI-CMI-IIT
  • 101

3 Answers3

1

The complete definition of inverse of a matrix is- if AB=BA=I, then B is inverse of A.

This actually is the definition of the inverse for any ring. If only one of $AB=I$ or $BA=I$ holds then we talk about partial inverses.

But l want to know that from AB=I, can we comment B is inverse of A? Is it possible that AB=I but BA is not equal to I?

Yes and no. Depends on the underlying ring.

Let $A,B\in\mathbb{M}_n(R)$ for some commutative ring $R$. If $AB=I$ then

$$1=\det(I)=\det(AB)=\det(A)\det(B)$$

showing that $\det(A)$ is invertible in $R$ and so $A$ is invertible. It's a theorem that $A$ is invertible if and only if $\det(A)$ is invertible. In particular $BA=I$.

For non-commutative rings the determinant is not well defined and moreover it may happen that $AB=I$ while $BA\neq I$. So over non-commutative rings there are partial inverses that are not inverses.

freakish
  • 42,851
1

Edit: as noted in the comments, this is not necessarily true for matrices with coefficients in non-commutative rings. However, the following is a positive result in the subcase where $A, B \in M_n(\mathbb{k})$ with $\mathbb{k}$ a field. In particular, this holds in $M_n(\mathbb{C})$

Let $A, B \in M_n(\mathbb{k})$. If $A$ is non singular, it has an inverse. Now, left multiplying by $A^{-1}$, we go from

$$ AB = I $$

to

$$ B = A^{-1} $$

Let's weaken the hypothesis and only assume that $AB = I$, $A$ not necessarily invertible. In other words, the linear mappings induced by $A$ and $B$, defined as $f(x) = Ax$ and $g(x) = Bx$ verify $fg \equiv id$, so $g$ must be injective. Since we are in finite dimension, a linear mapping is injective iff it is bijective. Therefore $g$ is invertible, or equivalently, $B$ is invertible. By the original argument, $B = A^{-1}$.

qualcuno
  • 17,121
  • 1
    This is a very clear way of looking at it. – littleO Apr 25 '18 at 06:37
  • And wrong. OP didn't say that matrices are over $\mathbb{C}$ or even over a field. – freakish Apr 25 '18 at 06:45
  • 1
    OP's title assumes $A$ invertible, so the first computation can be easily shown to hold in any group. Since he then in the post has weakened the hypothesis, I figured it'd be appropriate to give a positive result over $\mathbb{C}$. Btw, there's no need to be so harsh! – qualcuno Apr 25 '18 at 06:48
  • @GuidoA. I'm not harsh. I'm simply stating fact that the answer is wrong because the thesis does not hold over non-commutative rings. You've made an assumption that is nowhere to be found. – freakish Apr 25 '18 at 06:49
  • In any case, this may not answer the original question (which again, assumes in the title A invertible so in that case it always holds), but since I have clearly stated the setting in which I'm working in, the argument is by no means factually incorrect. – qualcuno Apr 25 '18 at 06:51
  • @GuidoA. You imply that you actually answer the question while you don't. This is wrong. You create a confusion. OP may now think that this is true for whatever ring he picks. It's like asking "are even numbers prime?" and answering "$2$ is prime". – freakish Apr 25 '18 at 07:01
  • 1
    I understand your point, but I think there is a distinction to be made. To be fair, OP should've specified the context of the matrices he's working with. You also assume that having no context implies working in the most general setting. For all we know, OP could not even know what a ring is. I think it is abundantly clear that this is (at least) a partial answer. Don't you think that comparing that with answering '$2$ is prime' is a bit bizarre? – qualcuno Apr 25 '18 at 07:05
0

It is not difficult to prove that if [B] is the inverse of [A] then [A][B]=[I} and ALSO (quite generally) [B][A]=[I]. That is, the operation of matrix multiplication of one square matrix by its inverse matrix inverse is always commutative - one of the few cases where this is generally true for matricies.