# Thread:

1. ## Conflicting defintions/explanations

My professor states in one instance that if $A$ is invertible, then $B=A^{-1}$ when $AB=BA=I$, leading to $AA^{-1}=A^{-1}A=I$. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that $B=A^{-1}$ implies that $B$ is invertible when $BA=AB=I$, leading to $A^{-1}A=AA^{-1}=I$. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is $AB=BA=I$ or $BA=AB=I$, yet he clearly contradicts himself if order is so important. A similar is in once instance, $B=A^{-1}$ leads to $AB=BA=I$, and then later on states that it is $A=B^{-1}$ that leads to $AB=BA=I$. Again, I understand fairly easily that is a matrix is an inverse of another matrix, then the inverse of that first matrix is the other matrix, but this professor is pedantic.

So I need clarification on the theorem. If $AB=BA=I$ then $A$ is invertible and $B=A^{-1}$, or does $B=A^{-1}$ imply $B$ is invertible and therefore $BA=AB=I$. My professor is a real hardass about following the proofs EXACTLY as they are stated, but like I said, the two claims that he made seem to contradict themselves if order is so important.

2. Originally Posted by Pinkk
My professor states in one instance that if $A$ is invertible, then $B=A^{-1}$ when $AB=BA=I$, leading to $AA^{-1}=A^{-1}A=I$. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that $B=A^{-1}$ implies that $B$ is invertible when $BA=AB=I$, leading to $A^{-1}A=AA^{-1}=I$. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is $AB=BA=I$ or $BA=AB=I$, yet he clearly contradicts himself if order is so important. A similar is in once instance, $B=A^{-1}$ leads to $AB=BA=I$, and then later on states that it is $A=B^{-1}$ that leads to $AB=BA=I$. Again, I understand fairly easily that is a matrix is an inverse of another matrix, then the inverse of that first matrix is the other matrix, but this professor is pedantic.

So I need clarification on the theorem. If $AB=BA=I$ then $A$ is invertible and $B=A^{-1}$, or does $B=A^{-1}$ imply $B$ is invertible and therefore $BA=AB=I$. My professor is a real hardass about following the proofs EXACTLY as they are stated, but like I said, the two claims that he made seem to contradict themselves if order is so important.
if $A$ is a square matrix, say $n \times n,$ then $A$ is invertible if and only if $AB=I_n$ or $BA=I_n,$ for some $n \times n$ matrix $B.$ the important point, which unfortunately is not mentioned in most elementary

linear algebra courses, is that for square matrices $A,B$ we have $AB=I$ if and only if $BA=I.$ anyway, in this case we call $B$ the inverse of $A$ and similarly $A$ would be the inverse of $B.$

so for square matrices the following four statements are equivalent, meaning any of them implies the other three:

1. $AB=I,$

2. $BA=I,$

3. $A=B^{-1},$

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

for non-square matrices things are different. an $n \times m, \ n \neq m,$ matrix $A$ could have left inverse, i.e. $BA=I_m,$ for some $m \times n$ matrix $B,$ but no right inverse, i.e. $AC \neq I_n,$ for any $m \times n$

matrix $C.$ it is also possible that $A$ has a right inverse but not a left inverse (or of course $A$ might have neither left nor right inverse). but if $A$ has both left and right inverse, then it has to be

a square matrix, i.e. $n=m$ and then the left and right inverse must be equal.

3. I understand that $A=B^{-1}$ implies $B=A^{-1}$ given the necessary conditions, but my professor used a proof to show this where he stated the definition I mentioned in my first post. But he is not consistent with that definition in the proof. The definition he first gave was $AB=BA=I$ implies $A$ is invertible and thus $B=A^{-1}$. He uses this defintion to show that $A=B^{-1}$, but as I mention in my first post, he somewhat "contradicts" himself because he put great emphasis on the order of those equations.

4. Originally Posted by Pinkk
I understand that $A=B^{-1}$ implies $B=A^{-1}$ given the necessary conditions, but my professor used a proof to show this where he stated the definition I mentioned in my first post. But he is not consistent with that definition in the proof. The definition he first gave was $AB=BA=I$ implies $A$ is invertible and thus $B=A^{-1}$. He uses this defintion to show that $A=B^{-1}$, but as I mention in my first post, he somewhat "contradicts" himself because he put great emphasis on the order of those equations.
in my previous post i actually ignored what your prof said because it's nonsense. i also said that the definition is equivalent to saying that $AB=I$ or $BA=I$ because for square matrices

$AB=I$ if and only if $BA=I.$ anyway, let's just stick to your prof's definition: then in order to show that $A=B^{-1}$ we need to show that $BA=AB=I,$ which is obviously the same as

$AB=BA=I.$ so, yes, the order is not important and you're right. another point that i mentioned is the "uniqueness", i.e. if $AB=BA=I$ and $AC=CA=I,$ then $B=C.$ now we can

5. Ah okay, thank you. And yes, I had a hunch that my professor was making it more convoluted than it has to be. The if and only if statement applies only to square matrices since performing $BA$ would result in an $m \times m$ and $AC$ results in an $n \times n$, and therefore $BA \ne AC$, and if $AB$ are $n \times n$ square matrices of equal size , then $AB=BA$ since $\sum_{r=0}^{n}a_{ir}b_{rj}=\sum_{r=0}^{n}b_{ir}a_{ rj}$ for square matrices.

6. see how an MIT prof explains the inverse of a matrix. he mentions what i said, i.e. for square matrices $AB=I$ if and only if $BA=I$ but he also says that the proof is not "easy",

(well, it's not terribly hard either! haha) and so he skips it. i like this guy! his lecture is very accurate, although he talks too much! haha watch the video from 21':30" if you like:

MIT OpenCourseWare | Mathematics | 18.06 Linear Algebra, Spring 2005 | Video Lectures | detail

7. No doubt someone can prove me wrong, but I don't think there is such a thing as a one-sided inverse for square matrices. If AB=I then BA=I.
But in a more abstract setting there are certainly monoids in which one can have AB=I but not BA=I. But this can only happen in rather restricted circumstances:
it can't happen for a finite monoid
if AB=I but BA != I then there is no C such that BC=I nor D such that DA=I. If there were then it is easy to show that AB=BA=I.

8. Originally Posted by alunw
No doubt someone can prove me wrong, but I don't think there is such a thing as a one-sided inverse for square matrices. If AB=I then BA=I.
But in a more abstract setting there are certainly monoids in which one can have AB=I but not BA=I. But this can only happen in rather restricted circumstances:
it can't happen for a finite monoid
if AB=I but BA != I then there is no C such that BC=I nor D such that DA=I. If there were then it is easy to show that AB=BA=I.
For instance, the Bicyclic Monoid: $<b,c|bc=1>$.
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