1. ## Conflicting defintions/explanations

My professor states in one instance that if $\displaystyle A$ is invertible, then $\displaystyle B=A^{-1}$ when $\displaystyle AB=BA=I$, leading to $\displaystyle AA^{-1}=A^{-1}A=I$. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that $\displaystyle B=A^{-1}$ implies that $\displaystyle B$ is invertible when $\displaystyle BA=AB=I$, leading to $\displaystyle A^{-1}A=AA^{-1}=I$. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is $\displaystyle AB=BA=I$ or $\displaystyle BA=AB=I$, yet he clearly contradicts himself if order is so important. A similar is in once instance, $\displaystyle B=A^{-1}$ leads to $\displaystyle AB=BA=I$, and then later on states that it is $\displaystyle A=B^{-1}$ that leads to $\displaystyle AB=BA=I$. 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 $\displaystyle AB=BA=I$ then $\displaystyle A$ is invertible and $\displaystyle B=A^{-1}$, or does $\displaystyle B=A^{-1}$ imply $\displaystyle B$ is invertible and therefore $\displaystyle BA=AB=I$. 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 $\displaystyle A$ is invertible, then $\displaystyle B=A^{-1}$ when $\displaystyle AB=BA=I$, leading to $\displaystyle AA^{-1}=A^{-1}A=I$. Yet later on in class, during a proof involving the multiplicity of invertible matrices, he makes the claim that $\displaystyle B=A^{-1}$ implies that $\displaystyle B$ is invertible when $\displaystyle BA=AB=I$, leading to $\displaystyle A^{-1}A=AA^{-1}=I$. I understand this theorem fairly easily, but my professor made a big deal about whether the order of the equation is $\displaystyle AB=BA=I$ or $\displaystyle BA=AB=I$, yet he clearly contradicts himself if order is so important. A similar is in once instance, $\displaystyle B=A^{-1}$ leads to $\displaystyle AB=BA=I$, and then later on states that it is $\displaystyle A=B^{-1}$ that leads to $\displaystyle AB=BA=I$. 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 $\displaystyle AB=BA=I$ then $\displaystyle A$ is invertible and $\displaystyle B=A^{-1}$, or does $\displaystyle B=A^{-1}$ imply $\displaystyle B$ is invertible and therefore $\displaystyle BA=AB=I$. 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 $\displaystyle A$ is a square matrix, say $\displaystyle n \times n,$ then $\displaystyle A$ is invertible if and only if $\displaystyle AB=I_n$ or $\displaystyle BA=I_n,$ for some $\displaystyle n \times n$ matrix $\displaystyle B.$ the important point, which unfortunately is not mentioned in most elementary

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

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

1. $\displaystyle AB=I,$

2. $\displaystyle BA=I,$

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

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

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

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

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

3. I understand that $\displaystyle A=B^{-1}$ implies $\displaystyle B=A^{-1}$ 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 $\displaystyle AB=BA=I$ implies $\displaystyle A$ is invertible and thus $\displaystyle B=A^{-1}$. He uses this defintion to show that $\displaystyle A=B^{-1}$, 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 $\displaystyle A=B^{-1}$ implies $\displaystyle B=A^{-1}$ 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 $\displaystyle AB=BA=I$ implies $\displaystyle A$ is invertible and thus $\displaystyle B=A^{-1}$. He uses this defintion to show that $\displaystyle A=B^{-1}$, 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 $\displaystyle AB=I$ or $\displaystyle BA=I$ because for square matrices

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

$\displaystyle AB=BA=I.$ so, yes, the order is not important and you're right. another point that i mentioned is the "uniqueness", i.e. if $\displaystyle AB=BA=I$ and $\displaystyle AC=CA=I,$ then $\displaystyle B=C.$ 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 $\displaystyle BA$ would result in an $\displaystyle m \times m$ and $\displaystyle AC$ results in an $\displaystyle n \times n$, and therefore $\displaystyle BA \ne AC$, and if $\displaystyle AB$ are $\displaystyle n \times n$ square matrices of equal size , then $\displaystyle AB=BA$ since $\displaystyle \sum_{r=0}^{n}a_{ir}b_{rj}=\sum_{r=0}^{n}b_{ir}a_{ rj}$ 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 $\displaystyle AB=I$ if and only if $\displaystyle BA=I$ 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: $\displaystyle <b,c|bc=1>$.