Prove B is invertible if AB = I

[Hall]

It seems to me that the thread went a little out of course. A fine approach to this problem would be to consider the matrices as connected with the Linear transformations.

If A (n x n dimension) is the coefficient matrix for T:V_n -> V_n , T is a Linear transformation and the columns of A represent the coefficients of the basis elements of range V_n when T is applied to one of the basis elements of domain V_n.

Prove that A is invertible if and only if T is invertible. And then prove that the inverse of A is the coefficient matrix of $T^{-1}$.

 

[nuuskur]

$\det AB\neq 0$ implies $\det B\neq 0$. Therefore, as was already pointed out, $B=BAB$ and therefore $BA$ is the identity.

 

[JFerreira]

Hi everyone,
I founded this discussion very interesting, so a search if someone did it before.

"A is linear transformation from a finite dimensional vector space to itself. AB=I amounts to saying A is surjective, hence it is bijective and has a left inverse C, so that CA=I. Of course, B=(CA)B=C(AB)=C, i.e. BA=I."

I founded this in the thread https://math.stackexchange.com/questions/152668/what-'s-the-short-proof-that-for-square-matrices-$ab-=-i$-implies-$ba-=-i$?
(searching for "\(AB = I\) for square matrices " on SearchOnMath) that can give some other contribuitions.

 

[Prof B]

You say that you learned about ranks. Did you learn the following?

1. rank(A) $\geq$ rank(AB)?
2. If A is nxn and rank(A) = n then A is invertible.

You could prove that A is invertible using 1 and 2.

 

[Delta2]

Using determinants and that $det(AB)=det(A)det(B)$ for square matrices+ A square matrix is invertible if its determinant is not zero, this problem is very easy. But OP says at post #4 that they haven't covered determinants yet.

 

[nuuskur]

Since $AB=I$, it means $A$ is surjective and $B$ is injective. Thus, $A,B$ are both bijective due to finite dimension.