# Prove B is invertible if AB = I

• songoku
In summary: B) + dim (nul B^T) = nMy bad, I was thinking of the matrix as a linear transformation and not as a matrix. It's been a while since I've done this stuff. So, if B is not invertible, then it's not surjective, and since it's not surjective it's not injective. So, the null space of B is non-empty. Take a vector v in the null space of B. Then, we have Bv = 0. Now suppose BA =/= I, then there's some vector w such that BAw =/= w. It follows then that B(Aw) =/= Bw. However, Bw is an
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}##.

songoku and Delta2
##\det AB\neq 0## implies ##\det B\neq 0##. Therefore, as was already pointed out, ##B=BAB## and therefore ##BA## is the identity.

songoku and Delta2
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.

songoku and Delta2
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.

Last edited:
songoku and 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.

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

songoku, PeroK and Delta2

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