If AB = 0 for square matrices, must det(A) and det(B) both be zero?

[zenn]

Need some help on how to prove this:

Let A and B be two non-zero square matrices of n x n, if AB = 0n, then det(A) and det(B) must both equal to 0.

I know if det(AB) = 0, then det(A) or det(B) must be 0, but here the question states that they must both be 0.

Appreciate any help, thanks.

 

[Count Iblis]

Suppose that not both are zero, let's take Det(A) to be nonzero. Then A has an inverse. In the equation:

A B = 0,


multiplying both sides by the inverse of A on the left then gives:

B = 0

But B was assumed to be different from the null matrix. So, we arrive at a contradiction. The assumption that Det(A) is nonzero has to be false!

 

[HallsofIvy]

Count Iblis used, without saying it explicitely, the fact that a matrix is invertible if and only if its determinant is non-zero.

 

[maze]

The condition is even stronger - in addition to having zero determinants, the matrices can't have the same column space (or row space) either.

 

[mathwonk]

when maps compose, the nullspace is at most as big as the sum of the dimensions of the individual nullspaces. you have a composition whose nullspace is maximal, yet neither individual one was. hence both nullspaces are positive dimensional.

i.e. N(A) was not everything, but N(AB) is everything, so the extra dimensions had to come from N(B).

this is the fundamental theorem of linear algebra, the inverse image under A of a finite dimensional space Y, is larger than Y at most by dimN(A).

it is applied here to Y = N(B), since the nullspace of BA is the inverse image of N(B) under A.

i just taught this topic today.