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To put it better, if A and B are non-invertible matrices (for whatever reason), can the matrix AB be invertible?

Just used to help understand a Linear Transformation assignment question, don't necessarily need an explanation as to why, but it would help. Thankyou.

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$A,B$ square: never See for example How to prove and interpret rank(AB)$\le$ min(rank(A),rank(B))? But... $$\pmatrix{1&0}\pmatrix{1\cr0} = (1).$$

EDIT: more general and abstract example. For $\dim V = n$, take $$A: V\times V\longrightarrow V,\qquad A(x,y) = x,$$ $$B: V\longrightarrow V\times V,\qquad B(x) = (x,0),$$ and $AB =$ identity.

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1) A square matrix is non-invertible if its determinant vanishes.

2) The determinant of a matrix product of two square matrices is the product of the determinants of the two individual matrices.

So, for square matrices the product is not invertible if one of the factors is not. For non-square matrices (which are non-invertible) I do not know if in some cases an invertible product can occur.

Urgje
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