Sylvester's Determinant identity proof

Asked Jun 22 '17 at 17:30

Active Jun 22 '17 at 17:31

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How to prove $$\det(I-AB)=\det(I-BA)$$ if one of the matrices is singular. I tried the following way: Let $A$ be singular. Then $$\det(AB)=0$$ and $$\det(BA)=0$$

Now since $AB$ be singular $\exists$ a non-zero vector $x$ such that $$ABx=0$$ Hence

$$(I-AB)x=x$$ and similarly $\exists$ non-zero vector $y$ such that

$$(I-BA)y=y$$

Any help here?

edited Jun 22 '17 at 17:31

Michael Hardy

asked Jun 22 '17 at 17:30

Ekaveera Gouribhatla

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I dont think there is a proof for singular matrices in the duplicate – Ekaveera Gouribhatla Jun 22 '17 at 17:54
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If I'm not mistaken, this identity should hold when $A$ and $B$ are not square matrices, as long as they are of such shapes that $AB$ and $BA$ both exist. – Michael Hardy Jun 22 '17 at 18:25
@EkaveeraKumarSharma Dubuque's proofs work for singular matrices, even for non-square matrices. – Robert Israel Jun 22 '17 at 18:59

Sylvester's Determinant identity proof

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