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Let $A$, $B$ and $C$ be three endomorphisms of a finite-dimensional vector space such that $AB=BA$. Prove that $$\det\left(A+BC\right)=\det\left(A+CB\right)$$

Fabrosi
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89085731
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    Did you make any effort to start the proof on your own? – Fabrosi Aug 20 '15 at 18:27
  • Replacing $A$ by $A + \epsilon \operatorname{id}$ for some $\epsilon$ in the ground field (or an extension, if necessary) allows you to WLOG assume that $A$ is invertible. Then you can use http://math.stackexchange.com/questions/844204/prove-that-detxi-m-ab-xm-n-detxi-n-ba?rq=1 . I think there is a better solution, though. – darij grinberg Aug 20 '15 at 18:27
  • This works for all fields: apply the determinant formula for block matrices to $\pmatrix{C&-I\ A&B}$ twice, once using the commutativity of the pair of matrices on bottom row and once using the commutativity of the two sunblocks on the right. – user1551 Aug 21 '15 at 10:05

2 Answers2

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First assume that $B$ is invertible. We then have $$A = B^{-1}AB$$ and

$$\det(A+BC) = \det(B^{-1}(A+BC)B) = \det(B^{-1}AB + CB) = \det(A + CB)$$

Since invertible matrices are dense and determinant is continuous, equality extends to non-invertible $B$'s by continuity.

Ennar
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To see this, note that:

1) $\det(A+BC)=\det(A)\det(I+A^{-1}BC)$,

2) $AB = BA$ implies $BA^{-1}=A^{-1}B$,

3) $\det(I+ABC)=\det(I+BCA)$ (this is true for any cyclic permutation of $ABC$).

Combining these gives us: $$\det(A+BC)=\det(A)\det(I+BA^{-1}C)=\det(A)\det(I+A^{-1}CB)= \det(A+CB).$$

Note: we have implicitly assumed that $A$ is invertible.

David Simmons
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