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Let $A_{n×n}$ be a complex matrix then prove that $ch_{AB}(x) = ch_{BA}(x)$

I saw somewhere that $ch_{AB}(x) = ch_{BA}(x)$ if $A$ is invertible and if $AB$ is nilpotent then $BA$ is nilpotent. What happened if $A$ is singular.

Largest Prime
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  • Can you explain what ch is supposed to represent? – Eric Aug 26 '21 at 01:57
  • Here $ch_A(x)$ denotes characteristic polynomial of matrix $A$ – Largest Prime Aug 26 '21 at 02:02
  • The reason it's easy if $A$ is invertible is that then $AB$ and $BA$ are similar via $A$. To prove the general case, you can either use a continuity argument (by first proving that the set of invertible matrices is dense), or you can use an algebraic argument. – Anonymous Aug 26 '21 at 02:02
  • @Anonymous, I don't know how can we use concept of real analysis in linear algebra. – Largest Prime Aug 26 '21 at 02:12
  • Fix the matrix $B$. Consider the set of $n \times n$ complex matrices to be $C^{n^2}$. It is easy to see that the coefficients of the polynomial $\operatorname{ch}{AB} - \operatorname{ch}{BA}$ are continuous functions of the $n^2$ coefficients of the matrix $A$. Each of these coefficients is zero for all invertible $A$. Since the invertible matrices form a dense set, these coefficients are identically zero. (I understand that if you haven't seen this kind of argument before, then you have a lot of things to think about.) – Anonymous Aug 26 '21 at 02:19

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