Find characteristic polynomial

Asked Jun 24 '14 at 08:57

Active Jun 24 '14 at 10:57

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Suppose $A$ and $B$ are $n \times n$ complex matrices such that $$AB-BA=aI+A,$$ where $a \in \mathbb{C}.$ Find the characteristic polynomial of $A$.
If $A$ happens to be a Jordan block, this would be easy:

Write $A=cI+N$ with $c \in \mathbb{C}$ and $N$ nilpotent. Then $$(cI+N)B-B(cI+N)=(a+c)I+N,$$ that is, $$NB-BN=(a+c)I+N.$$ Take trace of both sides to get $c=-a$. So the characteristic polynomial would be $(x+a)^n$.
So I guess for a general $A$, the answer is the same?

asked Jun 24 '14 at 08:57

AaronS

1,248

Note that if $A=XJX^{-1}$, then $X^{-1}(AB-BA)X=X^{-1}(XJX^{-1}B-BXJX^{-1})X=J\tilde{B}-\tilde{B}J$ and $X^{-1}(aI+A)X=aI+J$, thus transforming the problem to the case where $A$ is the matrix from the Jordan form. However, $J$ is generally a direct sum of elementary Jordan blocks. – Algebraic Pavel Jun 24 '14 at 09:11
@AlgebraicPavel But what if $\tilde{B}$ cannot be decomposed into blocks that match the sizes of the blocks of $J$? – AaronS Jun 24 '14 at 09:20
You might still work with $\tilde{B}$ partitioned according to the dimensions of the small Jordan blocks? – Algebraic Pavel Jun 24 '14 at 09:27
possible duplicate of If A and (AB-BA) commute, show that AB-BA is nilpotent – AaronS Jun 24 '14 at 10:49

Find characteristic polynomial

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