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The following stumps me:

Let $\mathbb K$ be a field. Let $A, B \in \mathbb K^{n \times n}$ where $B$ is nilpotent and commutes with $A$, i.e., $A B = B A$. Show that $$ \det(A+B)=\det(A) $$

I have no idea how to approach this I thought perhaps raise both sides to a power but nothing works. Thanks for all help.

darij grinberg
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Croc2Alpha
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    Hint: First prove it when $A$ is invertible (in which case $BA^{-1}$ is nilpotent and thus $\det\left(I_n + BA^{-1}\right) = 1$). Then, argue that in the general case, you can replace $A$ by an invertible matrix $xI_n + A$ over the ring of Laurent power series in an indeterminate $x^{-1}$ over $\mathbb{K}$. This new matrix $xI_n + A$ still commutes with $B$, and turns into $A$ if you substitute $x = 0$ (which you can do, because all entries are actually polynomials in $x$). I did something like that in https://math.stackexchange.com/questions/1514408//1847767#1847767 (albeit with $xI_n - A$). – darij grinberg Mar 23 '19 at 20:57
  • This question is a special case of MSE question 19464 "Spectrum of the sum of two commuting matrices". – Somos Mar 24 '19 at 19:30
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    @Somos: Not clear. We're talking about spectra with multiplicities, while the question you are citing is talking about spectra as sets. And the "WLOG all eigenvalues distinct" trick doesn't work here, since the problem space is not an affine variety (I don't think it is even an irreducible variety). – darij grinberg Mar 24 '19 at 22:54

1 Answers1

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Here's a way to do it without the machinery of JNF or triangulization, although that machinery is absolutely worth learning. You really only need to know that $\det(M) = 0$ if and only if there exists $x\ne 0$ such that $Mx = 0$ (for an $n\times n$ matrix $M$), that $\det$ is multiplicative, and that $\det(M)$ is the product of the roots (in an algebraic closure) of the characteristic polynomial of $M$.

Let $m>0$ be such that $B^m = 0$. Since $A$ and $B$ commute, we may expand as follows \begin{equation}\tag{$*$}(A+B)^m = \sum_{i=0}^m \binom{m}{i} B^{m-i} A^{i}.\end{equation}

Suppose that $\det(A) = 0$, and let $v\ne 0$ be such that $Av = 0$. Observe that $(A+B)^m(v) = 0$, since each term in the expansion $(*)$ kills $v$: for $0< i \le m$, we have $B^{m-i}(A^i(v)) = B^{m-i}A^{i-1}(Av) = 0$, and for $i=0$, we have $B^m(v) = 0$ since $B^m = 0$. Therefore $0\ne v\in \ker (A+B)^m$, so $(\det(A+B))^m = \det((A+B)^m) = 0$, so $\det(A+B)=0=\det(A)$.

Now suppose that $\det(A) \ne 0$. Let $C = A^{-1}B$. It suffices to show that $\det(I+C) = 1$. Let $\lambda \in \bar K$ be a root of the characteristic polynomial of $I+C$, so that $\det((1-\lambda) I + C) = \det(I + C - \lambda I) = 0$. Since $A$ commutes with $B$, so does $A^{-1}$, and thus $-C = -A^{-1}B$ is nilpotent. Now since $-C$ is nilpotent and commutes with $(1-\lambda)I + C$, which has determinant $0$, by the above paragraph we have $$(1-\lambda)^n = \det((1-\lambda)I) = \det(((1-\lambda)I+C)+(-C)) = \det((1-\lambda)I+C) = 0,$$ so $\lambda = 1$. Therefore, the characteristic polynomial of $I+C$ has only $1$ as a root, and thus $\det(I+C) = 1$.

user555203
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