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Let $V$ be a finite dimensional vector space over a field $K$. An operator $T:V\to V$ is called semi-simple if every $T$-invariant subspace of $V$ has a $T$-invariant complement(for algebraically closed fields these are exactly diagonalizable operators). Is it true that the sum (or the product) of two commuting semi-simple operators is semi-simple?

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Sorry for the false answer below where I forgot the commutation. Indeed the answer is Yes, because $V$ is the direct sum of the eigenspaces $V_\mu$ under $A$. By commutation, $BV_\mu\subset V_\mu$. Hence we are down to look at the restrictions of $A+B$ to each $V_\mu$, which is nothing but $\mu{\rm id}+B|_{V_\mu}$. The latter is semis-imple, hence $A+B$ is too. – Denis Serre Dec 3 '12 at 16:04
@Denis: $K$ is not algebraically closed, so you cannot speak about diagonalizability and eigenspaces... – Ilya Bogdanov Dec 3 '12 at 16:34
You have no direct sum decomposition of eigenspace here. – M. Shahryari Dec 3 '12 at 16:51
up vote 8 down vote accepted

The answer is still No, but for not that obvious reason. To show that, let us start with a positive claim.

1. FIrst of all, an operator $T$ is semisimple iff all the factors in the prime expansion of its minimal annihilating polynomial $\mu$ are distinct. Actually, the algebra $K[T]$ is isomorphic to $K[X]/(\mu)$; so, if there are no multiple factors, then this algebra is a direct sum of fields, and each its finitely generated module is semisimple. Otherwise, if $\mu=p^2q$ with $p$ nonconstant, then the annihilator space of $p(T)$ has no complement.

2. Now assume that the extension of $K$ generated by all the roots of characteristic polynomials of $T_1$ and $T_2$ is separable. Then they should be diagonalizable over this extension by the reasons of the minimal polynomial. Since they commute, they are simultaneously diagonalizable. Hence their sum and product are also diagonalizable, and their minimal polynomials have no multiple roots (from separability!) and hence no multiple factors. So $T_1+T_2$ and $T_1T_2$ are both semisimple.

3. And here is a counterexample for non-separable case. Let $K=F_2(t)$ be the field of rational fractions over $F_2$. Set $T_1=\begin{pmatrix}0&0&1&0\cr 0&0&0&1\cr t&0&0&0\cr 0&t&0&0\end{pmatrix}$ and $T_2=\begin{pmatrix}0&1&0&0\cr t&0&0&0\cr 0&0&0&1\cr 0&0&t&0\end{pmatrix}$; their common minimal polynomial is $X^2-t$, so they are semisimple (this polynomial is irreducible although not separable). But their sum is $S=T_1+T_2=\begin{pmatrix}0&1&1&0\cr t&0&0&1\cr t&0&0&1\cr 0&t&t&0\end{pmatrix}$ with $S^2=0$, and their product is $P=T_1T_2=T_2T_1=\begin{pmatrix}0&0&0&1\cr 0&0&t&0\cr 0&t&0&0\cr t^2&0&0&0\end{pmatrix}$ with the minimal polynomial $(X+t)^2$. Hence both are not semisimple. One may present a direct example of a spaces that cannot be complemented as $\langle e_2+e_3,e_1+te_4\rangle$ in both cases.

In fact, this example was obtained from the action of algebra $K[X,Y]/(X^2-t,Y^2-t)$ on its regular module; $T_1$ and $T_2$ correspond to $X$ and $Y$, respectively.

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Thank you. I satisfied by your answer. – M. Shahryari Dec 3 '12 at 16:49
@Ilya: A variant of your argument applies in any characteristic $p > 0$. Let $F = K(a)$ be finite purely inseparable of degree $p$ with $a^p=c\in K^{\times}$, $V = F^{\otimes p}$ (tensor over $K$), and $T_i:V \rightarrow V$ the commuting endomorphisms given by $a$ on the $i$th tensor factor. The $T_i$-stable $K$-subspaces of $V$ are $F$-subspaces for the $i$th tensor structure, so each is semisimple over $K$. But $\sum T_i$ is nilpotent ($p$th power vanishes) and nonzero, so it isn't semisimple, and $T_1 \cdots T_p$ is a root of $X^p - c^p = (X-c)^p$ but not $X-c$ so it isn't semisimple. – user28172 Dec 3 '12 at 16:51
One more question: Is there any global condition on $K$ implying the semisimplicity of the sum (the product) of "any" pair of commuting semisimple operators? The separability condition you gave depends on $T_1$ and $T_2$. (for example $char=0$ is a sufficient condition, because it implies separability). – M. Shahryari Dec 3 '12 at 16:59
@Shahriyari: You need all finite extensions of $K$ to be separable, that is --- you need $K$ to be a perfect field. This holds also, for instance, for finite fields. Otherwise, a version of the same argument should work. – Ilya Bogdanov Dec 3 '12 at 17:06
@noar: Yes, my example is just a particular case of this construction. A more direct example in char p is just the same. If $K=F_p(t)$, then we may consider a multiplication by $X$ and $Y$ in $K[X,Y]/(X^p-t,Y^p+t)$. Then $(X+Y)^p=0$ and $(XY+t^2)^p=0$. – Ilya Bogdanov Dec 3 '12 at 17:12

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