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We know that

$$\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)^2=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$

Why cannot we say that the square root of the later matrix is the former one? Can we somehow consistently define square root from nilpotent matrices?

I know that the equation $X^2=A$ would probably have more than one solution but why we cannot define the "main branch" exactly this way?

Anixx
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  • The non-uniqueness is the problem. What should be "the" solution ? In the case of a diagonal matrix, it is easy , but how should we handle the general case , in particular singular matrices ? If you know a good possibility, please mention it here. – Peter Dec 11 '22 at 10:21
  • @Peter I thought only about diagonal matrices, I think, in this case the square root should be defined... – Anixx Dec 11 '22 at 10:24
  • @Peter I think, singular matrices are not nilpotents?... – Anixx Dec 11 '22 at 10:25
  • Hm, isn't being singular not necessary for being nilpotent ? But it is not sufficient, right ? – Peter Dec 11 '22 at 10:32
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    It should be noted that in $\Bbb C^{n\times n}$ (or in $\overline {\Bbb F}^{n\times n}$, for that matter) the equation $X^2=A$ (for a generic matrix $A$, not necessarily nilpotent) has a solution only under some conditions on the sequence ${\dim\ker A^n}_{n\in\Bbb N}$. I'll see if I can find a link. – Sassatelli Giulio Dec 11 '22 at 10:36
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    The non-uniquenes can be pretty bad. For example all the matrices $$\pmatrix{0&x&0&0\cr 0&0&1/x&0\cr 0&0&0&x\cr0&0&0&0\cr}$$ with $x$ arbitrary non-zero element of $\Bbb{F}$ work in your example. My point is that you will have difficulties selecting a main branch when there is a continuum of nearby matrices that also work. – Jyrki Lahtonen Dec 11 '22 at 11:01
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    @Peter matrix $\left( \begin{array}{cc} 1 & 1 \ 1 & 1 \ \end{array} \right)$ is singular but not nilpotent. – Anixx Dec 11 '22 at 11:11
  • @SassatelliGiulio what is $\mathbb F$?.. What is $\text{ker}$? – Anixx Dec 11 '22 at 11:13
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    @JyrkiLahtonen but for formal power series the operation is well defined and they are isomorphic to those diagonal matrices. – Anixx Dec 11 '22 at 11:16
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    I do think that for matrices with simple eigenvalues the only source of non-uniqueness is, roughly speaking, the choice of square roots of the eigenvalues. That is somewhat manageable (choice of branch). When some roots of the characteristic polynomial aren't simple, then the trouble starts. Particularly in characteristic two. – Jyrki Lahtonen Dec 11 '22 at 11:20
  • Your point about formal power series is interesting... My first impression is that there are many ways to define a ring homomorphism from formal power series to upper triangular matrices. Some formal power series may have the same matrix as their homomorphic image under different homomorphisms in such a way that the square root of the power series is mapped to distinct matrices. – Jyrki Lahtonen Dec 11 '22 at 11:22
  • Instead of power series I was thinking about linear transformations on the space of sequences $(a_n)_{n\in\Bbb{N}}$. Fix a non-zero constant $x$. Consider the transformation $T_x$ that multiplies the odd numbered entries by $x$, multiplies the even numbered entries by $1/x$, and then shifts the entire sequence by one position. All the transformations $T_x$ are square roots of the transformation that simply shifts the sequence by two positions. That is also what I meant by using different homomorphisms from the ring $\Bbb{F}[[T]]$ of formal power series. You can map $T\mapsto T_x$ and extend. – Jyrki Lahtonen Dec 11 '22 at 11:34
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    We could ignore transition matrices and work with Jordan forms only no ? This way the $x,\frac 1x,x$ matrix proposed for instance become $1,1,1$. – zwim Dec 11 '22 at 11:34
  • @JyrkiLahtonen formal power series are isomorphic to upper triangular matrices where the first row is the coefficients and each following row is shifted by one element to the right. – Anixx Dec 11 '22 at 11:54
  • @zwim I would be glad if someone could construct a formula, and preferably Mathematica code for this... – Anixx Dec 11 '22 at 12:00
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    My point is that the map you describe is not the only way to map the formal power series into upper triangular matrices. – Jyrki Lahtonen Dec 11 '22 at 12:30
  • @JyrkiLahtonen definitely. But I would consider it canonical. Also, regardless of mapping, square root in formal power series is unique. – Anixx Dec 11 '22 at 12:32
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    @Anixx Here it is: given a matrix over an algebraically closed field of characteristic $\ne 2$, the equation $X^2=A$ has a solution if and only if there aren't $j\ge1$ and $k\in\Bbb N$ such that $\dim\ker A^{j+1}-\dim \ker A^j=\dim\ker A^j-\dim\ker A^{j-1}=2k+1$. https://math.stackexchange.com/questions/65227/under-what-conditions-does-a-matrix-a-have-a-square-root I was wrong in assuming that it was the same in characteristic $2$. I think in characteristic $2$ you need the same condition on all $\dim \ker(A-\lambda I)^k$, not just $\lambda=0$, but I'm not sure. – Sassatelli Giulio Dec 11 '22 at 14:28

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