Do all complex operators $T: \mathbb{C}^n \rightarrow \mathbb{C}^n$ admit a square root $S$ such that $S \circ S = T$?

It is clear that all normal operators admit this: Write the matrix $T$ in the eigenbasis (that is guaranteed from spectral decomposition of normal operators), which gives us a diagonal representation, with $t_i$ as the diagonal elements. That is, $T = \text{diag}(t_1, t_2, \dots t_n)$.

Let $S$ be the diagonal matrix (in the eigenbasis of $T$) with entries $\sqrt t_i$. is always possible since $\mathbb{C}$ is algebraically closed. That is, $S = \text{diag}(\sqrt t_1, \dots, \sqrt t_n).$ Clearly, $S \circ S = T$.

I assumed the existence of an eigenbasis of $T$ for this. What happens in general, for all operators? What about other fields? More general an answer, the better :)

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    $\begingroup$ This has been answered here many times. Let me see if I can find it, you could search better too. This is one place but not the only one. $\endgroup$ – blueInk Mar 23 '18 at 10:51
  • $\begingroup$ Ah, I kept searching about "operator", not "matrix". The questions are not equivalent, since being an operator would allow one to synthesize many matrices by picking different bases, correct? So the matrix question is more specific. $\endgroup$ – Siddharth Bhat Mar 23 '18 at 10:58
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    $\begingroup$ Incorrect. Fixed one basis the correspondence between matrices and linear operators is a ring isomorphism in your finite-dimensional space. $\endgroup$ – blueInk Mar 23 '18 at 10:59
  • $\begingroup$ Damn, you're right. Thank you :) What happens in the infinite dimensional case? $\endgroup$ – Siddharth Bhat Mar 23 '18 at 11:00
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    $\begingroup$ This can happen $\endgroup$ – blueInk Mar 23 '18 at 11:03

For the case $T$ singular, see the Marc van Leeuwen's post in

$A = B^2$ for which matrix $A$?

In the sequel, we assume that $K$ is a perfect field.

When $T\in M_n(K)$ is invertible, if you want explicitly (that is, not an approximation) a square root, then practically the Jordan form is useless.

EDIT 1. There is a method, with polynomial complexity, that reduces the problem to the case where $T$ is semi simple. Yet, to obtain an explicit square root in this last case, there remains a gap because we must work in a field that contains the square roots of the eigenvalues of $T$.

Calculate the Jordan Chevalley decomposition: $T=D+N$ where $D$ is semi simple, $N$ is nilpotent, $DN=ND$ and $D,N$ are polynomials of $T$ of degree $<n$ with coefficients in $K$. There exists a pseudo-Newton algorithm to do that; cf. (in french)


Clearly, $D$ is invertible and $D^{-1}N$ is nilpotent; now $T=D(I+D^{-1}N)$ and, (formally and rigorously)


Note that the second term of RHS is a matrix $M\in K[T]$. We want only square roots that are polynomials in $T$; yet the coefficients are no more in $K$ but in an algebraic extension $L$ of $K$ (cf. below).

Note that $D=Pdiag(\lambda_1I_{n_1},\cdots,\lambda_pI_{n_p})P^{-1}$ over $\overline{K}$, the algebraic closure of $K$ (the $(\lambda_i)$ being distinct). Then there are $2^p$ values for $D^{1/2}\in L[T]$: $D^{1/2}=Pdiag(\sqrt{\lambda_1}I_{n_1},\cdots,\sqrt{\lambda_p}I_{n_p})P^{-1}$.

Let $q$ be the minimal polynomial of $D$ (of degree $p$). Then a solution as above for $D^{1/2}$ can be written $r(D)$ where $r(x)=a_0+\cdots+a_{p-1}x^{p-1}$ satisfies $r(x)^2=x \;mod(q(x))$; it's a system of $p$ equations of degree $2$ in the $p$ unknowns $(a_i)$.

EDIT 2. Using Grober basis theory, we can eliminate $p-1$ among the $p$ unknowns; in fine, the software gives a polynomial $s\in K[x]$ of degree $2^p$ that is canceled by the last unknown (note that its coefficients are very large for $n\geq 5$). If $L$ is the decomposition field of $s$, then $r\in L[x]$ and $D^{1/2}\in M_n(L)$. Note that $r_{\epsilon}$ is the Lagrange interpolation polynomial that sends the $(\lambda_i)$ to the $(\epsilon_i\sqrt{\lambda_i})$; that explains why $L$ contains all the $(\sqrt{\lambda_i})_i$; then the Galois group of $s$ has (in the generic case where $p=n$) $n!2^n$ elements. Finally, the problem "find explicitly some square roots of $D$" is solvable (by radicals) IFF the characteristic polynomial of $T$ is solvable. For a generic $T$, that can be done only if $n\leq 4$.

$\textbf{Proposition.}$ Let $T\in GL_n(K)$ be s.t. the minimal polynomial $m_D$ of $D$ has degree $p$. Then there are $2^p$ square roots of $T$ that are polynomials in $T$; each square root is in the form $r(D)M$ where $M\in K[T]$ is calculable and $r(D)\in M_n(L)$ where $L$ is the decomposition field of a polynomial $s\in K[x]$ of degree $2^p$ which is explicitly calculable. Moreover, $s$ is solvable iff $m_D$ is solvable.

$\textbf{Remark.}$ We can do similar calculations for the $p^{th}$ roots of $T$.


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