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Let $S_{>0}$ be the space of symmetric positive definite real $n \times n$ matrices.

Is there a closed-form formula for the positive square root function $\sqrt \cdot :S_{>0} \to S_{>0}$?

I want an explicit formula (using roots, algebraic operations, and perhaps some other natural operators) for $\sqrt A$ in terms of the entries of $A$.

So, using the fact that $A$ is orthogonally diagonalizable is not explicit enough for me:

We can write $A=Q \Sigma Q^T$, and so $\sqrt A=Q \sqrt\Sigma Q^T$, but there are no explicit formulas for the eigenvalues - since they are roots of a high-order polynomial (when $n \ge 5$).

However, a-priori the fact we do not have a formula for the eigenvalues does not rule out that the specific combination $Q \sqrt\Sigma Q^T$ could be expressed in a formula.

A related question is whether there is a formula for the positive factor of a matrix:

Given an $n \times n$ matrix $A$, it can be written as $A=OP$, where $O$ is orthogonal and $P$ is symmetric positive definite. $P$ is given by $P=\sqrt{A^TA}$.

Is there a formula for $P$?

(If $A=U\Sigma V^T$ is the SVD of $A$, then $P=V\Sigma V^T$ but again this is not really explicit).

Of course, if we had a formula for the square root, we had also a formula for $P$, but maybe more can be said on this particular case - in particular $P(A)$ behaves homogeneously with $A$, i.e. $P(\lambda A)=\lambda P(A)$, while the square root behaves differently of course.

I guess I am rather pessimistic about the existence of nice formulas, but I haven't seen an explicit discussion of these anywhere. (Except for when $n=2$).

Asaf Shachar
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1 Answers1

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$\textbf{Proposition}.$ Let $A\in S_n^{++}(\mathbb{Q})$. Then $\chi_A$, the characteristic polynomial of $A$, is not solvable by radicals IFF $\sqrt{A}$ is not calculable by radicals.

We consider a generic matrix $A$; in particular, its eigenvalues $(\lambda_i)$ are distinct and the Galois group of $\chi_A$ is the total permutation group. Note that $\sqrt{A}$ is a polynomial in $A$ and decomposes uniquely on $\{I,\cdots,A^{n-1}\}$. Thus, if we can calculate $\sqrt{A}$, then we can calculate the coefficients $(u_i)$ in $\sqrt{A}=\sum_{j=0}^{n-1}u_jA^j$.

First point of view. $\sqrt{A}=P(A)$ where $P$ is the Lagrange interpolating polynomial (of degree $n-1$) that sends the $\lambda_i$'s on the $\sqrt{\lambda_i}$'s. We see that the $(u_j)$'s are in $\mathbb{Q}(\sqrt{\lambda_1},\cdots,\sqrt{\lambda_n})$, an algebraic extension of $\mathbb{Q}$ of degree $2^nn!$.

In particular, if $\chi_A$ is solvable, then we can calculate $\sqrt{A}$ by radicals.

Otherwise, one may wonder if these elements are in a solvable extension of $\mathbb{Q}$, for example, of dimension $2^n$. The answer is no as we will see.

Second point of view. We write the system of $n$ equations in the $n$ unknowns $(u_j)$; $(\sum_{j=0}^{n-1}u_jx^j)^2=x \;mod( \;\chi_A(x))$.

This system reduces (in general) to solving a polynomial of degree $2^n$ with Galois group containing $2^nn!$ elements, that is not solvable when $n\geq 5$.

That follows is a randomly chosen example for $n=5$

$A = \begin{pmatrix}21& 2& -5& -9& 1\\2& 10& -5& -6& 6\\-5& -5& 14& -6& -9\\-9& -6& -6& 24& 9\\1& 6& -9& 9& 17\end{pmatrix}>0$.

$\chi_A(x)=x^5-86x^4+2491x^3-27698x^2+102394x-61296$ has $S_5$ as Galois group. $u_0$ is some root of a polynomial of degree $32$ with a Galois group containing $2^55!$ elements and $u_1,u_2,u_3,u_4$ are polynomials in $u_0$ with rational coefficients.