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 $n=2$).


$\textbf{Proposition}.$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.