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Let $P\in{\mathbb R}[X]$ be a monic polynomial with roots on the unit circle. For the problem below, we may assume wlog that the roots are simple and distinct from $\pm1$. It can be shown that there exists a matrix $M\in{\bf SO}_n({\mathbb R})$, whose characteristic polynomial is $P$ (an orthogonal companion matrix of $P$, in short OCM). See for instance Exercise 99 on my list . Regretfully, this exercise uses the square root of Hermitian positive definite matrices, which cannot be computed in finitely many operations.

Does there exist a construction of an OCM that uses only finitely many elementary operations (including the square root of complex numbers) ?

Thanks to the reduction to Hessenberg form, which can be done in finite time and which preserves the orthogonal group, we may restrict our attention to a Hessenberg orthogonal matrix $M$. It writes $$\left( \begin{array}{ccccc} c_1 & s_1c_2 & s_1s_2c_3 & s_1s_2s_3c_4 & \ldots \\\\ -s_1 & c_1c_2 & c_1s_2c_3 & c_1s_2s_3c_4 & \ldots \\\\ 0 & -s_2 & c_2c_3 & s_2s_3c_4 & \ldots \\\\ 0 & 0 & -s_3 & c_3c_4 & \ldots \\\\ 0 & 0 & 0 & -s_4 & \ldots \end{array} \right)$$ where $(c_j,s_j)$ are cosine/sine pairs.

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This sounds like the "inverse unitary eigenproblem"; see for instance. – J. M. Sep 16 '10 at 10:07
Nice paper, but my question is different: $n$ is even and the data consists of $n/2$ real numbers, the arguments of the roots (associated pairwise by c.c.). Whereas there are $n$ parameters left (the angles of the cosines). The solution exists but must be highly non-unique. I wish an algorithm, involving elementary operations and square roots of complex numbers, giving such a matrix. Such a trick is well-known in the Inverse Symmetric Eigenvalue Problem. – Denis Serre Sep 16 '10 at 16:11
up vote 3 down vote accepted

I'd do this in three steps:

  1. Find any $2n \times 2n$ matrix $A$ whose eigenvalues are $e^{\pm i \theta}$.

  2. Find a positive definite quadratic form preserved by $A$. In equations, we want $A P A^T = P$.

  3. Find an orthonormal basis for $P$, using the Gram-Schimdt algorithm. In equations, we want $S P S^T = \mathrm{Id}$.

Then $S A S^{-1}$ is orthogonal and has the required eigenvalues.

I can think of two ways to do step 2. The first is more purely algebraic, the second I think would be much easier to implement.

Algebra: Let $f(x) = \prod_{j=1}^{n} (x-e^{i \theta_j}) (x - e^{-i \theta_j}) = \prod (x^2 - 2 \cos \theta_j + 1)$ be your characteristic polynomial. Let $V$ be the ring $\mathbb{R}[x]/f(x)$. Note $1$, $x$, ..., $x^{2n-1}$ is a basis for this ring, in which multiplication can be written down algebraically in terms of the coefficients of $f$. Also, multiplication by $x$ has the desired eigenvalues, so that accomplishes part 1.

For $y \in T$, let $T(y)$ be the trace of multiplication by $y$. Also, let $y \mapsto \overline{y}$ be the automorphism of $V$ induced by $x \mapsto x^{-1}$. Again, both of these can be written down, in the monomial basis, algebraically in terms of the coefficients of $f$.

Then $\langle y,z \rangle = T(y*\overline{z})$ is the desired positive definite quadratic form. Namely, observe that the ring $V$ is isomorphic to $\mathbb{C}^{\oplus n}$. In terms of this isomorphism, $\langle (z_1, \ldots, z_n), (z_1, \ldots, z_n) \rangle = 2 \sum |z_i|^2$.

In practice: The condition that $APA^T = P$ is a linear condition on $P$. Let $W$ be the subspace of the vector space of symmetric matrices where this condition is satisfied; finding $W$ is just algebra. Now, our goal is to find a positive definite element of $W$. For large $N$, $(1/N) \left( \mathrm{Id} + AA^T + A^2 (A^{T})^2 + \cdots + A^{N-1} (A^T)^{N-1} \right)$ is positive definite and is near $W$. I would guess that the orthogonal projection of this matrix onto $W$ would probably be positive definite for large $N$.

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In second paragraph of Algebra, did you mean For $y \in V$, let $T(y)$ ... automorphism of $V$ ... ? Is this a fair summary: Let $A_0$ be the companion matrix of $f$ (in the usual sense). Let $SA_0$ be the induced action $Q \to A_0 Q A_0^t$ on the vector space of quadratic forms. Then $1$ is an eigenvalue of multiplicity $n$. The eigenspace $Q_0$ has a basis consisting of symmetric integer matrices. The set of positive definite quadratic forms form an open convex cone within $Q_0$ defined by positivity of various determinants, so a positive definite integer matrix can be readily found. – Bill Thurston Sep 17 '10 at 1:21
Thanks for catching those typos. And, yes, that's a very fair summary. – David Speyer Sep 17 '10 at 11:11

You can use the Cayley transform and reduce this problem to generating a skew-symmetric matrix with a prescribed characteristic polynomial. For example, this works in the $3\times 3$ case (although when $n$ is odd, $1$ is always an eigenvalue). I have not thought about it thoroughly, but presumably, methods used in Inverse Symmetric Eigenvalue Problem should apply in the skew-symmetric case.

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You can go from the skew-symmetric to the self-adjoint problem by multiplication by $i$. I think this approach would work, one just needs to carefully check that one always stays in the right set... – Helge Sep 16 '10 at 22:08
I didn't understand the comment about multiplication by $i:$ both symmetry and skew symmetry are linear conditions, therefore, they are invariant under scalar multiplication. – Victor Protsak Sep 16 '10 at 23:37
Self-adjointness isn't, since it's $a_{ij} = \overline{a_{ij}}$. It's only REAL linear. – Helge Sep 17 '10 at 11:30
And it's "well-known" that self-adjointness is the "correct" condition for complex matrices ... (There is some room for debate here, that's why the "). – Helge Sep 17 '10 at 11:31
The problem is not about "correct conditions", it is manifestly about reconstructing REAL orthogonal matrices with prescribed eigenvalues, which become real skew-symmetric after Cayley transform. Replacing "real skew-symmetric" with "complex skew-hermitian" corresponds to replacing special orthogonal group with special unitary group. There are other ways to generalize the question, too, but is there any evidence that these generalizations can be solved more easily than the original question? No matter how many times you say "halva", your mouth wouldn't become sweet. – Victor Protsak Sep 17 '10 at 15:07

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