Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If a finite dimensional matrix $A$ is positive semidefinite, how can we write $A$ as a polynomial in $A^{2}$? Thanks.

share|improve this question
    
Interesting question. How does it arise? –  user1551 Nov 30 '11 at 13:25
add comment

1 Answer 1

up vote 4 down vote accepted

$A = q(A^2)$ for polynomial $q$ if and only if $\lambda = q(\lambda^2)$ for each eigenvalue $\lambda$ of $A$. Use Lagrange interpolation. If $A$ is $n \times n$, there is a solution of degree at most $n-1$.

Alternatively: knowing there is such a solution, write $q(t) = \sum_{j=0}^{n-1} a_j t^j$. Let $p(t)$ be the minimal polynomial of $A$ (which can be obtained from the squarefree factorization of the characteristic polynomial of $A$). We want $q(t^2) - t$ to be divisible by $p(t)$. If $r_j(t)$ is the remainder of $t^{2j}$ on division by $p(t)$, this says $\sum_{j = 0}^{n-1} a_j r_j(t) - t = 0$. Set to $0$ the coefficients of this for each power of $t$ from $t^0$ to $t^{n-1}$, and you have $n$ equations to solve for the $n$ coefficients $a_0, \ldots, a_{n-1}$. This version of the solution does not require solving for the eigenvalues, and in fact if the entries of $A$ are rational the computation uses only rational arithmetic and the result will have rational coefficients.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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