# Determinant perturbation approximation

The following problem comes from a max likelihood calculation for gaussian families, but is of independent interest.

Is it possible to find a closed-form approximation for small values of $x$ for

$\text{det}(B + xI)$

where I is the identity matrix and B is hermitian rank-deficient positive semidefinite?

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 can I assume you know the eigenvalues of $B$? If so, no approximation is needed. – cardinal Feb 16 '11 at 15:05

I'll assume that you already know the eigenvalues of $B$. Since $B$ is symmetric positive semidefinite, it can be decomposed as $$B = U D U^T$$ where $U$ is an orthogonal matrix and $D$ is the diagonal of nonnegative eigenvalues (some of which may be exactly zero).

Now $$B+xI = U D U^T + x U U^T = U (D + x I) U^T$$ and since the determinant of a matrix is the product of its eigenvalues and the determinant is distributive over matrix products, then $$|B+xI| = |D+xI| = \prod_n (d_n + x)$$ where $d_n$ is the $n$th diagonal entry of $D$.

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Of course! Thanks. – gappy Feb 16 '11 at 15:35
@gappy, sure. Regards. – cardinal Feb 16 '11 at 15:37

I second @cardinal's answer, but provide a simple trick: If $p(z)$ is a polynomial (with integer powers), and $\mathbf{v}, \lambda$ are eigenvector and corresponding eigenvalue of matrix $M$, then $\mathbf{v}, p(\lambda)$ are eigenvector and corresponding eigenvalue of $p(M)$. The proof is a simple exercise. The polynomial $p$ may contain negative powers of $z$ and a constant term, which in the case of $p(M)$ corresponds to adding the constant times the identity matrix.

Since the determinant is the product of the eigenvalues, the determinant of $A = p(B)$, where $p(z) = z^1 + x$ is the product $\prod_i \left(\lambda_i + x\right)$, where $\lambda_i$ are the eigenvalues of $B$. You can also use this trick to find the trace of $p(B)$, of course, but it is overkill!

This polynomial trick is a classic in numerical analysis, used, for example, to prove convergence of the Gauss-Seidel method. See Cheney & Kincaid, or my answer to another question involving this trick.

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 The trick applies to non-integer powers, but the proof is a little different. – shabbychef Feb 16 '11 at 18:07