Taylor Expansion of Logarithm of Determinant near Identity for Non-Diagonalizable Matrix

Question

I have been working on a problem where I need to Taylor expand an expression of the form $\log \det(I-A)$ in terms of traces of the matrices $A^m$ for $m \in \mathbb N$, where $A$ is a general $n \times n$ matrix.

I did notice that if the eigenvalues of $A$ are $\lambda_1, \cdots , \lambda_n$ then those of $I-A$ are exactly $1 - \lambda_1, \cdots , 1- \lambda_n$, so we may write $$\log \det(I-A) = \sum_{i=1}^n \log (1 - \lambda_i) = \sum_{i=1}^n \sum_{m=1}^\infty \frac{(-1)^m \lambda_i^m}{m} = \sum_{m=1}^\infty \frac{(-1)^m}{m} \sum_{i=1}^n \lambda_i^m \hspace{10mm} \cdots (1)$$

At this point, I noticed that if $A$ were diagonalizable then I could say that $P^{-1}AP = diag\{\lambda_1, \cdots , \lambda_n\}$ (the diagonal matrix with entries $\lambda_1, \cdots , \lambda_n$ along the principal diagonal), so for every $m \geq 1$, I could write $P^{-1}A^mP = diag\{\lambda_1^m, \cdots , \lambda_n^m \}$ and $tr(A^m) = tr(P^{-1}A^mP) = \sum_{i=1}^n \lambda_i^m$ and (1) would then give us $$\log \det(I-A) = \sum_{m=1}^\infty \frac{(-1)^m}{m} tr(A^m)$$

which is what I want. But I couldn't get around the case when $A$ was non-diagonalizable. I was wondering what happens in that case. Can we still give the same (or maybe similar) expansions? Would the Smith Normal Form, Rattional Canonical Form etc. be of any help?

P.S.: I didn't find any standard reference containing the kind of expansion I wanted. I would appreciate if I would come to know what is the best thing one can say in the diagonalizable case, and/or if i were pointed out to some reference containing with these kinds of result(s).

Edit: I think my argument for $tr(A^m) = \sum_{i=1}^n \lambda_i^m$ doesn't require diagonalizability of $A$: eigenvalues of $A^m$ should exactly be $\lambda_1^m, \cdots , \lambda_n^m$ because if $\lambda$ is an eigenvalue of $A$ with eigenvector $v$ then $Av = \lambda v$ implies by induction that $A^kv = \lambda^k v$ for every $k \in \mathbb N$. So I believe this result should hold in general (by the same proof as above). Can someone confirm? I decided not to delete this post as I haven't found this question or anything similar to this asked on this site.

Answer 1

Denote the eigenvalues of $X$ by $\lambda_k$, then using the basic definitions of the $(\det,{\rm trace},\log)$ operations, and assuming $X$ is non-singular $$\eqalign{ \log\left(\prod_k\lambda_k\right) &= \sum_k\,\log(\lambda_k) + 2n\pi i \\ \log(\det{X}) &= \operatorname{tr}(\log{X}) + 2n\pi i \\ }$$ which is actually just another form of Jacobi's formula.

Set $X=(I-A)$ and assume that $\|A\|\lt 1$
then substitute the well-known Taylor series $$\eqalign{ \log(I-A) &= -\sum_{k=1}^\infty \frac{A^k}{k} \\ }$$ into Jacobi's formula to obtain $$\eqalign{ \log\!\big(\det(I-A)\big) &= 2n\pi i-\operatorname{tr}\left(\sum_{k=1}^\infty\frac{A^k}{k}\right) \\ &= 2n\pi i-\sum_{k=1}^\infty \frac{{\rm tr}\left(A^k\right)}{k} \\ }$$

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NB:$\;$ The $(2n\pi i)$ term is usually zero but not always, e.g. $$\eqalign{ X &= \pmatrix{2-i&0\\0&-2-i} \quad\implies\quad \log(\det X) - {\rm tr}(\log X) = 2\pi i \\ \\ }$$ I'm not sure if the conditions $\big\{X=I-A,\,\,\|A\|<1\big\}$ will ensure that the term is zero.

The details involve (un)winding numbers and the evaluation of complex functions.

To summarize Higham's excellent paper, the unwinding number is a function $\,{\cal U}:{\mathbb C}\to{\mathbb Z}$
which measures the discrepancy between $\log(e^z)$ and $z$. $$\eqalign{ z &= \log(e^z) + 2\pi i\,{\cal U}(z) \\ }$$ It can be defined in various ways, e.g. $$\eqalign{ {\cal U}(z) &= \frac{z-\log(e^z)}{2\pi i} = \left\lceil\frac{Im(z)-\pi}{2\pi}\right\rceil \\ }$$ For the current problem, Lemma 2.5 (on the third page of the paper) says that $$\eqalign{ \log\left(\prod_k\lambda_k\right) &= \left(\sum_k\,\log(\lambda_k)\right) - 2\pi i\;{\cal U}\!\left(\sum_k\,\log(\lambda_k)\right) \\ }$$ One of the many interesting properties of the unwinding function is that its derivative is zero, and this leads to one of the most useful formulas in all of matrix calculus: $$\eqalign{ \frac{d\log(\det{X})}{dt} \;=\; \frac{d\operatorname{tr}(\log{X})}{dt} - 2\pi i\,(0) \;=\; \operatorname{tr}\left(X^{-1}\frac{dX}{dt}\right) \\ }$$

Answer 2

To the OP; you are not professional enough -remember that amateurism doesn't pay- You should be asking yourself $4$ questions, not just one.

i) What is the log of a complex number?

ii) Is $\log(\Pi \mu_k)=\sum \log(\mu_k)$?

iii) For which $\lambda_k$ does your double series converge ?

iv) Is the complete spectrum of $A^m$ the image of the complete spectrum of $A$ by $x\mapsto x^m$ ?

ANSWERS. i) The principal log is defined by

$\log:z\in\mathbb{C}\setminus (-\infty,0]\mapsto \log(r)+i\theta$ where $z=re^{i\theta}$ with $\theta\in (-\pi,\pi)$.

ii) No. Consider $\log((1+i)^5)$.

iii) That works when $|\lambda_k|<1$, that is, $\rho(A)<1$.

iv) The result is true but your proof is false (because your method says nothing about the multiplicity of eigenvalues). The simplest method is to triangularize $A$ over $\mathbb{C}$ (then $A^m$ becomes triangular too).

The simplest case is when $A\in M_n(\mathbb{R})$ and $\rho(A)<1$. Then the eigenvalues $1-\lambda_k$ of $I-A$ are $>0$ or conjugate and the $\log(1-\lambda_k)$ are well defined; thus the imaginary parts cancel by pairs and, in this case, ii) is true. The sequel is straightforward.