Let $$ u(T)=\sum\_{n = 0}^\infty a\_nT^n$$ be a formal power series over a field $K$. Then why does $u(T)$ lie in $K(T)$ (i.e. is the Taylor expansion of a rational function) if and only if there is an $N > 0$ such that the Hankel determinants $$\det(a\_{i + j + M}){\_{0 \le i, j \le N}} = \det \begin{pmatrix} a\_M & a\_{M + 1} & \ldots a\_{M + N} \\ a\_{M + 1} & a\_{M + 2} & & \\ \vdots & & & \\ a\_{M + N} & \ldots & & a\_{M + 2N} \end{pmatrix}$$vanish for all $M\gg0$?



Formal power series is Taylor expansion of rational function iff Hankel determinants vanish?
Call your displayed matrix $H\_{M,N}$.


**Theorem** The following are equivalent:


1. $u$ is the Taylor series of a rational function.
2. There is a finite sequence $q\_0,\ldots, q\_N$, not all zero, such that for all $m\gg0$, $a\_mq\_N+a\_{m+1}q\_{N-1}+\cdots+a\_{m+N}q\_0=0$.
3. There exists $N$ and $M$ such that $|H\_{m,N}|=0$ for all $m>M$.
4. There exists $M$ and $N$ such that $|H\_{m,n}|=0$ for all $m>M$ and $n>N$.


**Proof that 1) is equivalent to 2):**


$u$ is rational iff there is a polynomial $q$ with $qu$ a polynomial. Taking the $q\_i$ to be the coefficients of $q$, we are done.


**Proof that 2) implies 3):** Suppose $a\_mq\_N+\cdots+a\_{m+N}q\_0=0$ for all $ m>M$. Then for all $m>M$, $H\_{m,N}$ kills the column vector $\pmatrix{q\_0&q\_1&\ldots&q\_N}^T$, so $|H\_{m,N}|=0$.


**Proof that 2) implies 4):** Exactly as above, noting that
$$H\_{m,N+p}\cdot\pmatrix{q\_0\cr\vdots\cr q\_N\cr 0\cr \vdots\cr 0}=0$$


**Proof that 4) implies 3):** Trivial.


**Proof that 3) implies 2):**


Let $N$ be the smallest number with the property that there exists $M$ with $|H\_{m,N}|=0$ for all $m>M$. If $M=1$, then $u$ is a polynomial and we are done. Otherwise there are arbitrarily large values of $m$ with $|H\_{m,N-1}|\neq 0$.


Claim: $|H\_{m,N-1}|\neq 0$ for *any* $m>M$. Proof of claim: Let $m>M$. It is pretty easy to check that
$$|H\_{m,N-1}||H\_{m+2,N-1}|=|H\_{m+1,N-1}|^2+|H\_{m,N}||H\_{m+2,N-2}|$$
(I'll come back and insert a proof of this displayed equation when I have a little more time.)
By assumption, $|H\_{m,N}|=0$, so $|H\_{m,N-1}|=0$ would imply $|H\_{m+1,N-1}|=0$, whence by induction $|H\_{m+p,N-1}|>0$ for all $p>0$, contradicting the last line of the preceding paragraph. This proves the claim.


Continue to assume $m>M$.


Note that $H\_{m,N}$ is an $(N+1)\times (N+1)$ matrix of determinant zero that contains a nonsingular $N\times N$ submatrix (namely $H\_{m,N-1}$) and hence has rank precisely $N$. Thus there is a unique (up to linear multiples) column vector $\overline{q}$ such that
$$H\_{m,N}\overline{q}=0$$


Then $\overline{q}$ is the unique (again up to scalar multiples) column vector killed by the top $n$ rows of $H\_{m,N}$ and also the unique column vector killed by the bottom $n$ rows. Now since $H\_{m+1,N}$ is also rank 1, and because its kernel must be killed by the bottom $n$ rows of $H\_{m,N}$,it follows that the kernel is again generated by the same vector $\overline{q}$, and by induction $\overline{q}$ is killed by $H\_{m+p,N}$ for all $p\ge 0$, so that in particular $a\_{m+p}b\_0+\cdots+a\_{m+p+N}b\_N=0$ for all $p>0$, which, modulo a change of notation, is 2).


**Edited to add:**


The displayed equation above is a special case of the following:


**Theorem**: Let $a,b,c,d$ be scalars. Let $\alpha,\beta$ be row vectors of length $n$. Let $\phi,\psi$ be columns of length $n$. Let $B$ be an $n\times n$ matrix. Let $A$ be the following $(n+2)\times(n+2)$ matrix:
$$A=\pmatrix{a&\alpha&b\cr\phi&B&\psi\cr c&\beta&d\cr}$$
Then
$$\det\pmatrix{a&\alpha\cr\phi& B\cr}\cdot \det\pmatrix{B&\psi\cr \beta &d\cr}
=\det\pmatrix{\alpha&b\cr B&\psi\cr} \det\pmatrix{\phi&B\cr c&\beta\cr}+\det(A)\cdot\det(B).$$


**Proof:** We can assume $B$ is invertible and then premultiply $A$ by
$$\pmatrix{1&0&0\cr 0&B^{-1}&0\cr 0&0&1\cr}$$
where the zeroes represent scalars, rows or columns as appropriate. This does not affect the truth of the theorem and allows us to assume $B$ is the identity, after which row and column operations render the desired equality trivial.


