Connection Between Bézout's Identity and Linear Algebra Today I looked at Bézout's Identity and I became reminded a bit of linear algebra. Let me explain.
Bézout's Identity says that for coprime integers $a, b$, there exists integers $x,y$ such that $ax+by=1$. So in essence $a$ and $b$ can generate any integer $n$ through the linear combination $a(nx)+b(ny)=n$. I thought this reminiscent of the notion of linear independence and a spanning set, in that coprime integers can be thought of as independent, and two coprime integers $\{a,b\}$ can be said to span the integers.
It seems to me that this bit of number theory generalizes to some algebraic structure. Is my intuition off? My knowledge of abstract algebra is to the extent of elementary group theory.
 A: Bezout's identity with polynomials is used in linear algebra when you want to decompose a vector space according to the action on it by a linear operator.
For example, we'll show a vector space is a direct sum of its generalized eigenspaces for different eigenvalues.  Let $V$ be a finite-dimensional complex vector space and $A \colon V \to V$ be linear with minimal polynomial $f(X) = \prod_{i=1}^m (X - \lambda_i)^{e_i}$: the $\lambda_i$'s are the eigenvalues of $A$. Set $V_i = \ker((A - \lambda_i)^{e_i}) = \{v \in V : (A - \lambda)^{e_i}(v) = 0\}$.  We want to show $V = \bigoplus_{i=1}^m V_i$.
Step 1: $V = \sum_{i=1}^m V_i$.
Set $g_i(X) = f(X)/(X - \lambda_i)^{e_i}$. Since $(X - \lambda_i)^{e_i}g_i(X) = f(X)$, substituting $A$ for $X$ gives us $(A - \lambda_i)^{e_i} g_i(A) = f(A) = O$, so $g_i(A) \colon V \to V$ has image in $\ker((A - \lambda_i)^{e_i})$.
The polynomials $g_i(A)$ are relatively prime as an $m$-tuple: $\gcd(g_1(X), \ldots, g_m(X)) = (1)$, so by Bezout some $\mathbf C[X]$-linear combination of them is 1: $g_1(X)h_1(X) + \cdots + g_m(X)h_m(X) = 1$ in $\mathbf C[X]$.  Therefore $g_1(A)h_1(A) + \cdots + g_m(A)h_m(A) = I$, so for each $v \in V$ we have
$$
v = g_1(A)(h_1(A)v) + \cdots + g_m(A)(h_m(A)v).
$$
The image of $g_i(A) \colon V \to V$ is inside $V_i$, so $V = \sum_{i=1}^m V_i$.
Step 2: The sum is direct.
Suppose $v_1 + \cdots + v_m = 0$ where $v_i \in V_i$.  We want to prove each $v_i$ is $0$. By symmetry, we will prove $v_1 = 0$.
The case $m = 1$ is trivial ($V = V_1$), so take $m \geq 2$. Apply $\prod_{i=2}^m (A - \lambda_i)^{e_i}$ to both sides of $v_1 + \cdots + v_m = 0$ to kill off all but the first term: we get $\prod_{i=2}^m (A - \lambda_i)^{e_i}(v_1) = 0$. Thus $v_1$ is killed by $\prod_{i=2}^m(A - \lambda_i)^{e_i}$. Also $(A - \lambda_1)^{e_1}(v_1) = 0$ from the definition of $V_1$.  The polynomials $\prod_{i=2}^m (X - \lambda_i)^{e_i}$ and $(X - \lambda_1)^{e_1}$ are relatively prime in $\mathbf C[X]$, so by Bezout some $\mathbf C[X]$-linear combination of them is $1$: $u(X)\prod_{i=2}^m (X - \lambda_i)^{e_i} + v(X)(X - \lambda_1)^{e_1} = 1$.  Replacing $X$ with $A$ and applying both sides to $v_1$, we get
$$
u(A)(\prod_{i=2}^m(A - \lambda_i)^{e_i}(v_1)) + v(A)((A - \lambda_1)^{e_1}(v_1)) = v_1.
$$
On the left side, $\prod_{i=2}^m(A - \lambda_i)^{e_i}(v_1) = 0$ and $(A - \lambda_1)^{e_1}(v_1) = 0$, so $0 = v_1$.
A: Here is a beautiful generalization mentioned in the comments.
Let $G(s)$ a $p\times q$ rational polynomials matrix (that is, with quotients of polynomials in $s$ as entries), then there exist polynomial matrices $N(s),M(s),\tilde N(s),\tilde M(s),
X(s),Y(s),\tilde X(s),\tilde Y(s)$ of suitable sizes such that
$$G(s)=N(s)M^{-1}(s)=\tilde N(s)\tilde M^{-1}(s)$$ and fulfill
$$
\left[
\begin{array}{cc}
\tilde X(s)&-\tilde Y(s)\\
-\tilde N(s)&-\tilde M(s)
\end{array}
\right]
\left[
\begin{array}{cc}
M(s)&Y(s)\\
N(s)&X(s)
\end{array}
\right]=
\left[
\begin{array}{cc}
1\!\!1_q&0\\
0&1\!\!1_p
\end{array}
\right].
$$
This relation is a generalized Bezout's identity.
