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As usual, when I pose a question here the answer I receive generates more questions. Today I posed myself a problem originating from this answer by Joel Cohen.

Let $V$ be a finite dimensional vector space over an arbitrary field. Let us agree to say that the linear operators $A, B$ verify a Bézout-like identity if

there exist linear operators $X, Y$ such that $I=XA+YB,$ where $I$ denotes the identity mapping.

Problem: Find necessary and sufficient conditions for $A$ and $B$ to verify a Bézout-like identity.


I believe the answer lies somewhere around $\ker(A), \ker(B)$. For example, if $A$ and $B$ are associated to the block $n \times n$ matrices

$A \equiv \begin{bmatrix} \mathbf{0} & \mathbf{0} \\ \mathbf{0} & P_{k \times k} \end{bmatrix}, \quad B \equiv \begin{bmatrix} Q_{h \times h} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{bmatrix}$

with nonsingular $P, Q$, then we can take

$X= \begin{bmatrix} \mathbf{0} & \mathbf{0} \\ \mathbf{0} & P^{-1} \end{bmatrix}, \quad Y=\begin{bmatrix} Q^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{bmatrix}$

which yield a Bézout-like identity if and only if $k +h=n$. Moreover, if $k+h < n$, then we can be sure that no Bézout-like identity is possible. This could suggest that the sought condition is

$\ker(A) \oplus \ker(B)=V.$

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    @Qiaochu: Of course, thank you. Next time I should think more before posting a conjecture! :-) Anyway, the problem stays. In fact, I was thinking at Joel's construction: if $A, B$ are matrices, then we can put $f(X_{1, 1}\ldots X_{n, n}, Y_{1, 1} \ldots Y_{n, n})= \det(XA+YB), $ and the problem is equivalent to find necessary and sufficient conditions for the polynomial $f$ to be non null.2011-06-08

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It's necessary and sufficient that the intersection of the kernels is zero. Necessity is obvious. To show sufficiency, choose a basis $v_1, ... v_n$ of $V$ such that the first $a$ vectors span $\ker A$ and the next $b$ vectors span $\ker B$. Then we can find $X, Y$ such that $XA$ is the projection onto the vectors $v_{a+1}, ... v_n$ and $YB$ is the projection onto the vectors $v_1, ... v_a$, hence $XA + YB = I$ as desired.

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    Nice argument. The direct approach above also gives the same result, I think.2011-06-08