# Bézout-like identities for linear operators

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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Don't think so. Take $A = B = I$. A necessary condition is that the intersection of the kernels is zero, and a sufficient condition is that either $A$ or $B$ is invertible. – Qiaochu Yuan Jun 8 '11 at 14:47
@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. – Giuseppe Negro Jun 8 '11 at 14:57

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.
Ok, I'm convinced! Good point, Qiaochu. And if we had wanted a decomposition like $$I=AX+BY?$$ Then, I say, the condition would be that the sum of the ranges of $A$ and $B$ to be the whole space. To prove it quickly I would appeal to duality: we have $$I=AX+BY$$ if and only if $$I^\star=X^\star A^\star+Y^\star B^\star$$ and the kernels of $A^\star, B^\star$ intersect only at zero iff the ranges of $A, B$ span the whole space. – Giuseppe Negro Jun 8 '11 at 16:06