Right inverses of the square of a right invertible operator.

Question

Let $A$ be a right invertible operator acting on a Hilbert space $\mathcal{H} $. Can every right inverse of $A^{2}$ be expressed as a product $B_{1}B_{2}$ where $B_{1}$ and $B_{2}$ are two right inverses of $A$ ?

Thank you !

Answer 1

The answer is yes. It holds for any vector space $X$ and is just linear algebra.

Lemma. If $B$ is a right-inverse for $A$ and $\operatorname{im}B = M$, then $A$ is surjective and

(a) $X = \ker A\oplus M$;

(b) $A|_M : M\to X$ is bijective and $B = (A|_M)^{-1}$;

(c) $BA = P_M$, where $P_M$ is the projection onto $M$ along $\ker A$.

Proof. It is clear from $AB = I$ that $A$ is surjective.

(a). If $x\in\ker A\cap M$, then $Ax = 0$ and $x = Bu$ with $u\in X$. Hence, $u = ABu = Ax = 0$ and thus $x=Bu = 0$. Assume that $\ker A\oplus M\neq X$. Then there exists $z\in X\setminus(\ker A\oplus M)$. Choose $m\in M$, $m\neq 0$, and put $y = A(m+z)$. As $A$ maps $M$ bijectively to $X$, there is also $m'\in M$ such that $Am' = y$. So, $m-m'+z\in\ker A$, in other words, $z\in\ker A\oplus M$, a contradiction.

(b). This follows directly from $(A|_M)B = I$.

(c). First, $(BA)^2 = B(AB)A = BA$, so $BA$ is indeed a projection. From (b), $B = (A|_M)^{-1}$. Hence, if $x\in M$, then $BAx = x$. Moreover, $BAx = 0$ for $x\in\ker A$. q.e.d.

In what follows, for a complementary space $M$ to $\ker A$ let us define $B_M = (A|_M)^{-1}$.

Theorem. If $C$ is a right-inverse to $A^2$ and $N := \operatorname{im}C$, then $AN$ is complementary to $\ker A$ in $X$ and $C = B_{N+M}B_{AN}$, where $M$ is any complementary subspace of $\ker A$ in $\ker A^2$.

Proof. To begin with, we shall prove that $$ X = AN\oplus\ker A. $$ If $x\in AN\cap\ker A$, then $x=Au$ for some $u\in N$ and $Ax=0$. Hence, $A^2u=0$. But $N\cap\ker A^2=\{0\}$ and so $u=0$ which implies $x=0$. Moreover, if $x\in X$, then $x=Au$ with some $u\in X$ since $A$ is surjective. We can write $u=z+w$ with $z\in\ker A^2$ and $w\in N$. So, $x = A(z+w)\in\ker A\oplus AN$.

Now, we have $I = A^2C$. Hence, since $\operatorname{im}(AC)\subset AN$, $$ B_{N+M}B_{AN} = B_{N+M}(B_{AN}A)AC = B_{N+M}P_{AN}AC = B_{N+M}AC = P_{N+M}C = C. $$