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In a Banach Algebra, can the product of two non-invertible element be invertible? Basically in linear algebra, we know that if the product of two matrix is invertible, then both matrix must be invertible. The proof is very simple if we use determinants. Now for general Banach space with infinite dimensions, there is no such thing as a determinant, so I was wondering if we have two elements whose product is invertible, do we know if both elements are invertible? And what if the elements commute?

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Consider the Hilbert space $l^2(\mathbb{N})$, the Banach algebra of all linear and continuous operators $T:l^2(\mathbb{N}) \to l^2(\mathbb{N})$ and the left and right shift operators, that is $L(x)=(x_2,x_3,\dots)$ and $R(x)=(0,x_1,x_2,\dots)$ for each $x=(x_1,x_2,\dots) \in l^2(\mathbb{N})$. Then $L(R(x))=x$, hence $L\circ R= I$ is invertible. But $L$ is not injective and $R$ is not surjective. If ${\cal A}$ is any Banach algebra with unit $e$, and $a,b \in {\cal A}$ with $ab=ba$ and $ab=:c$ invertible, then $ca=ac$, $cb=bc$, hence $ac^{-1}=c^{-1}a$, $bc^{-1}=c^{-1}b$ and therefore $$ e=a(bc^{-1})=(bc^{-1})a, $$ thus $a$ is invertible. By the same way $b$ is invertible.

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Take the space $X=\ell^2$ and let $A$ be the algebra of all bounded linear operators on $X$. Define $f,g\in A$ like this:

$f(x_1,x_2,x_3,...)=(x_2,x_3,x_4,...)$

$g(x_1,x_2,x_3,...)=(0,x_1,x_2,x_3,...)$

Both are clearly not invertible, as $f$ is not injective and $g$ is not surjective. However, note that $fg=id_X$, which is clearly invertible.

If the elements commute then they are indeed both invertible. Suppose that $x,y\in A$ commute and $xy$ is invertible. Then there is some $a\in A$ such that:

$xya=axy=1$

Since $xy=yx$ it follows that $x(ya)=(ay)x=1$. So $x$ has both a left and right inverse. This is an easy exercise to show that both one sided inverses must be equal in this case, and so $x$ is invertible.

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