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?
In a Banach Algebra, can the product of two elements be invertible if we know one of the element is non invertible
functional-analysislinear algebra
Best Answer
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.