To find an element in $A$ which is invertible in $B$ but not in $A$. Let $A$ and $B$ be two Banach Algebra, $A \subset B$. 
To find an element in $A$ which is invertible in $B$ but not in $A$.
I think we can get examples in Matrices, but I am not able to do it.
 A: An examle was almost given in one of the comments: Let $H^\infty(\Bbb T))$ be the set of all bounded functions on $\Bbb T$ which arise as boundary values of holomorphic functions in the unit disk. Then $H^\infty(\Bbb T)\subset L^\infty(\Bbb T)$, and the function $e^{it}\in H^\infty$ is invertible in $L^\infty$ but not in $H^\infty$.
There is no example using matrices. In fact there is no finite-dimensional example:


Lemma Suppose $B$ is a finite-dimensional algebra over $F$ and $x\in B$ is invertible. Then there exists a polynomial $p$ (with coefficients in $F$) such that $x^{-1}=p(x)$. (And $p$ is non-trivial: $\deg(p)>0$.)


Proof: The elements $1,x,x^2,\dots$ cannot be independent. So there exists a polynomial $P$ with $P(x)=0$.
Say $m$ is a polynomial of minimal positive degree such that $m(x)=0$.
Let $c=m(0)\in F$. If $c=0$ then there exists a polynomial $q$ with $m(t)=tq(t)$. Hence $xq(x)=0$; since $x$ is invertible it follows that $q(x)=0$, contradicting the minimality of $m$.
So $c=m(0)\ne0$. Now there certainly exists a polynomial $p$ with $m(t)=tp(t)+c$. So $$xp(x)=-c,$$hence $$x^{-1}=-c^{-1}p(x).$$


Corollary If $A\subset B$ are finite-dimensional $F$-algebras and $x\in A$ is invertible in $B$ then $x$ is invertible in $A$.


Proof: Since $A$ is an algebra, $x\in A$ implies $x^{-1}=p(x)\in A$.
