Let $A$ be a unital $C^\ast$ algebra and let $B$ be a $\ast$ subalgebra such that $B \oplus \mathbb C = A$ and such that the unit in $B$, $1_B$, is not equal to the unit in $A$. I am trying to show:
If $\lambda \in \mathbb C$ is non-zero then $b-\lambda\cdot 1_B$ is invertible in $B$ if and only if $b-\lambda \cdot 1_A$ is invertible in $A$.
(see Murphy's book at the top of page 45).
I started the proof like this: Let $b-\lambda\cdot 1_B$ be invertible in $B$. Then it is also invertible in $A$. Let $a \in A$ denote its inverse. Then
$$ ba - \lambda 1_B a = 1_A$$
Now the goal is to find $c \in A$ such that $(b -\lambda 1_A)c = 1_A$. Somehow I have to show that $1_B a = 1_A c$ but I can't seem to do it. So this leads nowhere.
Can someone help me prove this please?
Assume $(b-\lambda 1_B)c=1_B$.
When we use $1_A=(0,1)$, $$ [(b, 0) - \lambda (0,1)](x,t)=(b,-\lambda)(x,t)=\left(bx-\lambda x+tb, -\lambda t\right). $$ Looking at the second coordinate, $t=-1/\lambda$. Then the equality in the first coordinate becomes $$ 0=bx-\lambda x -\frac1\lambda b=(b-\lambda 1_B)x-\frac1\lambda b, $$ So $x=\frac1\lambda cb=\frac1\lambda 1_B+c $. That is, in $A$ we have $(b-\lambda 1_A)^{-1}=\frac1\lambda cb=\frac1\lambda 1_B+c-\frac1\lambda 1_A$.
The computations above also allow us to do the converse: if $b-\lambda 1_A$ is invertible in $A$, then $x$ above exists. Now take $c=x-\frac1\lambda 1_B$, and a straightforward calculation shows that $(b-\lambda 1_B)c=1_B$.