Spectral Permanance in $C^{*}$ algebras Problem:
Let $A$ be a unital $C^{*}$ algebra and let $B$ be sub $C^{*}$ algebra of A with possibly a different unit. Then I need to prove that for any $b$ $\in$ $B$
spectrum($b,B$)$\cup$ {$0$}  $\subset$ spectrum($b,A$) $\cup$ {$0$} .
Attempt for commutative algebras:
I was wondering that when we consider the unitization of a $C^{*}$ algebra $A$ to $A \bigoplus$$\mathbb{C}$ with the product defined as $(a,\lambda). (b,\gamma)=(ab+\lambda b +\gamma a, \lambda \gamma)$  with the unique $C^{*}$ norm then the spectrum of any element in $A$ gets augmented by the element $(0,0)$ in $A \bigoplus$$\mathbb{C}$
and if two $C^{*}$ algebras have same unit then spectrum of every common element is same.
How to prove for non commutative case?
 A: I think you can go one step further and say that equality holds.  The inclusion $i:B\hookrightarrow A$ induces a unital $*$-homomorphism $\phi:B^+\to A^+$ (here $B^+$ and $A^+$ denote the unitizations of $B$ and $A$ respectively) with $\phi\mid_B=i$.  Then $B^+$ is $*$-isomorphic to $\phi(B^+)$ and we have
$$\sigma_B(b)\cup\{0\}=\sigma_{B^+}(b)=\sigma_{\phi(B^+)}(b)=\sigma_{A^+}(b)=\sigma_A(b)\cup\{0\}.$$
As stated in the comments, I am not sure whether the first and last equalities hold in the noncommutative case.  I will think about this and let you know if I come up with anything.

EDIT  I've finally found time to come back to this, and it turns out that $\sigma_A(a)\cup\{0\}=\sigma_{A^+}(a)$ when $A$ is unital.  Let $1_A$ and $1_{A^+}$ denote the units of each.   
If $a\in A$ and $a-\lambda1_{A^+}$ is invertible in $A^+$, clearly $\lambda\neq0$, and $(a-\lambda1_{A^+})a_0=a_0(a-\lambda1_{A^+})=1_{A^+}$ for some $a_0\in A^+$.  We have $1_Aa_0=a_01_A$ is an inverse to $a-\lambda1_A$ in $A$, so $\lambda\in(\sigma_A(a)\cup\{0\})^c$.
Now suppose $\lambda\neq 0$ and $a-\lambda1_A$ is invertible in $A$. Put $f=1_{A^+}-1_A$.  If $(a-\lambda1_A)a_0=a_0(a-\lambda1_A)=1_A$, then
$$(a-\lambda1_{A^+})(a_0+\lambda^{-1}f)=(a_0+\lambda^{-1}f)(a-\lambda1_{A^+})=1_{A^+},$$
and thus $\lambda\in\sigma_{A^+}(a)^c$.
