Let a,b be 2 elements in a Banach Algebra.Let Spec(x) denote the spectrum of an element x. If a,b commute with each other, then by Gelfand Transformation, we have Spec(a+b) is a subset of Spec(a)+Spec(b)& Spec(ab) is a subset of Spec(a)Spec(b) (Where the definition of addition and multipication of two sets is obvious). My Questions, if we drop the the condition "a,b commute", then to what extent is the above relations about spectra still hold?

In general Banach algebras the spectral radius is neither subadditive nor submultiplicative; in particular, neither of the two properties you mention holds. $2\times 2$ (real or complex) matrices should suffice to give examples, so this is not to do with any subtleties of infinitedimensional algebras. For example, take $ a= \left(\begin{matrix} 0 & 1 \\\\ 0 & 0 \end{matrix} \right) $. Note that this is nilpotent, so the only point in the spectrum is zero, and hence $\sigma(a)\sigma(b)= \{0\}$ for any other matrix $b$. On the other hand, we can find $b$ for which $ab$ is not nilpotent, so that $\sigma(ab)\not\subseteq \{0\}$. A simple choice which works is $b=\left(\begin{matrix} 0 & 0 \\\\ 1 & 0 \end{matrix} \right)$, since then $ab$ is a nonzero projection (=idempotent) and so contains $1$ in its spectrum. The same pair also works as a counterexample for the "additive question". For since $\sigma(a)=\sigma(b)=\{0\}$, we have $\sigma(a)+\sigma(b)=\{0\}$. On the other hand, $a+b$ is a reflection and hence its spectrum is $\{1,1\}$ (edited 110210 for a couple of typos/omissions) Jonas Meyer points out in comments that one can pose the following converse question: let $A$ be a Banach algebra with identity, and suppose that we have ($*$) $\sigma(a+b)\subseteq\sigma(a)+\sigma(b)$ and $\sigma(a)\sigma(b)\subseteq\sigma(ab)$ for all $a,b\in A$. Must $A$ be commutative? As Jonas also pointed out in comments, the answer is in general `no': the algebra of strictly uppertriangular matrices (or, more precisely, the algebra of scalar+strictly upper triangular $m\times m$ matrices, for some fixed $m$) gives a counterexample, at least when $m\geq 3$. A more careful version of this argument shows, I think, that if $A$ is a finitedimensional algebra with identity, such that $A/{\rm Rad}(A)$ is commutative, then $A$ will satisfy condition $(*)$. I suspect that the same might be true for any unital Banach algebra that is commutative modulo its radical, i.e. that the finitedimensional hypothesis is unnecessary; but at present I'm a bit too tired to check this properly. We could therefore modify the question yet further, and ask if a Banach algebra that satisfies condition $(*)$ must be commutative modulo its radical. The answer turns out to be yes, after a wander down memory lane and a forage on MathSciNet: MR0461139 (57 #1124) J. Zemánek. Spectral radius characterizations of commutativity in Banach algebras. Studia Math. 61 (1977), no. 3, 257268. The MR is short and informative enough to give in full, for those without access: It is standard that the spectral radius is subadditive and submultiplicative on any commutative complex Banach algebra. The author proves that, for a complex Banach algebra $A$, the following three conditions are equivalent: (1) the spectral radius is subadditive on $A$, (2) the spectral radius is submultiplicative on $A$, (3) $A$ is commutative modulo its radical. Some applications of this result to other problems in Banach algebras are given, along with references to a number of related papers. 

