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Suppose I have a bounded linear operator $K$. Let's say the spectrum of $K$ is the union of the spectrum of two other operators, that is

$$\sigma(K) = \sigma(A) \bigcup \sigma(B)$$

Now if I know the operator $B$ is it possible to find the operator $A$?

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    No, for many reasons. Let $U$ be a unitary then $\sigma(UAU^*)=\sigma(A)$ for example. – 2017-01-18
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    If the underlying (presumably Hilbert) space is separable, it is possible to find *an* operator $A$ satisfying this, but not a unique one. Just pick a compact set $S$ such that $\sigma(K) = S \bigcup \sigma(B)$, choose a dense subsequence $\{\lambda_n\}$ in $S$, and let $A$ be the diagonal operator with respect to some o.n. basis $\{e_n\}$ whose diagonal is $\{\lambda_n\}$. – 2017-01-18

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