Please help me with the following problem( give some hints or references):
Let $X$ be a Banach space and $B(X)$ be the algebra of bounded linear operators on $X$. Suppose that $A$ and $B$ are two operators in $B(X)$ such that for every $T \in B(X)$ we have $\sigma(AT)=\sigma(BT)$. Show that $A=B$.
Here $\sigma(A)$ denotes the spectrum of $A$, i.e., the the set of all complex numbers $\lambda$ such that $\lambda I − A$ has no inverse in $B(X)$.
Thanks in advance.
For a given $x\in X$ and $f\in X^*$ consider rank one operator $$ x\bigcirc f:X\to X: z\mapsto f(z)x $$ One can check that $\sigma(x\bigcirc f)=\{f(x)\}$. Also note that $S(x\bigcirc f)=S(x)\bigcirc f$ for $S\in \mathcal{B}(X)$.
Now we turn to the original problem. Fix $x\in X$ and $f\in X^*$ and consider $T=x\bigcirc f$, then $\sigma(AT)=\sigma (A(x)\bigcirc f)=\{f(A(x))\}$ and $\sigma(BT)=\sigma (B(x)\bigcirc f)=\{f(B(x))\}$. From assumption it follows that $f(A(x))=f(B(x))$. Since $f\in X^*$ is arbitrary, by corollary of Hahn-Banach theorem we have $A(x)=B(x)$. Since $x\in X$ is arbitrary we get that $A=B$.