Cayley-Hamilton theorem for Operator

[zetafunction]

let be $$f(x)=det(xI-T)$$ for some operator 'T'

then does Cayley-Hamilton theorem apply so $$f(T)=0$$ in the sense of operator

 

[Eynstone]

Good question , Zetafunction.
I think we need restrictions on T to ensure convergence in the corresponding Banach/Hilbert space( or else f(T) may not make sense.)