Every linear bounded self-adjoint operator $T : {\scr H} \to {\scr H} $can be written in terms of its eigenvalues and their associated projectors (see spectral theorem): $$ T = \sum_{{\frak spec}(T) }\lambda \ P_\lambda$$

For finite-dimensional spaces, there's an equivalent matrix formulation:

Every hermitian matrix is similar to a diagonal one via unitary transformations.

In this context, we can also consider "functions of operators": $$f(T) = \sum_{{\frak spec}(T)} f(\lambda) P_\lambda$$

It's simply the function acting on each of its eigenvalues. But what about a "Taylor expansion"?

What does it mean $$f(T) = \sum _n \frac {f^{(n)} (0)}{n!} T^n?$$

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    $\begingroup$ Maybe writing it down helps, because the answer is almost there with your second formula. Note that $P^2_\lambda = P_\lambda$ since it is a projection operator. Rest you can fill in. $\endgroup$
    – exp ikx
    Feb 17, 2021 at 14:53
  • 1
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 17, 2021 at 16:33
  • $\begingroup$ Tip: Try an exponential of a matrix, and then an operator. I assume you get it about Syvester's formula. $\endgroup$ Feb 17, 2021 at 20:33

1 Answer 1


Consider, $$ T=\sum_{\lambda\in\sigma\left(T\right)}\lambda P_{\lambda} $$ and, $$ P_{\lambda}P_{\lambda'}=\delta\left(\lambda,\lambda'\right)P_{\lambda}$$ Do the expansion and you'll see immediately that, $$ f\left(T\right)=\sum_{\lambda\in\sigma\left(T\right)}f\left(\lambda\right)P_{\lambda} $$

  • $\begingroup$ Made one mistake that I think you spotted immediately: $P_{\lambda}P_{\lambda'}=\delta\left(\lambda,\lambda'\right)P_{\lambda}$. Edited and corrected omission. $\endgroup$
    – joigus
    Feb 17, 2021 at 18:46

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