# Function of an operator on a Hilbert space

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

• 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. – Abhay Hegde Feb 17 at 14:53
• Would Mathematics be a better home for this question? – Qmechanic Feb 17 at 16:33
• Tip: Try an exponential of a matrix, and then an operator. I assume you get it about Syvester's formula. – Cosmas Zachos Feb 17 at 20:33

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, $$f\left(T\right)=\sum_{\lambda\in\sigma\left(T\right)}f\left(\lambda\right)P_{\lambda}$$
• Made one mistake that I think you spotted immediately: $P_{\lambda}P_{\lambda'}=\delta\left(\lambda,\lambda'\right)P_{\lambda}$. Edited and corrected omission. – joigus Feb 17 at 18:46