power series of a matrix well-defined I am working on a seminar lecture and have found the following lemma without a proof:
Given a convergent power series $f(z)=\sum_{n=0}^\infty a_nz^n$  and a diagonalizable matrix $M$ with diagonal matrix $\tilde M$ of eigenvalues $\lambda_i$ and a regulare matrix $L$ such that $M=L\tilde M L^{-1}$.
Then $f(M)= LKL^{-1}$ with $K:=\pmatrix{f(\lambda_1)&&\\&\ddots&\\&&f(\lambda_k)}$ is well-defined.
So my question is why is the mapping well-defined? Thanks a lot!
 A: "Well-defined" just means it exists. It certainly exists, since you can write it down! Your notation is a little muddled, but $M=L\tilde{M}L^{-1}$, $\tilde{M}=\mathrm{diag}(\lambda_i)$ is defined by $f(M)=L \,\,\mathrm{diag}(f(\lambda_i)) L^{-1}$.
The reason why this makes good sense is that we also have
$$\sum_{n=0}^\infty a_n M^n = \sum_{n=0}^\infty a_n (LKL^{-1})^n = \sum_{n=0}^\infty a_n LKL^{-1}\cdots LKL^{-1} = \sum_{n=0}^\infty a_n LK^nL^{-1} = L\left(\sum_{n=0}^\infty a_n K^n\right)L^{-1} = L\left(\sum_{n=0}^\infty a_n \mathrm{diag}(\lambda_i)^n\right)L^{-1} = L\left(\sum_{n=0}^\infty a_n \mathrm{diag}(\lambda_i^n)\right)L^{-1} = L\left(\mathrm{diag}(\sum_{n=0}^\infty a_n\lambda_i^n)\right)L^{-1} = L\left(\mathrm{diag}(f(\lambda_i)\right)L^{-1} = f(M)$$
Other than that, I'm not sure what to say!

Edit: Actually, I guess one thing you might consider 'well-defined' is that the above means that polynomials $f(M)$ defined this way agree with their definition via matrix multiplication - because of the above argument.
A: $$f(M) = \sum_{n=0}^\infty a_nM^n =  \sum_{n=0}^\infty a_n(LKL^{-1})^n = \sum_{n=0}^\infty a_nLK^nL^{-1}  = L(\sum_{n=0}^\infty a_nK^n)L^{-1} = Lf(K)L^{-1}  $$
A: If $f_k(z)=\sum_{n=0}^k a_n z^n$ then
$f_k(M)=\sum_{n=0}^k a_n (L\tilde M L^{-1})^n=L(\sum_{n=0}^k a_n \tilde M^n)L^{-1}=Lf_k(\tilde M)L^{-1}$.
If you are on $\mathbb C$ with euclidean norm, product by a matrix is continuous and $f_k(\tilde M)\to K$ (a matrix converges if and only if each component converges), thus $f_k(M)\to LKL^{-1}$.
