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If $f$ is a complex analytic function, one can define a matrix function $F$ using the Taylor series of $f$ by

$$ F(A) = f(0) + f'(0)\cdot x + f''(0)\cdot \frac{A^2}{2!} + \cdots $$

If the radius of convergence of $f$ is $r$, then the above matrix series will also converge for $\|A\|<r$, where $\|\cdot\|$ is a submultiplicative matrix norm.

Is this still valid if $A$ is a bounded linear operator instead of a matrix?

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Yes it is still valid . An operator can be represented as a matrix . Assuming it's a normal matrix ,it must have an eigenbasis and if we transform it into that basis then it is a diagonal matrix.

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    $\begingroup$ but every operator can be represented as a matrix? $\endgroup$
    – shamisen
    Commented Nov 8, 2016 at 16:37

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