1
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ but every operator can be represented as a matrix? $\endgroup$
    – shamisen
    Nov 8, 2016 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.