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Suppose we have a convergent power series of matrices $$A=\sum_{n=0}^\infty a_nX^n,$$ for $X\in M_n(\mathbb{C})$. Is it true that if $v\in\mathbb{R}^n$ then $$Av=\sum_{n=0}^\infty a_n(X^nv)?$$ If no, under what conditions is it true?

I am asking this question because I am trying to prove that if $\lambda$ is an eigenvalue of $A$, then $e^\lambda$ is an eigenvalue of $\exp(A)$. Indeed, if $Av=\lambda v$ then $$\exp(A)v=\left(\sum_{n=0}^\infty\frac{1}{n!}A^n\right)v\stackrel{?}{=}\sum_{n=0}^\infty\frac{1}{n!}(A^nv)=\sum_{n=0}^\infty\frac{1}{n!}\lambda^nv=e^\lambda v.$$ But I am not sure if the second equality is justified. Otherwise, we can try to show that $$\det(\exp(A)-e^\lambda I)=0,$$ but this seems much harder.

user1551
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Grain
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1 Answers1

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$\sum\limits_{n\geq0}X_n$ denotes the limit of $\sum\limits_{n\geq0}^kX_n$ as $k\to\infty$. Since matrix multiplication is continuous, this means that $\bigl(\sum\limits_{n\geq0}X_n\bigr)v$ is the limit of $$\bigl(\sum\limits_{n\geq0}^kX_n\bigr)v\tag{1}$$ as $k\to\infty$.

On the other hand, $\sum\limits_{n\geq0}X_nv$ denotes the limit of $$\sum_{n\geq0}^kX_nv\tag{2}$$ as $k\to\infty$.

Check that both sequences (1) and (2) are in fact the same.

Mariano Suárez-Álvarez
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