# Eigenvalues of a sum of power of a matrix [duplicate]

$A$ is defined as a real $n×n$ matrix. $B$ is defined as: $$B=A+A^2+A^3+A^4+ \dots +A^n$$ What's the relation between eigenvalues of $A$ and eigenvalues of $B$? Can anyone give me some materials?

## marked as duplicate by Arnaud D., Xam, A. Salguero-Alarcón, Misha Lavrov, The PhenotypeFeb 24 '18 at 0:06

• Hint : if $X$ is an eigenvector of $A$ associated with the eigenvalue $\lambda$, then $AX=\lambda X$. What is $BX$ ? – krirkrirk Feb 23 '18 at 13:40
• @krirkrirk Thanks! – Julian Feb 23 '18 at 13:41

Let $\lambda$ be an eigenvalue of $A$ and $v$ a corresponding eigenvector. Then $$Bv = (A+A^2+\cdots + A^n)v\\ = Av+A^2v + \cdots +A^nv\\ = \lambda v + \lambda^2v + \cdots + \lambda^nv\\ = (\lambda + \lambda^2+\cdots + \lambda^n)v$$

Try to prove that if $\lambda$ is an eigenvalue of $A$ then $\phi(\lambda)$ is an eigenvalue of $\phi(A)$ for every $A\in \mathbb{R}^{n\times n}$ and $\phi(x)\in\mathbb{R}[x]$

I want to add a bit with gt6989b that if you consider it in complex $\operatorname{GL}(n, \Bbb C)$ then it will always be equivalent with a Jordan canonical block form.

Now for your answer for any polynomial $p(x)$; $p(A)v=P(\lambda)v$ holds if $Av=\lambda v$.

HINT

If $A$ is diagonalizable, say $A = VDV^{-1}$ then $$B = \sum_{k=1}^n \left(VDV^{-1}\right)^k = \sum_{k=1}^n VD^kV^{-1} = V \left(\sum_{k=1}^n D^k \right)V^{-1}$$ and $D$ is a diagonal matrix. Can you take it from here?

• Do you mean $V^{-1}$ instead of $V^T$? – Arthur Feb 23 '18 at 13:38
• @Arthur fixed thanks – gt6989b Feb 23 '18 at 16:32