TITLE: Eigenvalues and Eigenvectors of Sum of Symmetric Matrix
QUESTION [5 upvotes]: Question:
Let A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}
Find all eigenvalues and eigenvectors of the martrix:
$$\sum_{n=1}^{100} A^n = A^{100} +A^{99} +...+A^2+A$$
I know that the eigenvectors of A are \begin{bmatrix} 1 \\ 1 \end{bmatrix} and \begin{bmatrix} 1 \\ -1 \end{bmatrix}
But I do not see any sort of correlation with the sum term and A's eigenvectors.

REPLY [3 votes]: By linearity, given any polynomial $p$ and matrix $A$, the eigenvectors of $p(A)$ are the same as the eigenvectors of $A$, and the associated eigenvalues are $p(\lambda)$; see this question.
For instance, in this case, if $Av=\lambda v$, then $A^nv=\lambda^nv$, and $(\sum_{n=1}^{100}A^n)v=\sum_{n=1}^{100}(A^nv )=\sum_{n=1}^{100}(\lambda^nv)=(\sum_{n=1}^{100}\lambda^n)v$. Thus, $v$ is an eigenvector with eigenvalue $\sum_{n=1}^{100}\lambda^n$. $A$ has eigenvectors, eigenvalues of $v=\begin{bmatrix} 1 \\ 1 \end{bmatrix} $ $\lambda=2$ and  $v=\begin{bmatrix} 1 \\ -1 \end{bmatrix} $ $\lambda=0$. $p(2)$ is a geometric series, so it is $2^{101}-1$. $p(0)$ is just zero. So $p(A)$ has eigenvectors, eigenvalues of  $v=\begin{bmatrix} 1 \\ 1 \end{bmatrix} $ $\lambda=2^{101}-1$ and  $v=\begin{bmatrix} 1 \\ -1 \end{bmatrix} $ $\lambda=0$