# Eigenvalues and Eigenvectors of Sum of Symmetric Matrix

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.

• Try evaluating $(\sum_{n=1}^{100} A^n) \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and do the same with the other eigenvector. What happens? – Giuseppe Negro Dec 11 '18 at 19:24

Hint: If $$A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}$$then we have $$A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}=\begin{bmatrix} 2 & 2 \\ 2 & 2 \\ \end{bmatrix}\\A^3=\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 2&2 \\ 2&2 \\ \end{bmatrix}=\begin{bmatrix} 4&4 \\ 4&4 \\ \end{bmatrix}\\A^4=\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 4&4 \\ 4&4 \\ \end{bmatrix}=\begin{bmatrix}8&8 \\ 8&8 \\ \end{bmatrix}\\.\\.\\.\\.$$and you can prove by induction that $$A^k=\begin{bmatrix} 2^{k-1}&2^{k-1} \\ 2^{k-1}&2^{k-1}\\ \end{bmatrix}$$can you finish now?

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 $$v=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$$ $$\lambda=0$$

• This is the simplest way to go. – Giuseppe Negro Dec 12 '18 at 22:14

Hint :

Recall the Cayley-Hamilton Theorem (by Wikipedia) :

For a general n×n invertible matrix $$A$$, i.e., one with nonzero determinant, $$A^{−1}$$ can thus be written as an $$(n − 1)$$-th order polynomial expression in $$A$$: As indicated, the Cayley–Hamilton theorem amounts to the identity : $$p(A) = A^n + c_{n-1}A^{n-1} + \dots + cA + (-1)^n\det(A)I_n = O$$ The coefficients ci are given by the elementary symmetric polynomials of the eigenvalues of $$A$$. Using Newton identities, the elementary symmetric polynomials can in turn be expressed in terms of power sum symmetric polynomials of the eigenvalues: $$s_k = \sum_{i=1}^n \lambda_i^k = \text{tr}(A^k)$$

You can explicitly compute $$\sum_{i=1}^{100}A^i$$. First diagonalize $$A$$, namely rewrite $$A$$ as $$A=PDP^{-1}$$.

Now \begin{align} \sum_{i=1}^{100}A^i&=\sum_{i=1}^{100}PD^iP^{-1}\\&=P\left(\sum_{i=1}^{100} D^i\right)P^{-1} \end{align}

Notice that $$(D-I)\left(\sum_{i=1}^{100}D^i\right)=D^{101}-I.$$ SInce $$D-I$$ is invertible (you can check it) $$\sum_{i=1}^{100}D^i=(D-I)^{-1}(D^{101}-I).$$ Therefore $$\sum_{i=1}^{100}A^i=P(D-I)^{-1}(D^{101}-I)P^{-1}.$$

It is easy to prove that for $$k\in \Bbb{N},$$ $$A^k=\begin{bmatrix} 2^{k-1} & 2^{k-1} \\ 2^{k-1} & 2^{k-1} \\ \end{bmatrix}.$$ The sum is $$\Sigma=\begin{bmatrix} 2^{100}-1 & 2^{100}-1 \\ 2^{100}-1 & 2^{100}-1 \\ \end{bmatrix},$$ from where the eigenvalues $$0$$ and $$(2^{101}-2).$$

Each matrix $$A^k, k=1,\dots,100$$ has eigenvalues $$0$$ and $$2^k,$$ the corresponding eigenvectors are those of $$A:$$ $$(1,-1)^T, (1,1)^T.$$
Thus $$(1,-1)^T, (1,1)^T$$ are eigenvectors of $$\Sigma.$$

Since you have 2 linear independent eigenvectors, $$A$$ is diagonalizable. You may find useful to replace $$A$$ in your polynomial expression by its diagonalization because this will simplify the operations you need to do.