Qfyilyi

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.

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?

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 :
$$p(A) = A^n + c_{n-1}A^{n-1} + dots + cA + (-1)^ndet(A)I_n = O$$
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:
$$s_k = sum_{i=1}^n lambda_i^k = text{tr}(A^k)$$

It is easy to prove that for $kin Bbb{N},$ $$A^k=begin{bmatrix} 2^{k-1} & 2^{k-1} \ 2^{k-1} & 2^{k-1} \ end{bmatrix}.$$ 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.

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}\&=Pleft(sum_{i=1}^{100} D^iright)P^{-1}
end{align}

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

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} $ $lambda=0$