$$ \left(\begin{matrix} 0 & 1 \\ x & 1 \\ \end{matrix}\right) $$ Suppose I have a matrix above to the power of n, is it possible to use eigendecomposition in order to find all the eigenvectors in terms of x?
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The title says $n \times n$ matrix, but you're asking about a $2 \times 2$ matrix to the power $n$, which is still a $2 \times 2$ matrix. – Robert Israel Aug 06 '18 at 07:34
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The title does not reflect the question. – lhf Aug 06 '18 at 11:02
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Hint
$$A^n = P D^n P^{-1}$$ where $P$ is the matrix of the eigenvectors and $D$ the matrix of the eigenvalues
Davide Morgante
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Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how? – king jigg Aug 06 '18 at 13:24
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To find eigenvectors you either solve the equation $$A\mathbf{v}=\lambda\mathbf{v}$$ for a generic $\mathbf{v}$ or $$ker(A-\lambda \mathcal{I})=0$$ – Davide Morgante Aug 06 '18 at 14:40
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Hint: An eigenvector of $A$ for eigenvalue $\lambda$ is an eigenvector of $A^n$ for eigenvalue $\lambda^n$.
The process of finding eigenvalues and eigenvectors is pretty much the same with a symbolic parameter $x$.
The only tricky part is if $x = -1/4$.
Robert Israel
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A more general result is true. It is a well known result that if $A \in M_n(\mathbb{C})$, there exists $P \in GL_n(\mathbb{C})$ such that $PAP^{-1}$ is upper triangular. In particular, all eigenvalues of $A$ are in this matrices' diagonal, counted with multiplicity (if not inmediately clear, recall that eigenvalues are preserved by conjugation, and write the expression of a characteristic polynomial for a general upper triangular matrix).
Now, if $A, B \in M_n(\mathbb{C})$ are upper triangular and $i \in [n]$,
$$ (\lambda A)_{ii} = \lambda A_{ii} $$
and
$$ (AB)_{ii} = \sum_{m= 1}^ma_{im}b_{mi} = a_{ii}b_{ii} $$
so if $q \in \mathbb{C}[X]$, then $q(A)_{ii} = q(A_{ii})$ for all $i$ (again, induction should do the trick). Moreover, if $q = \sum_{j = 1}^{d}c_jX^{j}$, then
$$ q(PAP^{-1}) = \sum_{j = 1}^{d}c_j(PAP^{-1})^{j} = \sum_{j = 1}^{d}c_jPA^jP^{-1} = P\left(\sum_{j = 1}^{d}c_jA^j\right)P^{-1} = Pq(A)P^{-1} $$
To sum up, since
$$ q(A) = q(P^{-1}PAP^{-1}P) = P^{-1}q(PAP^{-1})P $$
the eigenvalues of $q(A)$ are the same of those of $q(PAP^{-1})$. Since the eigenavalues of $q(PAP^{-1})$ are the eigenvalues of $PAP^{-1}$ via $q$ and $PAP^{-1}$ has the same eigenvalues than $A$,
$$ \operatorname{Spec}(q(A)) = q(\operatorname{Spec}(A)). $$
In your particular case, $q = X^{n}$. Thus, it will suffice to find the eigenvalues of $A$ and take their $n$-th power.
qualcuno
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