The wikipedia page Eigenvalues gives an example of how to compute eigenvectors of a matrix if already given the eigenvalues. The page claims this is an application of the Cayley–Hamilton theorem. However, when I read the Cayley–Hamilton theorem page, I don't see how it applies?
Can someone help me understand which part of the Cayley–Hamilton theorem gives me the formula for eigenvectors, and in particular how to apply the theorem in the special cases of repeated eigenvalues or if there is a zero eigenvalue?
There is an example on the eigenvalues page of how to compute some eigenvectors of a $3\times 3$ matrix, given its eigenvalues, but I'm not 100% sure how this applies to larger $n\times n$ matrices.
Which part is unclear to me? Apparently the kth eigenvector is the kth column of the matrix formed by multiplying $(A-\lambda_k)^{-1}\prod_{i =1}^{n}(A-\lambda_i I) = \prod_{i \ne k}(A-\lambda_i I)$ in the case that $\lambda_k$ is not a repeated eigenvalue.
How is this a consequence of the Cayley–Hamilton theorem, and how do I modify this formula to account for repeated eigenvalues and zero-eigenvalues?
Let $\lambda$ be an eigenvalue of $A$ of multiplicity $m$. The characteristic polynomial of $A$ is then in the form of $(x-\lambda)^mg(x)$ for some polynomial $g$ such that $g(\lambda)\ne0$. By Cayley-Hamilton theorem, we have $(A-\lambda I)^mg(A)=0$.
We haven’t calculated any eigenvector for the eigenvalue $\lambda$ yet, but its existence is theoretically guaranteed. Let $v$ be such an eigenvector. Note that $$ g(A)v=g(\lambda)v\ne0. $$ Therefore $g(A)$ must be a nonzero matrix. Let’s say its $j$-th column is nonzero, so that $x:=g(A)e_j\ne0$ (where $e_j$ denotes the $j$-th vector in the standard basis of $\mathbb F^n$). Then $$ (A-\lambda I)^mx =(A-\lambda I)^mg(A)e_j =0e_j=0. $$ Therefore, there exists a least positive integer $k\le m$ such that $$ (A-\lambda I)^{k}x=0. $$ Now let $v:=(A-\lambda I)^{k-1}x$. Then $v\ne0$ and $(A-\lambda I)v=0$. Hence $v$ an eigenvector of $A$ corresponding to the eigenvalue $\lambda$.
In particular, if the algebraic multiplicity of $\lambda$ is $m=1$, any nonzero column of $g(A)$ will be a corresponding eigenvector.