Can eigenvectors be found without finding eigenvalues? Given a matrix $A$ and the set of all of its eigenvectors, it is possible to find all of the matrix’s eigenvalues by solving $A\vec v = \lambda\vec v$.  Given the set of all eigenvalues, it is possible to find the corresponding eigenspaces by finding $\mathrm N(A-\lambda I)$.
It is also possible to find all of the eigenvalues independently of the eigenvectors by finding the zeroes of the characteristic polynomial $\lvert A – \lambda I\rvert$.
This makes me wonder, is it possible to find all of the eigenvectors or eigenspaces independently of the eigenvalues? If so, how?
My linear algebra instructor’s assistant directed me to this video, which I found unhelpful.
 A: Some iterative methods find eigenvectors (approximately, at least) without going through the eigenvalues.
As a simplest possible approach, given your matrix $A$, you can pick an arbitrary starting vector $\mathbf x$, and repeatedly multiply it by $A$, scaling the result down whenever it gets too big (or scaling it up whenever it gets too small). This doesn't necessarily converge (you're in trouble if $A$ is a rotation matrix, for instance), but if it converges up to a scaling factor, it converges to an eigenvector.
(For most starting $\mathbf x$, you'll get the eigenvector associated to the eigenvalue that's largest in absolute value. You can also replace $A$ by $(A - t I)^{-1}$ to find the eigenvector associated to the eigenvalue closest to $t$.)
Once you have the approximate eigenvector, you can estimate the associated eigenvalue, but that's inescapable. In any case, the eigenvector comes first in this method.
A: Some words about the reference -given by the OP and that I don't know- taken from the AMM. 
i) If you randomly choose $u$ and if $char(F)=0$, then, with probability $1$, your polynomial $a_0+\cdots+a_kt^k$ is the minimal polynomial of $A$ (and not only the one of $u$). cf. my post in 
A better algorithm to find the minimum polynomial of a matrix?
ii) The title of this paper “Computing Eigenvalues and Eigenvectors Without Determinants" seems to be a joke; indeed, the decomposition of $A^ku$ as a linear combination of the previous calculated vectors needs the calculation of the reduced row echelon form (which is a hidden determinant).
iii) Your factorization $(t-\lambda)Q(t)$ over $F=\overline{\mathbb{Q}}$ (for example) can only be done in an approximate way; moreover, since $\lambda\in spectrum(A)$ , you obtain the eigenvalue before the eigenvector $Q(A)u$!!!
Everything looks tautological in your posts. 
iv) About the Misha's good answer. It is false to say that  the eigenvalue does not appear during the calculation; at each step, if we scale the result, then we obtain an approximation (at least of the absolute value) of the considered eigenvalue.
A: I am convinced that it is more likely than not that it is impossible to isolate the eigenvectors without referencing the eigenvalues $\lambda$, without referencing the zeroes of the characteristic polynomial, and without referencing any immediately recognisable derivative thereof.
Nevertheless, my instructor sent me an article that is as close to a solution as I expect is possible in that it does not require the eigenvalues from the beginning, and it is a single process that returns them both, rather than sticking together an eigenvalue-finding process and an eigenvector-finding process and calling it “one process.”
The algorithm is described by William A. McWorter, Jr., and Leroy F. Meyers in their paper “Computing Eigenvalues and Eigenvectors Without Determinants” in Mathematics Magazine, vol. 71, no. 1 (Feb. 1998):

Let $\newcommand{\u}{\vec{u}} \u$ be any nonzero vector in $\newcommand{\F}{\Bbb{F}} \F^n$ [the same field that matrix $A$ comes from]. Since $\F^n$ has finite dimension $n$, the $n+1$ vectors $\u, A\u, A^2\u, \dots, A^n\u$ are linearly dependent. Let $k$ be the smallest positive integer such that $a_0\u+a_1A\u+a_2A^2\u+\cdots+a_kA^k\u=\vec0$, for some $a_0,\dots,a_k$ in $\F$ with $a_k\neq0$. Algebraic closure ensures that the polynomial $a_0+a_1t+a_2t^2+\cdots+a_kt^k$ in $\F[t]$ is factorable as $(t-\lambda)\,Q(t)$ for some $\lambda$ in $\F$ and some polynomial $Q(t)$ in $\F[t]$. Hence $(A-\lambda I)\,Q(A)\,\u=\vec0$ [this is part of the proof, not the process]. The minimality of $k$ implies that the vector $Q(A)\,\u$ is nonzero and so is an eigenvector. . . .

If you can eyeball the factorisation $(t-\lambda)\,Q(t)$, then at least you don’t have to start with the eigenvalues.
