how to find eigenvectors of a 3by 3 matrix in terms of elements of one eigenvector? I want to find eigenvectors of a 3 by 3 matrix, $N_1$, $N_2$, $N_3$, based on elements of $N_1$, $n_1, n_2 , n_3$.
all I could find online was analytical (iterative methods obviously are not usefull) methods for finding eigenvectors. but I dont know how to find the eigenvectors in terms of one of them.
Assume the matrix to be symmetric positive definite.
The link of the paper I mentiond is below:
https://arxiv.org/pdf/physics/0610206.pdf 
 A: It cannot be done.  Given one eigenvector $N_1$ of a symmetric positive definite matrix, all you can say about the other eigenvectors is that they lie in the plane orthogonal to $N_1$, and are orthogonal to each other as well.  
The problem is that any vectors $N_2$, $N_3$ which are orthogonal to each other and to $N_1$ are the eigenvectors of some symmetric positive definite matrix with given eigenvalues $n_{1,2,3}$.  Here's how you construct that matrix: 
$$
\frac{n_1 N_1 \otimes N_1}{||N_1||^2} + \frac{n_2 N_2 \otimes N_2}{||N_2||^2} + \frac{n_3 N_3 \otimes N_3}{||N_3||^2}
$$
Where $\otimes$ is the outer product.
In the basis formed by the orthonormal vectors $N_i/||N_i||$, this matrix has the form
$$
\begin{bmatrix}
n_1 & 0 & 0 \\
0 & n_2 & 0 \\
0 & 0 & n_3
\end{bmatrix}
$$
Since it is symmetric and positive definite in this basis, it is symmetric and positive definite in any other basis related by a orthogonal transformation (including the canonical basis).  

EDIT: An example may clarify:  Suppose $N_1=e_1=(1,0,0)^T$.  For any unit vectors $N_2=(0,a,b)^T$ and $N_3=(0,-b,a)^T$ (note that the unit vector requirement means $b=\sqrt{1-a^2}$), we can form a symmetric matrix with eigenvalues $n_{1,2,3}$ and eigenvectors $N_{1,2,3}$.  That matrix is given by 
$$
\begin{bmatrix}
n_1 & 0 & 0 \\
0 & n_2 a^2 + n_3 b^2 & (n_2-n_3) a b \\
0 & (n_2-n_3) a b & n_2 b^2 + n_3 a^2 
\end{bmatrix}
$$
You can verify by hand that this matrix has eigenvalues and eigenvectors $n_{1,2,3}$ and $N_{1,2,3}$.
A: Let $V=[u,v,w]^T$ be an eigenvector of a symmetric real matrix $A$. You can reduce the research of the other $2$ eigenvectors to solving an equation of degree $2$.


*

*You write an orthonormal basis of the orthogonal of $V$, that is $X=\dfrac{x}{||x||},Y=\dfrac{y}{||y||}$ where $x=[v,-u,0]^T,y=[v,\dfrac{v^2}{u},\dfrac{-(u^2+v^2)v}{uw}]^T$ (we assume $uw\not= 0$, otherwise you multiply $y$ by $uw$)).

*You write the matrix $B=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ where  $a=X^TAX,b=Y^TAX,c=Y^TAY$. $B$ is the matrix of the restriction of $A$ in the plane $span(X,Y)$ for the basis $X,Y$.

*You seek the eigenvalues $\lambda,\mu$ of $B$. Then the eigenvectors of $B$ are  $[b,\lambda-a]^T,[b,\mu-a]^T$ (if $b=0$, then there is nothing to do).
Conclusion. the required two eigenvectors of $A$ are $bX+(\lambda-a)Y,bX+(\mu-a)Y$.
