Finding approximate common eigenvector of two matrices Let $A$ and $B$ be two square matrices without any common eigenvectors. Is it possible to find an approximate solution, i.e., a vector $\vec{v}^*$, satisfying both of the following equations
$$\left\lbrace\begin{matrix}A\vec{v}\approx \lambda_A\vec{v}\\
B\vec{v}\approx \lambda_B\vec{v}\end{matrix}\right.$$ 
in some manner, e.g., with the least squares error criterion? I'm looking for nonzero $\vec{v}$ and $\lambda$.
If least squares is not good for this problem what criterion can simplify the problem?
Maybe a variation of the power method can be useful. Don't have any idea though.
 A: The problem is not easy because the eigenvectors of a matrix $U$ are not (in general) continuous functions of the entries of $U$, except, for example, if the eigenvalues are simple.
Assume that there is an approximate solution $v$ of $Av=\lambda v,Bv=\mu v$.
Intuitively, $v$ is close to an eigenvector of $BA^{-1}$ (we may assume that $A$ is invertible; change $A$ with $A+\alpha I$); then we calculate approximations of the eigenvalues of $A:(\lambda_i);B:(\mu_i);BA^{-1}:(\nu_i)$ and we search $i,j,k$ s.t. $\mu_i/\lambda_j\approx \nu_k$; finally $v$ is close to the vector space $\ker(BA^{-1}-\nu_k I)$. In a similar way , you can also consider a matrix $A+\alpha B$ where $\alpha\in\mathbb{C}$. This method works when the eigenvalues of $A,B$ are distinct.
Otherwise, it can be twisted. for example, consider these $2$ matrices close to $I_2$:
$A=diag(1-t,1+t),B=\begin{pmatrix}1&t\\t&1\end{pmatrix}$ where $t>0$ is small. Any vector $v$ satisfies $Av\approx Bv\approx v$. Yet, a basis of eigenvectors of $A$ is $[1,0]^T,[0,1]^T$ and, of $B$ is $[1,1]^T,[1,-1]^T$. 
EDIT. Answer to @SMA.D. If you consider some matrices $A_1,\cdots,A_k$, then randomly choose coefficients $(u_i),(v_i)$ and let $U=\sum_i u_iA_i,V=\sum_i v_iA_i$. Calculate the eigenvectors $(x_j)$ of $U$ and let $\alpha_j=|<\dfrac{Vx_j}{||x_j||},\dfrac{x_j}{||x_j||}>|$. If $\alpha_j$ is close to $1$, then $x_j$ is close to  eigenvectors of $U,V$ and, with a strong probability, is close to eigenvectors of the $(A_i)$.
