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.

  • $\begingroup$ $v = 0$ $\quad$ $\endgroup$
    – Exodd
    Jun 1 '17 at 14:56
  • $\begingroup$ @Exodd I meant nonzero. $v=0$ is trivial. $\endgroup$
    – SMA.D
    Jun 1 '17 at 19:24
  • 1
    $\begingroup$ What exactly do you mean by the verb "to find"? You can certainly set up some minimization problem like $\|Av-(Av,v)v\|^2+\|Bv-(Bv,v)v\|^2 \to\min$ under the condition $\|v\|=1$ but how "solvable" it is depends of your definition of "find". The least squares approach in its standard form is out of question because the minimum is not unique (how do you propose to use the least squares to treat the case $A=B$, which is a pure eigenvalue problem?) | $\endgroup$
    – fedja
    Jun 2 '17 at 13:17
  • $\begingroup$ @fedja Thank you for your comment. I do not insist on least squares. I was thinking of least squares since I can't apply the solving method of a pure eigenvalue problem here. I think the minimization would work but I think it is not the best possible solution. The only thing that I need is to find a vector such that $A\vec{v}$ is parallel to $B\vec{v}$. If the square of error is not a good measure in this problem, what criterion can lead to a solution? (possibly avoiding that minimization) if not how can I setup a minimization problem in this case? $\endgroup$
    – SMA.D
    Jun 2 '17 at 14:51
  • 1
    $\begingroup$ If you just want $Av$ parallel to $Bv$, solve the eigenvalue problem for $BA^{-1}$, say. $\endgroup$
    – fedja
    Jun 2 '17 at 22:40

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 $(A_i)$.

  • $\begingroup$ Thanks for your helpful answer. I'm also interested in the generalization of this problem with more than 2 matrices. ِDo you know any reference for such generalization? $\endgroup$
    – SMA.D
    Sep 29 '17 at 9:51

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