First I'll recap the normal eigenvalue problem to help explain what I'm asking. Say we have an $n\times n$ matrix $A$. Then $det(\lambda I-A)$ is its characteristic polynomial and its zeroes are the eigenvalues $A$. Counting multiplicity there are $n$ of them. For each eigenvalue there's a right eigenvector $\vec v_r$ which fulfils $(\lambda I-A)\vec v_r=\vec 0$ and also a left eigenvector $\vec v_l$ which fulfils $\vec v_l(\lambda I-A)=\vec 0$. If $A$ is diagonalizable we have the eigendecomposition:

$$A=Q\Lambda Q^{-1}$$

Here $\Lambda$ has the eigenvalues on the diagonal, $Q$ has the right eigenvectors as its columns, and $Q^{-1}$ has the left eigenvalues as its rows. Thus we can get the left eigenvectors from the right eigenvectors (and vice versa) by doing a matrix inversion and we don't need to know the eigenvalues to do it.

Now the actual problem I'm considering is the following. Suppose we have $k$ matrices that are $n\times n$ and we define a matrix polynomial a.k.a. matrix pencil with them:

$L(\lambda)=I\lambda^k+A_{k-1}\lambda^{k-1}+\ldots+A_1\lambda+A_0$

I'm only considering pencils for which the leading coefficient is the unit matrix here. Now we can get a similar characteristic polynomial as $det(L(\lambda))$. Its degree is $kn$ so counting multiplicity we can get $kn$ eigenvalues and for each eigenvalue we have right and left eigenvectors which fulfil $L(\lambda)\vec v_r=\vec 0$ and $\vec v_lL(\lambda)=\vec 0$.

Now my question is, can we determine the left eigenvectors from the right eigenvectors without reference to the eigenvalues like we could in the case of a normal eigenvalue problem? If it makes things easier, we can assume that each eigenvalue is distinct. If the case for arbitrary $k$ is too difficult, can we at least get a formula for the case $k=2$?

First I’ll recap the normal eigenvalue problem to help explain what I’m asking. Say we have an $n\times n$ matrix $A$. Then $\det(\lambda I-A)$ is its characteristic polynomial and its zeroes are the eigenvalues $A$. Counting multiplicity there are $n$ of them. For each eigenvalue there's a right eigenvector $\vec v_r$ which fulfils $(\lambda I-A)\vec v_r=\vec 0$ and also a left eigenvector $\vec v_l$ which fulfills $\vec v_l(\lambda I-A)=\vec 0$. If $A$ is diagonalizable we have the eigendecomposition:

$$A=Q\Lambda Q^{-1}.$$

Here $\Lambda$ has the eigenvalues on the diagonal, $Q$ has the right eigenvectors as its columns, and $Q^{-1}$ has the left eigenvalues as its rows. Thus we can get the left eigenvectors from the right eigenvectors (and vice versa) by doing a matrix inversion and we don’t need to know the eigenvalues to do it.

Now the actual problem I’m considering is the following. Suppose we have $k$ matrices that are $n\times n$ and we define a matrix polynomial a.k.a. matrix pencil with them:

$$L(\lambda)=I\lambda^k+A_{k-1}\lambda^{k-1}+\ldots+A_1\lambda+A_0.$$

I’m only considering pencils for which the leading coefficient is the unit matrix here. Now we can get a similar characteristic polynomial as $\det(L(\lambda))$. Its degree is $kn$ so counting multiplicity we can get $kn$ eigenvalues and for each eigenvalue we have right and left eigenvectors which fulfill $L(\lambda)\vec v_r=\vec 0$ and $\vec v_lL(\lambda)=\vec 0$.

Now my question is, can we determine the left eigenvectors from the right eigenvectors without reference to the eigenvalues like we could in the case of a normal eigenvalue problem? If it makes things easier, we can assume that each eigenvalue is distinct. If the case for arbitrary $k$ is too difficult, can we at least get a formula for the case $k=2$?