rank $r$ lambda matrix and relationship of its minors of order $\le r$ Could anyone help me to prove the following or at least make me understand with an example?
$A(\lambda)$ be any $\lambda$ matrix (A matrix whose entries are polynomial in variable $\lambda$) of rank $r$
$D_j(\lambda)$ be the greatest common divisor of all the minors of order $j$ in $A(\lambda), (j=1,2,\dots,r)$
I need to show $D_{i-1}(\lambda)|D_i(\lambda)\forall i=1,2,\dots,r$
I inderstand that $D_0(\lambda)\equiv 1$
Thanks!
 A: Think about how you compute minors. Each $i$-minor is a linear combination of $i-1$ minors. Since $D_{i-1}$ divides each $i-1$ minor, it will then divide each $i$-minor. Thus $D_{i-1} | D_i$.
Per request an example:
$$A(\lambda) = \begin{pmatrix}
\lambda & \lambda^2+\lambda & \lambda^3 -1\\
1 & 5\lambda & \lambda^2 \\
\lambda^3-3\lambda^2 & \lambda^3 & \lambda
\end{pmatrix}$$
$D_1(\lambda) = \textrm{ gcd of entries of } A(\lambda) = 1$.
Now to get $D_2(\lambda)$ we need to compute the gcd of determinants of all $2\times 2$ submatrices of $A(\lambda)$. Looking at the particular submatrix obtained from deleting the middle column and bottom row,
$$\begin{pmatrix}
\lambda & \lambda^3 - 1\\
1 & \lambda^2
\end{pmatrix}$$
we calculate that it's determinant is 1. Therefore $D_2(\lambda) = 1$.
$D_3(\lambda)$ will just be the determinant of $A(\lambda)$ since there is only one $3\times 3$ submatrix, namely $A(\lambda)$ itself. So you see that it works out in this case, but in a somewhat unilluminating way. 
The key observation is that when you compute a determinant using cofactor expansion, you get a linear combination of determinants of smaller matrices. These determinant of smaller matrices are lower order minors. So the divisibility pulls through.
For instance, in this example, $$D_3(\lambda) = \det(A(\lambda)) = \lambda \ \det\begin{pmatrix}
5\lambda & \lambda^2 \\
\lambda^3 & \lambda
\end{pmatrix} -(\lambda^2 + \lambda) \ \det\begin{pmatrix}
1 & \lambda^2\\
\lambda^3-3\lambda^2 & \lambda
\end{pmatrix} + (\lambda^3-1) \ \det\begin{pmatrix}
1 & 5\lambda\\
\lambda^3 - 3\lambda^2 & \lambda^3
\end{pmatrix}_.$$
Since each of these $2\times 2$ determinants are $2$-minors, and we know that $D_2(\lambda)$ divides all $2$-minors, then $D_2(\lambda)$ divides $\det(A(\lambda) =  D_3(\lambda)$. Notice that we didn't even need the fact that $D_2(\lambda) = 1$! 
