I am having issues figuring out how to approach this problem:
Let $\mathbf F$ be a field. For every matrix $A\in \mathbf F^{n\times n}$, $\textbf{Com}(A)$ is the comatrix of $A$. It is clear for me that the coefficients of $\textbf{Com}\left(A-XI_n\right)$ are elements of $\mathbf F[X]$. Prove that: $$\text{gcd}\left(\textbf{Com}\left(A-XI_n\right)_{i,j}, i,j=1,\ldots,n\right) = \frac{\chi_A(X)}{\pi_A(X)}.$$ Where $\chi_A$ is the characteristic polynomial of $A$, $\pi_A$ is its minimal polynomial and $\text{gcd}$ is taken in the ring $\mathbf F[X]$.
I am wondering if this is correct? And how can I approach it?
I don't want to give a full proof of the Rational Canonical Form theorem, but only a sketch.
Setup: We make the underlying $n$-dimensional $F$-vector space $V$ with basis $\{v_1,\dots,v_n\}$ into an $F[X]$-module by defining $X.v:=Av$ and extending this to the whole of $F[X]$ in the obvious way. We also have the obvious $F[X]$ module $F[X]^n$ on which $X$ acts on each component as multiplication by $X$; let $\{e_1,\dots,e_n\}$ be the natural "basis". The map $\sigma_A:(\phi_1(X),\dots,\phi_n(X))\mapsto \sum_{i=1}^n \phi_i(A)v_i$ is an $F[X]$-surjection, and so we get $V\simeq F[X]^n/\ker\sigma_A$. The kernel is generated by the $n$ elements $Xe_i-\sum_{i=1}^n a_{ji} e_j$.
The Theorem Basically the RCF Theorem asserts that we can choose a new basis $\{f_1,\dots,f_n\}$ of $F[X]^n$ such that the kernel $\sigma_A$ is generated by the $n$ elements $d_i(X) f_i$ where for each $i$ we have $d_i|d_{i+1}$. That is, $V\simeq \bigoplus_{i=1}^n F[X]/\langle d_i(X)\rangle$. Choosing in each summand the basis $\{1,X,\dots,X^{\deg d_i -1}\}$ we see that after a suitable change of basis we can write $A$ as the block-diagonal sum $A=\bigoplus_{i=1}^nC(d_i)$, where $C(d_i)$ is the companion matrix of $d_i$.
Corollaries. It is routine to check that $C(d_i)$ satisfies the polynomial $d_i$, and no polynomial of smaller degree: so the characteristic polynomial and minimal polynomial of $C(d_i)$ are both $d_i(X)$. We then have $\chi_A(X)=\prod_{i=1}^n d_i(X)$. Since the $d_i$ all divide $d_n$ we have that $d_n$ annihilates $A$, and since it is minimal on the last summand we must have that $d_n(X)$ is the minimal polynomial of $A$.
Smith Normal Form Algorithm How do we actually find this new "basis" $\{f_1,\dots,f_n\}$ and the simplified set of spanning relations? Clearly we may perform on the columns of $XI-A$ any of (a) elementary column operations of the sort (i) multiply a column by an element of $K$ (ii) swap two columns (iii) add to one column an $F[X]$-multiple of a different column; or (b) the corresponding row operations. It is easily proved that there is a sequence of such operations transforming $XI-A$ to the form $\text{diag}(d_1(X),\dots,d_n(X))$.
Corollaries It is routine to check that each of the elementary operations actually induces an elementary operation on the matrix $A^{(k)}$ of $k\times k$ minors. Hence the ideal generated by the $k\times k$ minors is unchanged: that is the highest common factor of the $k\times k$ minors of $XI-A$ is the same as the highest common factor of the $k\times k$ minors of $\text{diag}(d_1(X),\dots,d_n(X))$. Since we have $d_i|d_{i+1}$ that means that the highest common factor of the $k\times k$ minors of $XI-A$ is $\prod_{j=1}^{k}d_i(X)$. The case $k=n-1$ is what you want.