Which matrices $A$ have $f_{1} \neq 1$? I've been stuck on this problem for a while. I found a way to do but it seems inappropriate. I'm very grateful if anyone can help me solve this:


Let $A$ be an $n \times n$ matrix with entries in the field $F$ and let $f_{1}, f_{2}, ..., f_{n}$ be the diagonal entries of the normal form of $xI - A$. For which matrices $A$ is $f_{1} \neq 1?$


The way I did it is: The normal form of $A$ has diagonal entries including $1,1,..., 1, p_{r}, p_{r-1}, ..., p_{1}$. Here $r$ is number of factors in cyclic decomposition of $A$. So $f_{1} \neq 1 $ where the number of factors in the cyclic decomposition of $A$ is $n$. In this case, $A$ has $n$ characteristic vectors.
Thanks so much for your help.
 A: Here's a link discussing invariant factors of a matrix, which here are (apparently) the $f_i$, polynomials whose product is the characteristic polynomial of $A$ (necessarily of degree $n$) and which sequentially divide one another:
$$ f_i | f_{i+1} $$
Now by a degree argument, the only way to avoid $f_1 = 1$ is for all the invariant factors to be equal.  That is, if $f_1 \neq 1$, then $\deg(f_1) > 0$ and by the divisibility conditions, all $deg(f_i) > 0$.  But $\sum_i \deg(f_i) = n$ because their product is the characteristic polynomial of $A$ (determinant of $xI-A$), and this means all $deg(f_i) = 1$.  But monic first-degree polynomials $f_i|f_{i+1}$ only if they are equal.  Hence all the invariant factors are equal.
Now the minimal polynomial of $A$ is the least common multiple of its invariant factors, which by the divisibilty condition means $f_n$.  Since $deg(f_n) = 1$, $A$ must have a single "characteristic value" $r$ whose geometric multiplicity is $n$.  Equivalently $A = rI$ for some scalar $r \in F$.
