If [A|B] is controllable, then $M_c=[B \quad AB \quad \dots \quad A^{n-1}B]$ has $n$ linearly independent colums. Then we consider the following table
$\begin{bmatrix} b_1& b_2& b_3 & \dots & b_m\\ Ab_1& Ab_2& Ab_3& \dots & Ab_m \\ \vdots & \vdots& \vdots & \vdots &\vdots\\ A^{n-1}b_1&A^{n-1}b_2&A^{n-1}b_3&\dots&A^{n-1}b_m \end{bmatrix}$
if we pick the vectors by going from the left to the right of the first row of the table, then the second row, and then the third row ..., and stop after $n$ linearly independent vectors are selected. Let the number of the selected vectors in the $j$th column is $k_j$. Then the bag $k = \{k1, k2, \dots,k_m\}$.
The Hermite indices are obtained by picking the vectors by going from the top to the bottom of the rst column of the table, then the second column, and then the third column ..., and stop after n linearly independent vectors are selected. Then we get $h = \{h1, h2, \dots,h_m\}$
Two questions on control theory
1)Why are the controllability indices,$k_j$, and the Hermite indices $h_j$, of [A|B] the same as those of [A + BF|B]? This is an exercise of Linear Systems CH4 3.21
2)Why the set of Hermite indices of [A|B], $h$, majorizes $k$, i.e. $k\prec h$
You can argue in terms of the Hautus test. Since
$$\begin{bmatrix}A+BF & B\end{bmatrix} = \begin{bmatrix}A & B\end{bmatrix} \begin{bmatrix}I &0 \\F & I\end{bmatrix} $$ and the rightmost matrix is full rank, the controllability properties are invariant.
The second is actually by definition or I didn't understand the question properly.