Let $A: V \to V$ be a linear transform over a field $F$, and the characteristic polynomial is $f(\lambda) = (\lambda-\lambda_1)^{r_1}\ldots(\lambda-\lambda_s)^{r_s}$. Prove that: $$\operatorname{rank}(A - \lambda_1I)^{r_1} = n - r_1$$
I think this essentially requires me to show that $r_1 = \operatorname{null}(A-\lambda_1I)^{r_1}$. But I am not sure how to use the knowledge of minimal polynomial to prove that. Could someone help me with it?
On
We know that matrix A will satisfy its characteristic polynomial, such that
$(A-\lambda_1I)^{r_1}\ldots(A-\lambda_sI)^{r_s} = 0$, Now every matrix is similar to a upper triangular matrix. Let B is the upper triangular matrix similar to A. Where the diagonal elements are the eigen values of A and arranged similarly from left to right as in the equation above (by grouping together the same eigenvalues in Schur decomposition). Then the above can be written as
$(B-\lambda_1I)^{r_1}\ldots(B-\lambda_sI)^{r_s} = 0$. Now note that B is an upper triangular matrix and $(B-\lambda_1I)$ is an upper triangular matrix, with diagonal elements 0 in first $r_1$ column only (no other diagonal elements are 0) also with the first column vector 0 (all elements in column one are 0).
Now we see that in $(B-\lambda_1I)^{2}$ first two columns (vectors) are surely zero, and this continues, for ex. in $(B-\lambda_1I)^{x}$ first x columns (vectors) ($x \leq r_1$) are surely zero.
We also note that in any such multiplication process described above the diagonal elements in rest of the columns can not be zero (they actually powered up). So if we perform $(B-\lambda_1I)^{r_1}$, only first $r_1$ columns (vectors) will be zero (and no other column vector). Now looking at the rest of the $n-r_1$ columns in $(B-\lambda_1I)^{r_1}$, we see that there diagonal elements are not zero but all elements below diagonal elements are zero. That makes these column vectors linearly independent.
And proof is completed as $(B-\lambda_1I)^{r_1}$ is similar to $(A-\lambda_1I)^{r_1}$.
I won't use any other eigenvalues explicitly, so set $\lambda=\lambda_1$ and $r=r_1$ for brevity.
Let $P=(X-\lambda)^r$ and let $Q=f/P$ be the remaining factor of the characteristic polynomial $f$. By assumption (certainly you meant the $\lambda_i$ to be distinct, although this is not said explicitly), $\lambda$ is not a root of$~Q$, and so $P$ and $Q$ are relatively prime in $F[X]$. Therefore there exist Bézout coefficients $S,T\in F[X]$ with $SP+TQ=1$. Since $(PQ)[A]=0$ (by Cayley-Hamilton) it is easy to see that $(SP)[A]$ and $(TQ)[A]$ are projectors on complementary subspaces $U,W$ of $V$. Then $P[A]$ acts invertibly on $U$ and vanishes on $W$, so $W=\ker(P[A])=\ker((A-\lambda I)^r)$, and $U$ is the image of $P[A]$, in particular $\operatorname{rank}(P[A])=\dim(U)$.
Also the characteristic polynomial $f$ of $A$ is the product of the characteristic polynomials of the restrictions of$~A$ to $U$ and $W$, the first of which does not have $\lambda$ as a root (as $P[A]$ acts invertibly on $U$) while the latter is a power of $X-\lambda$ (since $(X-\lambda)^r$ is an annihilating polynomial). But then the characteristic polynomial of the restriction of$~A$ to$~W$ is $(X-\lambda)^r$, and $\dim(W)$ is its degree$~r$, and $\dim(U)=\dim(V)-r$, completing the proof.
Note that my second paragraph only depends of $PQ$ being an annihilating polynomial with $P$ grouping all its factors $X-\lambda$; in particular it could have been the minimal polynomial, and the exponent could have been different from $r$. This is in fact a bit confusing in a proof of something that eventually says something about$~r$. Eventually $r$ is gotten as $\dim(W)$, not as $\deg(P)$.