Degree of linear factors in minimal polynomial.

$V$ is a finite dimensional vector space over the field $F$.

$T$ a is linear operator on $V$. If $T$ is diagonalizable and if the distinct eigenvalues of $T$ are $a,b,...k$, then the minimal polynomial of $T$ is $(x-a)(x-b)...(x-k)$.

If $T$ is non diagonalizable(but has atleast one eigenvalue) and if $a,b,....,k$ are the distinct eigenvalues of $T$, then $(x-a)(x-b)...(x-k)$ divides $P$, where $P$ is the minimal polynomial of $T$.

Therefore $P= (x-a)(x-b)...(x-k)Q$, where $Q$ is a polynomial in $F$. Does this imply that none of the elements $a,b...,k$ are roots of $Q$?

In fact, if $Q$ is not equal to $1$, all roots of $Q$ are roots of $P$.
The distinct roots of the minimal polynomial are the same as the distinct roots of the characteristic polynomial, so $Q$ can boost the multiplicity, but can't add new roots.
For example, if $\displaystyle{A = \begin{bmatrix} 0& 1\\ 0& 0\\ \end{bmatrix} }$, then the minimal polynomial of $A$ is $x^2$, so $P = Q = x$.
Also, $A$ is not diagonalizable, since a diagonal matrix with the same eigenvalues as $A$ would have to be the $2{\,\times\,}2$ zero matrix, but $A$ is nonzero, hence can't be similar to the zero matrix.