Showing its minimal polynomial has a multiple root, given some facts about a linear operator Let $T$ be a linear transformation on $V$ over field $F$. Let $c\in F$ be a characteristic value of $T$, and $W_c$ the characteristic subspace of $T$ associated with $c$. Suppose a proper $T$-invariant subspace $W \supset W_c$, and there is a vector $\alpha\in V\setminus W$, such that $(T-cI)\alpha \in W$. Prove that the minimal polynomial $p_T$ of $T$ is in the form of $(x-c)^2q$ for nonzero polynomial $q\in F[x]$
My progress:
My first thought is to use $p_\alpha\mid p_T$, and I need to show $(T-cI)^2\alpha=0$. for a certain vector $\alpha$ in $V\setminus W$, $(T-cI)\alpha$ is not zero because otherwise $\alpha$ is in $W_c$ and thus is in $W$ (contradiction).
So if I can show $W=W_c$, then it's done. But I fail to show $W=W_c$. Or am I in the wrong way?
 A: You should start realising how little you are asked to do: it is given that $c$ is an eigenvalue (characteristic value), so a root of$~p_T$, and the (monic) factor $q$ can absorb anything; therefore all that is asked is to show that $c$ is a multiple root of the minimal polynomial $p_T$. Although I initially thought that "characteristic subspace" meant "generalized eigenspace" (blame my French connection, since that is precisely what "sous espace caractéristique" means), I will take it that $W_c$ here is just the ordinary eigenspace for$~c$ of$~T$, as with the other interpretation the hypothesis of question becomes absurd as we shall see.
It is given that for some $T$-stable subspace $W$ containing the characteristic subspace$~W_c$, there exists $\alpha\in V\setminus W$ with $(T-cI)v\in W$. This means that the image of $\alpha$ in the quotient space $V/W$ is an eigenvector with eienvalue$~c$ for the linear map $T_{/W}$ induced by $T$ in $V/W$. The existence of such a vector shows that the algebraic multiplicity of$~c$ as eigenvalue of$~T$ (on all of $V$) is strictly larger than its algebraic multiplicity as eigenvalue of the restriction$~T|_W$ of$~T$ to$~W$. This is because the characteristic polynomial of$~T$ is the product of the characteristic polynomials of $T|_W$ and of $T_{/W}$, the latter having$~c$ as root. But since $W$ already contains the whole eigenspace $W_c$, the geometric multiplicities of$~c$ as eigenvalue of $T|_W$ and of $T$ are the same. Hence the generalised eigenspace of$~T$ cannot be$~W_c$ (and in particular $T$ cannot be diagonalisable), nor indeed can it be contained in$~W$. Then there is some vector $v\in V$ with $(T-cI)^2(v)=0$ but $(T-cI)(v)\neq0$, which shows that $c$ is indeed a multiple root of$~p_T$.
