Show that $T(W_j) \subset W_j$ (T-invariant) We have a linear operator $T: V \to V$.
Suppose $W_j$ is the nullspace of $(T - \lambda_j I)^{r_j}$ for some $0 < r_j$, where $\lambda_j$ is a root of the minimal polynomial of $T$ of multiplicity $r_j$.
I would like to show that $T(W_j) \subset W_j$.
Here is my attempt:
We want to show that $\forall \gamma \in T(W_j)$, $\gamma \in W_j$ (i.e. $\gamma \in null(T - \lambda_jI)^{r_j}$ )
We know that since $\gamma \in T(W_j)$ there exists $\alpha \in W_j$ such that $T(\alpha) = \gamma$. This implies that $\alpha \in null(T - \lambda_jI)^{r_j}$.
I don't know how to proceed from here. Could I get some help?
Thanks.
 A: There's a more general fact. Fix any polynomial $f$; and consider the kernel of $f(T)$, i.e. those $w$ such that $f(T)w=0$. Then this is $T$ invariant, for $$f(T)(Tw)=Tf(T)w=T(0)=0$$
This is simply because $T$ and $f(T)$ commute, for $T$ commutes with any linear operator that is a polynomial in $T$. Your case is when $f(X)=(X-\lambda)^r$. The importance of this is that if we write $m_T(X)=f(X)g(X)$ with $f,g$ coprime, then we have a direct sum decomposition $$V=\ker f(T)\oplus \ker g(T)$$
Indeed, we can write $k(X)f(X)+h(X)g(X)=1$ for some polynomials $k,h$ since $f,g$ are coprime. This means that if $v$ is in both kernels, $$0=k(T)f(T)v+h(T)g(T)v=v$$
So the intersection is indeed trivial. Now pick any vector $w$, and consider $k(T)f(T)w=w_1$ and $h(T)g(T)w=w_2$. The equation above shows $w=w_1+w_2$, and $g(T)w_1=0$ and $f(T)w_2=0$, so indeed the sum is the whole space. 
One can do this more generally if we factor $m_T=f_1^{p_1}\cdots f_r^{p_r}$ with the $f_i$ irreducible, by iterating the construction above. This gives the first step into the proof of the Jordan decomposition if the $f_i$ are lineal, for example.
