Invariance of generalized eigenspace

Question

I have a lemma saying that each of the generalized eigenspaces of a linear operator $T$ is invariant under $T$. This means that if $E_j$ is a generalized eigenspace then $T:E_j \rightarrow E_j.$

The proof of this goes like this.

Take a $v\in E_j$ so that $(T-\lambda_j I)^{n_{j}}v=0$.

Now we are going to show that the same holds for $T(v)\in E_j$.

$(T-\lambda_j I)^{n_{j}}T(v)=(T-\lambda_j I)^{n_{j}-1} T(T-\lambda_j I)v=T(T-\lambda_j I)^{{n_j}}v= T(0)=0$.

Can someone help me with the proof?

Answer 1

Consider the polynomials $p(x)=(x-\lambda_j)^{n_j}$ and $q(x)=x$. Note that $p(T) = (T-\lambda_jI)^{n_j}$ and that $q(T)=T$. Then, as $p(x)q(x) = q(x)p(x)$, we have $p(T)q(T) = q(T)p(T)$, that is, $$(T-\lambda_jI)^{n_j}T = T(T-\lambda_jI)^{n_j}.$$ Then, for $v \in E_j$ we have $(T-\lambda_jI)^{n_j}v = 0$, and then $Tv \in E_j$ since $$(T-\lambda_jI)^{n_j}Tv = T(T-\lambda_jI)^{n_j}v = T0 = 0.$$

Answer 2

Notice that $T$ commutes with $T-\lambda_jI$. So pull one factor out of $(T-\lambda_jI)^{n_j}$ to get the first equality.