Why is $\left.(T-\lambda I)\right|_{G(\lambda,T)}$ nilpotent

Question

Axler defines a nilpotent operator on p. 248 0f "Linear Algebra Done Right" 3rd ed. as

some power of it equals $0$.

Then in his discussion of Generalized Eigenspaces on p. 252 he claims for $V$ a finite dimensional, complex vector space with $T\in \mathcal L(V)$ and $\lambda$ an eigenvalue of $T$ where ${G(\lambda,T)}$ is the generalized eigenspace of $\lambda$:

$\left.(T-\lambda I)\right|_{G(\lambda,T)}$ is nilpotent.

The first part of my question is why is the operator $T-\lambda I$ restricted? I thought you just need a power of the operator to equal $0$. Also, I would appreciate help seeing, as he state in his proof of the assertion, that it follows from the definitions.

Is it because $\left.(T-\lambda I)\right|_{G(\lambda,T)}v =0$ for any $v\in G(\lambda,T)$ so its eigenvalue equals $0$ which implies it is nilpotent? But that still doesn't answer my first question regarding the definition of a nilpotent operator, $N$ such that $N^m=0$ for $m$ some positive integer power.

Thanks

Answer 1

Take for simplicity a triangular matrix $$ T= \begin{bmatrix} \color{red}1 & \color{red}2 & \color{red}3 & * & *\\ \color{red}0 & \color{red}1 & \color{red}4 & * & *\\ \color{red}0 & \color{red}0 & \color{red}1 & * & *\\ 0 & 0 & 0 & \color{blue}5 & \color{blue}6\\ 0 & 0 & 0 & \color{blue}0 & \color{blue}5 \end{bmatrix}. $$ It has two eigenvalues: $1$ and $5$. Then $$ T-1\cdot I= \begin{bmatrix} \color{red}0 & \color{red}2 & \color{red}3 & * & *\\ \color{red}0 & \color{red}0 & \color{red}4 & * & *\\ \color{red}0 & \color{red}0 & \color{red}0 & * & *\\ 0 & 0 & 0 & \color{blue}4 & \color{blue}6\\ 0 & 0 & 0 & \color{blue}0 & \color{blue}4 \end{bmatrix}. $$ This operator is not nilpotent because the nonsingular block $$ \begin{bmatrix} \color{blue}4 & \color{blue}6\\ \color{blue}0 & \color{blue}4 \end{bmatrix} $$ in whatever power is never zero. The generalized subspace $G(1,T)$ is that spanned by the first three coordinates, so the restriction of $T-1\cdot I$ to it is $$ T_1=\begin{bmatrix} \color{red}0 & \color{red}2 & \color{red}3\\ \color{red}0 & \color{red}0 & \color{red}4\\ \color{red}0 & \color{red}0 & \color{red}0 \end{bmatrix}. $$ This operator is nilpotent.

He claims it by definition. I guess (maybe wrong) that he defines $G(\lambda,T)$ by chains of iterated kernels $\ker(T-\lambda I)^k$, and the result of this construction is that there exists $k_0$ such that $$ (T-\lambda I)^{k_0}v=0\quad\forall v\in G(\lambda,T). $$ It is the same ($G$ is invariant subspace) as $\left((T-\lambda I)|_{G(\lambda,T)}\right)^{k_0}=0$.

Answer 2

1. The operator $(T-\lambda I)$, when considering it as an element of $L(V)$ is generally not nilpotent.
2. $G(\lambda,T)$ is a subspace of $V$. You can show that this subspace is invariant under $T$ and also under $(T-\lambda I)$. Therefore you can restrict the operator $(T-\lambda I)$ to the subspace $G(\lambda,T)$ which gives you an operator in $L(G(\lambda,T))$. This restricted operator then is nilpotent.