I don't understand the proof below from Artin's Algebra. Can someone please explain? My specific questions are below.
Suppose $T$ is diagonalizable, so the matrix $\Lambda = [T]_\mathbf{B}$ with respect to $\mathbf{B}$ is diagonal with each $v_i \in \mathbf{B}$ an eigenvector with eigenvalue $\lambda_i$. Now let $v$ be a generalized eigenvector, so $(T – \lambda I_n) \circ (T – \lambda I_n) \circ \dots \circ (T – \lambda I_n) (T(v) – \lambda v)=0$, i.e. $(T – \lambda I_n)^k(v) = 0$ for some $k > 0$. Then Artin write "we replace $T$ by $T – \lambda$ to reduce to the case that $T^k v = 0$." I do not understand this step at all.
The coordinates of $T^kv$ are $[T^kv]_\mathbf{B} = [T^k]_\mathbf{B} [v]_\mathbf{B} = \Lambda^k (x_1, \dots, x_n)^t$, so I agree that if $T^k v = 0$, then the coordinates of $T^kv$ are $\lambda_i^kx_i = 0$ with scalars $\lambda_i \in \mathbb{C}$, so $\lambda_i = 0$ or $x_i = 0$ for each $i \in \{1, \dots, n\}$. How do we then conclude that $Tv = 0$? I think it relates to the "we replace $T$ by $T – \lambda$" step…
Best Answer
Simply name $T-\lambda I_n$ to something, say, $S$. (They kept calling it $T$, and it seems it might cause confusions.)
$S$ is still diagonalizable in basis $B$, with same generalized eigenvectors but with $0$ as the corresponding eigenvalue.
(Simply $(T-\lambda I_n)^kv=0$ becomes $S^kv=0$.)
Assuming $S={\rm diag}(\lambda_1,\dots,\lambda_n)$ and $v=(x_1,\dots,x_n)^T$, we have $$S^r(v)=(\dots,\, {\lambda_i}^rx_i,\dots)^T$$ for any $r\in\Bbb N$, in particular for $r=k$ and for $r=1$ as well.
So, if $S^kv=0$, then for each index $i$, either $x_i=0$ or $\lambda_i=0$, but in this case $S(v)$ is also $0$ by the above.