I am stuck at one step of the proof.
Let $G_{T}(\lambda)$ be the dimension of the generalized eigenspace of an upper triangular matrix $T$ with eigenvalue $\lambda$, and let $\#_T(\lambda)$ be the number of times $\lambda$ appears on the diagonal.
The author has already proved that
$$G_T(\lambda) \ge \#_T(\lambda)$$
and so the equality can be established by showing that
$$\#_T(\lambda) \ge G_T(\lambda)$$
The author uses MI on the dimensional $m$ of $T$, of which the base case $m=1$ is trivial.
Then assume the inequality is true for all $1$ to $m-1$, and then consider an upper triangular matrix $T$ with dimension $m$.
Suppose $\{v_1,v_2,\cdots,v_m\}$ is a basis for the vector space $V$ with diagonal entries $\lambda_1,\lambda_2,\cdots,\lambda_m$. Then $U=\langle\{v_1,v_2,\cdots,v_{m-1}\}\rangle$ is a subspace of $V$ that is invariant relative to $T$. Then the restriction $T_U:U\to U$ with basis $\{v_1,v_2,\cdots,v_{m-1}\}$ has an upper triangular representation with diagonal elements $\lambda_1,\lambda_2,\cdots,\lambda_{m-1}$, for which one can apply the induction assumption.
Now, the author's induction step starts by this: Suppose that $\lambda$ is any eigenvalue of $T$. Then suppose that $v\in Ker((T-\lambda I_V)^m)$.
As an element of $V$, we can write $v$ as a linear combination of the basis elements of $\{v_1,v_2,\cdots,v_m\}$, or more compactly, there is a
vector $u \in U$ and a scalar $\beta$ such that $v=u+\beta v_m$.
Then,
$$\beta(\lambda_m – \lambda)^m v_m =\beta(T-\lambda I_V)^m(v_m)$$
$$=-(T-\lambda I_V)^m (u) + (T-\lambda I_V)^m (u) + \beta(T-\lambda I_V)^m(v_m)$$
$$=-(T-\lambda I_V)^m (u) + (T-\lambda I_V)^m (u+\beta v_m)$$
$$=-(T-\lambda I_V)^m (u) + (T-\lambda I_V)^m (v)$$
$$=-(T-\lambda I_V)^m (u)$$
The final expression is an element of $U$ because $U$ is invariant relative to both $T$ and $I_V$. Hence
$$\beta(\lambda_m – \lambda)^m v_m=0$$
$$\cdots$$
I am stuck at the first step. The author says this step uses the theorem that if $\lambda$ is an eigenvalue of a matrix $A$, then for integer $s \ge 0$, $\lambda^s$ is an eigenvalue of $A^s$. But
$$(T-\lambda I_V)v_m \ne (\lambda_m-\lambda)v_m$$
Thanks in advance for any help! Regards!
Best Answer
You have $(T - \lambda I_v)(v_m) = (\lambda_m - \lambda) v_m + \tilde{u}$ where $\tilde{u} \in U$. By induction you know that $(T- \lambda I)^{m-1} (\tilde{u}) = 0$ for any eigenvalue $\lambda$ of $T_U$ and $\tilde{u} \in U$.
Then $(T - \lambda I_v)^m (v_m) = (\lambda_m - \lambda)^m v_m$