In Lang’s third edition of Linear Algebra on page $264$, I don’t understand the first line of his proof that any nonzero vector space over the complex numbers can be decomposed into the direct sum of $A$ invariant cyclic subspaces, where $A$ is a linear operator.
Here’s Theorem $4.2$, which he references:
Let $V$ be a vector space over $\mathbb{C}$, and let $A:V\to V$ be a linear operator. Let $P(t)$ be a polynomial such that $P(A)=0$, and let $P(t)=(t-\alpha_1)^{m_1}...(t-\alpha_r)^{m_r}$ be its factorization. Then $V=W_1\oplus W_2 \oplus ... \oplus W_r$, where $W_i=\mathrm{Ker}(A-\alpha_iI)^{m_i}.$
And the first line of his proof is:
By Theorem $4.2$, we may assume without loss of generality there exists a number $\alpha$ and an integer $r\geq 1$ such that $(A-\alpha I)^r=0$.
Where $A$ is the linear operator used in the proof. I just have no clue how this theorem is used at all to show the existence of such numbers. Any help would be appreciated.
Summary of a discussion from the comments:
The key is that because each $W_i$ is an invariant subspace, we can put $A$ in Jordan form by separately putting each restriction $A|_{W_i}$ into Jordan form. Because these invariant subspaces are given by $W_i = \ker(A - \alpha_i I)^{m_i}$, it must hold that $T = A|_{W_i}$ will satisfy $(T - \alpha I)^r = 0$ with $\alpha_i = \alpha$ and $r = m_i$.
For the purposes of the proof, we consider $A$ to be one of these maps over an invariant subspace rather than taking $A$ to be an arbitrary transformation.