Why must every vector in V belongs to one of the generalised eigenspaces of $T: V \to V?$

Question

Why must every vector in V belongs to one of the generalised eigenspaces of $T: V \to V?$ Is there a simple proof for this? Can someone provide me with an intuition behind it?

Note that V is an algebraically closed field.

Answer 1

Assume that all the eigenvalues of $T$ are in the field of scalars of $V$ ( true if $k=\mathbb{C}$. Then $V$ is a direct sum of the generalized eigenspaces $V_{\lambda}$. That is, every vector can be written uniquely as $$v = \sum_{\lambda} v_{\lambda}$$

where $(T-\lambda)^{n_{\lambda}}v_{\lambda} = 0$. Note that any vector for which more than one component $v_{\lambda}$ is nonzero cannot belong to any $V_{\mu}$.

A very simple example is $T= \left( \begin{array}{cc} 1 &0 \\ 0 & 0 \end{array} \right)$. Consider $v = e_1 + e_2 = (1,0)+(0,1) = (1,1)$.

Answer 2

I think this comes down to understanding the characteristic polynomial.

Suppose that $PQ = 0$, where $P$ and $Q$ are relatively prime polynomials of $T$. Then $V = PV \oplus QV$. This is basically the Chinese Remainder Theorem, but we can see it directly: if $AP + BQ = 1$, then $v = APv+BPv$, and if $w \in PV \cap QV$, then $Pw = Qw = 0$, so $w = (AP+BQ)w = 0$.

Note that $P$ is invertible on $PV$, and $Q$ is invertible on $QV$.

So, if we completely factor the characteristic polynomial, we can decompose $V$ into summands $V_\lambda$ corresponding to the eigenvalues. Furthermore, each $V_\lambda$ has only $\lambda$ as an eigenvalue, because each $T-\tau$ is invertible on $V_\lambda$ when $\lambda \neq \tau$.

My claim is that $V_\lambda$ is exactly the generalized eigenspace, and we can go home happy.

One standard definition is that the generalized eigenspace of $\lambda$ is the set of vectors $v$ which lie in the kernel of some power of $T-\lambda$. Clearly any such vector lies in $V_\lambda$, because, again, $T-\lambda$ is invertible on the other factors. So we just need to show that every vector in $V_\lambda$ has this property.

But this is straightforward: on $V_\lambda$, the only eigenvalue of $T$ is $\lambda$. There are many ways to see, from this fact, that $T-\lambda$ is nilpotent, but one way is to look at its characteristic polynomial. Another is to notice that each successive application must lower the dimension.