Prove that $T$ can be written as $T=S \oplus R $ such that $\lambda$ is the only eigenvalue of $S$ and is not an eigenvalue of $R$

Question

$T: V\to V$ a linear transformation with an eigenvalue $\lambda$. V is a finite-dimensional space. Prove that $T$ can be written as $T=S \oplus R $ such that $\lambda$ is the only eigenvalue of $S$ and is not an eigenvalue of $R$.

Answer 1

Let me sketch the proof for the case $\lambda = 0$. Let $T \colon V \rightarrow V$ be a linear operator on a finite dimensional vector space. Define the sequence of subspaces $W_i = \ker(T^i), V_i = \operatorname{im}(T^i)$ and note that $$\{ 0_V \} = W_0 \subseteq W_1 \subseteq \dots, \\ V = V_0 \supseteq V_1 \supseteq \dots $$

Since $V$ is finite dimensional, the sequence $\{ W_i \}$ must stabilize at some index so let $i$ be the minimal index such that $W_i = W_{i+1}$. Then:

1. We have $V = W_i \oplus V_i$ because $$\dim \ker(T^i) + \dim \operatorname{im}(T^i) = \dim W_i + \dim V_i = \dim V$$ and if $v \in W_i \cap V_i$ then $v = T^i(u)$ for some $u \in $ and $Tv = T^{i+1}(u) = 0$ so $u \in W_{i+1} = W_i$ but then $v = T^i(u) = 0$.
2. The spaces $W_i,V_i$ are $T$-invariant.
3. The operator $T|_{W_i}$ is nilpotent (because if $u \in W_i$ then $T^i(u) = \left( T|_{W_i} \right)^i(u) = 0$). In particular, the only possible eigenvalue of $T|_{W_i}$ is zero.
4. The operator $T|_{V_i}$ is invertible because if $v \in V_i$ such that $T(v) = 0$ then $v \in V_i \cap W_{i+1} = V_i \cap W_i = \{ 0 \}$. Equivalently, $0$ isn't an eigenvalue of $T|_{V_i}$

To show the result for general $\lambda$, consider $T - \lambda I$.

Answer 2

You don't need to completely decompose $T$ for a similar argument to work. Let $S$ be the eigenspace of $\lambda$. Threre is a natural way to find $R$: Take a basis of $S$ and complete it to be a basis for the entire space. $R$ can be defined as the span of the vectors in the completion (the vectors that were not in the basis of $S$).