📂Linear Algebra

The Relationship between Invariant Subspaces and Eigenvectors

Theorem¹

Let $V$ be a $n$-dimensional vector space, $T : V \to V$ be a linear transformation, and $W$ be $T$-invariant. Let $v_{1}, \dots, v_{k}$ be the eigenvector of $T$ corresponding to different eigenvalues. If $v_{1} + \cdots + v_{k} \in W$, then for all $i$, $v_{i} \in W$ holds.

Proof

We prove this by mathematical induction.

• For $k=1$

  It is trivially $v_{1} \in W \implies v_{1} \in W$.

• Assume it holds for $k = m-1$

  Now, let $k = m$ and suppose $v = v_{1} + \cdots v_{m} \in W$ holds. Since $W$ is $T$-invariant,

  $$T(v) = T(v_{1} + \dots + v_{m}) = T(v_{1}) + \cdots T(v_{m}) = \lambda_{1}v_{1} + \cdots + \lambda_{m}v_{m} \in W$$

  Here, $\lambda_{i}$ are distinct eigenvalues. Since ($W$ is a subspace) $\lambda_{m}v \in W$ and the following holds:

  $$T(v) - \lambda_{m}v = (\lambda_{1} - \lambda_{m})v_{1} + \cdots + (\lambda_{m-1} - \lambda_{m})v_{m-1} \in W$$

  Then, by the assumption that it holds for $k=m-1$, we have:

  $$(\lambda_{1}-\lambda_{m})v_{1}, \dots, (\lambda_{m-1}-\lambda_{m})v_{m-1} \in W \implies v_{1}, \dots, v_{m-1} \in W$$

  Therefore, since $W$ is a subspace, the following holds:

  $$v_{m} = v - v_{1} - \cdots - v_{m-1} \in W$$

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