📂Linear Algebra

The Relationship between Invariant Subspaces and Eigenvectors

Theorem¹

Let $V$ be a $n$-dimension vector space, $T : V \to V$ a linear transformation, and $W \le V$ a $T$-invariant subspace. Let $v_{1}, \dots, v_{k}$ be $T$’s eigenvectors corresponding to distinct eigenvalues. If $v_{1} + \cdots + v_{k} \in W$, then for every $i$ we have $v_{i} \in W$.

Explanation

Since $W$ is a subspace, if $v_{i} \in W$ then $\sum_{i}v_{i} \in W$ holds. However, the converse does not hold in general. This theorem states that when the $v_{i}$ are eigenvectors, the converse also holds.

Proof

We prove this by mathematical induction.

• Case $k=1$

  Clearly, $v_{1} \in W \implies v_{1} \in W$.

• Assume it holds for $k = m-1$.

  Now suppose $k = m$ and that $v = v_{1} + \cdots v_{m} \in W$ 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. Moreover, 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 induction hypothesis that the statement holds when $k=m-1$, we obtain

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

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

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

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