📂Linear Algebra

Diagonalizable Linear Transformations Also Diagonalize When Restricted to Invariant Subspaces

Theorem¹

Let $V$ be a vector space, $T : V \to V$ be a diagonalizable linear transformation, and $W \le V$ be a non-empty $T$-invariant subspace. Then, the restriction $T|_{W}$ is also diagonalizable.

Proof

Since the characteristic polynomial of a diagonalizable linear transformation splits, there exist $n = \dim(V)$ eigenvalues $\lambda$. Let $E_{\lambda}$ be the eigen space corresponding to $\lambda$.

$$ E_{\lambda} = \left\{ v \in V : Tv = \lambda v \right\} $$

Let it be $W_{\lambda} = E_{\lambda} \cap W$, since $W$ is $T$-invariant, it becomes the eigen space of $T|_{W}$ corresponding to $\lambda$.

$$ W_{\lambda} = \left\{ v \in W : T|_{W}v = \lambda v \right\} $$

Let $\beta_{\lambda}$ be the basis of $W_{\lambda}$. We will show that $\beta = \bigcup\limits_{\lambda} \beta_{\lambda}$ becomes the basis for $W$.

• $\beta$ is linearly independent.

  Since the union of linearly independent sets from different eigen spaces is linearly independent, $\beta$ is linearly independent over $W$.

• $\beta$ generates $W$.

  Since $T$ is diagonalizable, every vector in $V$ can be expressed as a linear combination of linearly independent eigenvectors of $T$. Since $W$ is a subspace of $V$, all vectors in $W$ can also be expressed in this manner.

  Subsidiary Lemma

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

  Thus, by the subsidiary lemma and the definition of $\beta$, all elements of $W$ can be expressed as a linear combination of $\beta$.

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See Also

• Alternative proof