📂Linear Algebra

대각화가능한 선형변환의 불변부분공간으로의 축소사상도 대각화가능하다

Theorem¹

Let $V$ be a vector space, and let $T : V \to V$ be a diagonalizable linear transformation. Suppose $\left\{ \mathbf{0} \right\} \ne W \le V$ denotes a non-trivial $T$-invariant subspace. Then the restriction map $T|_{W}$ is also diagonalizable.

A trivial $T$-invariant subspace refers to the zero vector set $\left\{ \mathbf{0} \right\}$, the entire set $V$, the range $R(T)$, the null space $N(T)$, and the eigenspace $E_{\lambda}$.

Proof

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

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

Assume $W_{\lambda} = E_{\lambda} \cap W$, and since $W$ is $T$-invariant, it becomes the eigenspace 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 $\beta = \bigcup\limits_{\lambda} \beta_{\lambda}$ forms a basis for $W$. Then, since $\beta$ is a set of eigenvectors, proving that $T|_{W}$ is diagonalizable is equivalent.

• $\beta$ is linearly independent.

  The union of linearly independent sets from different eigenspaces is also linearly independent, so $\beta$ is linearly independent on $W$.

• $\beta$ generates $W$.

  Since $T$ is diagonalizable, every vector in $V$ can be expressed as a linear combination of the (linearly independent) eigenvectors of $T$. Since $W$ is a subspace of $V$, the same holds for every vector in $W$.

  Lemma

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

  By the lemma and the definition of $\beta$, every element of $W$ can be expressed as a linear combination of $\beta$.

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See Also

• Alternative Proof