📂Linear Algebra

Simultaneous Diagonalization

Definition¹

Let $V$ be a finite-dimensional vector space, and let $\mathcal{A}$ be a collection of linear operators on $V$.

$$ \mathcal{A} = \left\{ A : V \to V \mid A \text{ is linear and }\right\} \subset \operatorname{End}(V) $$

If there exists a basis $\left\{ v_{1}, \dots, v_{n} \right\}$ of $V$ consisting only of simultaneous eigenvectors $v_{k}$ of $\mathcal{A}$, then $\mathcal{A}$ is said to be simultaneously diagonalizable.

Explanation

That $\mathcal{A}$ is simultaneously diagonalizable means that at least one simultaneous eigenvector exists, and hence the elements of $\mathcal{A}$ commute with each other. Theorem 1 below tells us the converse: if each of them is diagonalizable and they commute with each other, then they are simultaneously diagonalizable. Theorem 2 is an extension of the proposition that eigenvectors with distinct eigenvalues are linearly independent.

If $\mathcal{A}$ is a vector space, then $\mathcal{A}$ being simultaneously diagonalizable is equivalent to $V$ being decomposable into the direct sum of the weight spaces of $\mathcal{A}$. Just as, even for a single linear transformation, being diagonalizable is equivalent to being expressed as the direct sum of eigenspaces.

$$ \mathcal{A} \text{ is simultaneously diagonalizable. } \iff V = E_{\mu_{1}} \oplus \cdots \oplus E_{\mu_{k}} $$

Theorem

• Theorem 1

  Let $\mathcal{A}$ be a collection of linear operators on a finite-dimensional vector space $V$. If the elements of $\mathcal{A}$ commute with each other and each of them is diagonalizable, then $\mathcal{A}$ is simultaneously diagonalizable.

• Theorem 2

  Let $V$ be a vector space, and let $\mathcal{A} \le \operatorname{End}(V)$ be a vector space of linear operators on $V$. Let $\mu_{1}, \dots, \mu_{m}$ be distinct weights of $\mathcal{A}$, and let $v_{1}, \dots, v_{m}$ be the corresponding simultaneous eigenvectors. Then $\left\{ v_{1}, \dots, v_{m} \right\}$ is linearly independent.