Schur's Theorem

2023. 11. 18. 16:07·Mathematics/Linear Algebra

Schur's Theorem

Theorem 1. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional inner product space. Then there exists an orthonormal basis $\beta$ for $V$ such that $[T]_{\beta}$ is upper triangular. 
Proof. Let $n = \dim(V)$. The proof is by the mathematical induction on $n$. If $n = 1$, the result is immediate. So suppose that the theorem is true for $n-1$ where $n-1 \geq 1$. 
Let $W$ be a $T$-invariant supspace such that $\dim(W)= n -1$, and let $\beta_0 = \{v_1, ..., v_{n-1}\}$ be an orthonormal basis for $W$ such that $[T_W]_{\beta_0}$ is upper triangular. 
Let $u \in V$ but $u \notin W$. Since $u \notin \beta_0$, $\beta_0 \cup \{u\}$ is linearly independent. Set $u' = u - \sum_{i=1}^{n-1} \langle u, u_i \rangle v_i$ and define $v_n = \frac{u'}{||u'||}$. Then $\beta = \{v_1, ..., v_{n-1}, v_n\}$ is orthonormal basis for $V$ by Corollary 3 - 2 (b). Thus clearly $[T]_{\beta}$ is upper triangular. $\blacksquare$

Gram-Schmidt Process를 통해 모든 벡터 공간은 orthonormal basis를 가진다는 것을 보장할 수 있었다. Schur의 정리는 나아가 선형 연산자 $T$의 행렬 표현이 항상 upper triangular가 되는 orthonormal basis의 존재성을 보장한다. 

Schur's Theorem (Matrix Version)

Corollary. Let $A \in M_{n \times n}(\mathbb{C}) $ [$M_{n \times n}(\mathbb{R})$] be a matrix whose characteristic polynomial splits over $\mathbb{C}$ [$\mathbb{R}$]. Then $A$ is unitarily equivalent [orthogonally equivalent] to a complex [real] upper triangular matrix. 
Proof. By Schur's Theorem, there exists an orthonormal basis $\beta$ for $\mathbb{C}^n$ [$\mathbb{R}^n$] such that $[L_A]_{\beta}$ is upper triangular. Then by Theorem 2, $B = [L_A]_{\beta} = Q^{-1}AQ$ where $Q = [I]_{\beta}^{\gamma}$ for the standard ordered basis $\gamma$ for $\mathbb{C}^n$ [$\mathbb{R}^n$]. Note that $Q$ is the matrix whose columns are the vectors in $\beta$. Since $\beta$ is orthonormal, by remark, $Q$ is unitary [orthogonal]. Thus $A$ is unitarily equivalent [orthogonally equivalent] to $B$. $\blacksquare$
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