标签：linear overline beta eta Let Triangularization alpha Ordinary

目录

• Ordinary Triangularization
  • Trick : "pass to the quotient"
  • Statement and proof

Ordinary Triangularization

Many results above are proved over \(\mathbb{C}\) using Schur triangularization (see this blog : [https://www.cnblogs.com/chaliceseven/p/17094280.html]). Their generalizations are made possible by ordinary triangularization.

Trick : "pass to the quotient"

Lemma. Let \(F\) be any field. Let \(V\) be a vector space over \(F\), and \(T\) a linear operator on \(V\).

Suppose that \(W\) is a nonzero \(T\)-invariant subspace of \(V\). Then it is easy to check that

\[\overline{T}:V/W\to V/W\quad\overline{T}(v+W):=T(v)+W \]

is a well-defined linear operator, and it is the unique linear operator such that \(\eta T=\overline{T}\eta\), where \(\eta:V\to V/W\) is the linear transformation defined by \(\eta(v):=v+W\).

To go further, assume that \(V\) is finite-dimensional and nonzero. Let \(\gamma=\{v_1,\cdots,v_k\}\) be an ordered basis for \(W\) and extend \(\gamma\) to an ordered basis \(\beta=\{v_1,\cdots,v_k,v_{k+1},\cdots,v_n\}\) for \(V\). Then \(\alpha=\{v_{k+1}+W,\cdots,v_n+W\}\) is an ordered basis for \(V/W\). Using the equation \([\eta]_{\beta}^{\alpha}[T]_{\beta}=[\overline{T}]_{\alpha}[\eta]_{\beta}^{\alpha}\), it is easy to show that

\[[T]_{\beta}= \begin{pmatrix} [T_W]_{\gamma}&*\\ O&[\overline{T}]_{\alpha} \end{pmatrix}.\]

Based on this fact, we have:
(1) The characteristic polynomials satisfy

\[p_T(\lambda)=p_{T_W}(\lambda)p_{\overline{T}}(\lambda). \]

(2) If \(T\) is diagonalizable, then \([T]_{\beta}\) is normal and so we must have

\[[T]_{\beta}= \begin{pmatrix} [T_W]_{\gamma}&O\\ O&[\overline{T}]_{\alpha}     \end{pmatrix}. \]

(See the remark at the end of Generalized Schur's Theorem: Proof.) Hence both \(T_W\) and \(\overline{T}\) are diagonalizable.
(3) If both \(T_W\) and \(\overline{T}\) is diagonalizable and \(\text{gcd}(p_{T_W},p_T)=1\) over \(F\), then by removal rule \(T\) is diagonalizable as well. \(\blacksquare\)

Statement and proof

THEOREM. Let \(F\) be any field. Let \(V\) be a nonzero finite-dimensional vector space over \(F\), and \(T\) a linear operator on \(V\). If the characteristic polynomial of \(T\) splits, then there exists a basis \(\beta\) for \(V\) such that \([T]_{\beta}\) is an upper triangular matrix.

Proof. Induction on \(n:=\dim V\). The theorem is trivial when \(n=1\), so we may assume that \(n\ge 2\). Assume the theorem is true whenever the dimension the space is less then \(n\). Since the characteristic polynomial of \(T\) splits, \(T\) has an eigenvector \(z\). Define \(W:=\text{span}(\{z\})\), then \(W\) is \(T\)-invariant and \(\dim(V/W)=n-1\). By the lemma above, \(p_{\overline{T}}\) divides \(p_T\) and hence splits as well. Applying the induction hypothesis to \(\overline{T}:V/W\to V/W\) finishes the proof. \(\blacksquare\)

标签：linear,overline,beta,eta,Let,Triangularization,alpha,Ordinary
From： https://www.cnblogs.com/chaliceseven/p/17094288.html