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node126.html
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<!DOCTYPE html>
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* revised and updated by: Marcus Hennecke, Ross Moore, Herb Swan
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<html>
<head>
<title>锁定</title>
<meta charset="utf-8">
<meta name="description" content="锁定">
<meta name="keywords" content="book, math, eigenvalue, eigenvector, linear algebra, sparse matrix">
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<br>
<!--End of Navigation Panel--><h4><a name="SECTION001357020000000000000">
锁定 <span class="math-inline">\theta </span></a>
</h4>
首先讨论的是锁定单个收敛的 Ritz 值。假设
<div class="math-display">Ty = y \theta, \; \Vert y\Vert = 1,</div>
其中 <span class="math-inline">e_k^* y = \eta</span>，且 <span class="math-inline">|\eta| \le \epsilon_D \Vert T \Vert</span>。这里，<span class="math-inline">\epsilon_M \le \epsilon_D \lt 1</span> 是一个指定的相对精度容差，介于 <span class="math-inline">\epsilon_M</span> 和 <span class="math-inline">1</span> 之间。
<p>
如果 <span class="math-inline">\theta </span> 是“需要的”，那么锁定 <span class="math-inline">\theta </span> 是可取的。然而，为了实现这一点，需要对当前的 Lanczos 分解进行变换，使其具有一个小次对角线以隔离 <span class="math-inline">\theta </span>。这可以通过构造一个 <span class="math-inline">k \times k</span> 的正交矩阵 <span class="math-inline">Q = Q(y)</span> 来实现，使用算法<a href="node124.html#alg-orthQ">4.9</a>：
<div class="math-display">Q e_1 = y \quad \mathrm{且} \quad e_k^* Q = ( \eta, \tau e_{k-1}^*),</div>
其中 <span class="math-inline">\eta^2 + \tau^2 = 1 </span>。
<p>
这些变换的最终结果是
<div class="math-display">
\begin{aligned}
Av_1 &= v_1\theta + r\eta, \quad \text{其中} \quad v_1^* r = 0, \\
A V_2 &= V_2 T_2 + r\tau e_{k-1}^*,
\end{aligned}
</div>
其中 <span class="math-inline">[v_1 , V_2] = V Q</span>。
<p>
这意味着随后的隐式重启发生在
<div class="math-display">A V_2 = V_2 T_2 + r\tau e_{k-1}^*</div>
所有与隐式重启相关的后续正交变换应用于 <span class="math-inline">T_2</span>，并且不会干扰关系 <span class="math-inline"> Av_1 = v_1 \theta + r \eta </span>。在随后的 Lanczos 步骤中，<span class="math-inline">v_1</span> 参与正交化，从而自动实现了 Parlett 和 Scott 推荐的择优正交化 [<a href="node421.html#pasc79">363</a>,<a href="node421.html#parl80">353</a>]。
<p>
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<address>
Susan Blackford
2000-11-20
</address>
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</html>