গৃহস্থের প্রতিচ্ছবি কেন একটি ম্যাট্রিক্সকে তির্যক করতে পারে না?

Question

অনুশীলনে কিউআর ফ্যাক্টেরাইজেশন গণনা করার সময়, একটি ম্যাট্রিক্সের নীচের অংশটি শূন্যের জন্য গৃহকর্তার প্রতিচ্ছবি ব্যবহার করে। আমি জানি যে প্রতিসম ম্যাট্রিক্সের ইগেনভ্যালুগুলি গণনা করার জন্য, গৃহস্থালীর প্রতিচ্ছবিগুলির সাথে আপনি সবচেয়ে ভাল যা করতে পারেন তা এটি ত্রিভুজাকৃতি আকারে পেয়ে যাচ্ছে। কেন এই উপায়টি সম্পূর্ণরূপে তির্যক করা যায় না তা দেখার কোনও সুস্পষ্ট উপায় আছে? আমি এটি সহজভাবে ব্যাখ্যা করার চেষ্টা করছি তবে আমি একটি পরিষ্কার উপস্থাপনা নিয়ে আসতে পারি না।

Answer 1

প্রতিসম ম্যাট্রিক্স eigenvalues কম্পিউটিং যখন সেরা আপনি গৃহকর্তাকে প্রতিফলক করতে পারেন ড্রাইভ এম একটি tridiagonal ফর্মে। পূর্বের উত্তরে যেমন উল্লেখ করা হয়েছিল যেহেতু এম প্রতিসাম্যযুক্ত সেখানে একটি অर्थোগোনাল মিলের রূপান্তর রয়েছে যার ফলস্বরূপ একটি তির্যক ম্যাট্রিক্স, অর্থাৎ, ডি = এস টি এম এস । এটি সুবিধামত হবে যদি আমরা অজানা অর্থকোনাল ম্যাট্রিক্স এস এর ক্রিয়াটি কঠোরভাবে গৃহস্থালীর প্রতিচ্ছবিগুলির সাথে প্রতিবিম্বগুলির ক্রম গণনা করে এবং বাম থেকে এম এবং এইচ- তে H টি প্রয়োগ করে ব্যবহার করতে পারি$M\in\mathbb{R}^{n\times n}$$M$$M$$D=S^TMS$$S$$H^T$$M$$H$ডান থেকে । তবে এটি সম্ভব নয় কারণ গৃহকর্তা প্রতিফলককে কলামগুলি শূন্য করার জন্য তৈরি করা হয়েছে। আমরা নীচে সংখ্যার সব আউট শুন্যতে গৃহকর্তাকে প্রতিফলক গনা হলে এম 11 আমরা খুঁজে এম = $M$$M_{11}$ তবে এখন এম 12 - এম 1 এন এন্ট্রিগুলিবামে প্রয়োগ করারিফ্লেক্টর এইচ টি 1 দ্বারা পরিবর্তন করা হয়েছে। সুতরাং আমরা যখনডানদিকে এইচ 1 প্রয়োগ করি তখনএটিএম এরপ্রথম সারিটিকেবল এম 11 রেখে যায় না। পরিবর্তে আমরা এইচ টি 1 এম=পাব

$$M=\left(\!\!{\begin{array}{ccccc} * &* & * & *&* \\ * &* & * & *&* \\ * &* & * & *&* \\ * &* & * & *&* \\ * &* & * & *&* \\ \end{array}}\!\!\right)\rightarrow H^T_1M=\left(\!\!{\begin{array}{ccccc} * &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ \end{array}}\!\!\right).$$ $M_{12}-M_{1n}$$H^T_1$$H_1$$M$$M_{11}$ যেখানে আমরা কেবল সারিটি শূন্য করিনি, তবে আমরা কেবল প্রতিফলক এইচ টি 1 এর সাথে প্রবর্তিত শূন্য কাঠামোটি ধ্বংস করতে পারি। $$H^T_1M=\left(\!\!{\begin{array}{ccccc} * &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ \end{array}}\!\!\right)\rightarrow H^T_1MH_1=\left(\!\!{\begin{array}{ccccc} * &*'' & *'' & *''&*'' \\ *' &*'' & *'' & *''&*'' \\ *' &*'' & *'' & *''&*'' \\ *' &*'' & *'' & *''&*'' \\ *' &*'' & *'' & *''&*'' \\ \end{array}}\!\!\right).$$ $H^T_1$

$M$$H^T_1$

$$M=\left(\!\!{\begin{array}{ccccc} * &* & * & *&* \\ * &* & * & *&* \\ * &* & * & *&* \\ * &* & * & *&* \\ * &* & * & *&* \\ \end{array}}\!\!\right)\rightarrow H^T_1M=\left(\!\!{\begin{array}{ccccc} * &* & * & *&* \\ *' &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ \end{array}}\!\!\right).$$ $$H^T_1M=\left(\!\!{\begin{array}{ccccc} * &* & * & *&* \\ *' &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ 0 &*' & *' & *'&*' \\ \end{array}}\!\!\right)\rightarrow H^T_1MH_1=\left(\!\!{\begin{array}{ccccc} * &*' & 0 & 0&0 \\ *' &*'' & *'' & *''&*'' \\ 0 &*'' & *'' & *''&*'' \\ 0 &*'' & *'' & *''&*'' \\ 0 &*'' & *'' & *''&*'' \\ \end{array}}\!\!\right).$$

Applied recursively this allows us to drive $M$ to a tridiagonal matrix $T$. You can complete the diagonalization of $M$ efficiently, as was mentioned previously, using Jacobi or Givens rotations both of which are found in the Golub and Van Loan book Matrix Computations. The accumulated actions of the sequence of Householder reflectors and Jacobi or Givens rotations allows us to find the action of the orthogonal matrices $S^T$ and $S$ without explicitly forming them.

Answer 2

As the Comments to other Answers clarify, the real issue here is not a shortcoming of Householder matrices but rather a question as to why iterative rather than direct ("closed-form") methods are used to diagonalize (real) symmetric matrices (via orthogonal similarity).

Indeed any orthogonal matrix can be expressed as a product of Householder matrices, so if we knew the diagonal form of a symmetric matrix (its eigenvalues), we could solve for a complete set of orthonormalized eigenvectors and represent the corresponding change of basis matrix as a product of Householder transformations in polynomial time.

So let's turn to Victor's parenthetical comment "other than Abel's theorem" because we are effectively asking why iterative methods should be used find the roots of a polynomial rather than a direct method. Of course the eigenvalues of a real symmetric matrix are the roots of its characteristic polynomial, and it is possible to go in the other direction as well. Given a real polynomial with only real roots, it is possible to construct a tridiagonal symmetric companion matrix from a Sturm sequence for the polynomial. See also that poster Denis Serre's Exercise 92 in this set. This is rather nice for showing the equivalence of those problems since we've seen (@AndrewWinters) the direct application of Householder matrices will tridiagonalize a real symmetric matrix.

Analysis of the arithmetic complexity for an iterative (root isolation) method is given in Reif (1999), An Efficient Algorithm for the Real Root and Symmetric Tridiagonal Eigenvalue Problems. Reif's approach improves on tailored versions of QR for companion matrices, giving $O(n \log^3 n)$ instead of $O(n^2)$ complexity.

The Abel-Galois-Ruffini Theorem says that no general formula for roots of polynomials above degree four can be given in terms of radicals (and usual arithmetic). However there are closed forms for roots in terms of more exotic operations. In principle one might base eigenvalue/diagonalization methods on such approaches, but one encounters some practical difficulties:

1. The Bring radical (aka ultraradical) is a function of one variable, in that respect like taking a square root. Jerrad (c. 1835) showed that solving the general quintic could be reduced to solving $t^5 + t - a = 0$, so that univariate function $t(a)$ (used in addition to radicals and other usual arithmetic) allows all quintics to be solved.

2. This breaks down with degree six polynomials and above, although various ways can be found to solve them using functions of just two variables. Hilbert's 13th Problem was the conjecture that general degree seven polynomials could not be solved using only functions of at most two variables, but in 1957 V.I. Arnold showed they could. Among the multivariable function families that can be used to get solutions to arbitrary degree polynomials are Mellin integrals, hypergeometric and Siegel theta functions.

3. Besides implementing somewhat exotic special functions of more than one argument, we need direct methods for solving polynomials which work for general degree $n$ rather than ad hoc or degree specific methods. Guàrdia (2002) gives "a very simple expression of the roots of a polynomial of arbitrary degree in terms of derivatives of hyperelliptic theta functions." However this approach requires making choices of Weierstrass points on hyperelliptic curve $C_f: Y^2 = f(x)$ where all roots of polynomial $f(x)$ are sought. A good choice leads to expressing less than half of those roots, and it appears this approach requires repeated trials to get all of them. Each trial involves solving a homogeneous linear system at $O(n^3)$ cost.

Therefore the indirect/iterative methods for isolating real roots (equiv. eigenvalues of symmetric matrices), even to high precision, currently have practical advantages over the known direct/exact methods for these problems.

Comments

Some notes: 1. a practical method for building the tridiagonal companion matrix from Sturm sequences was outlined in papers by Fiedler and Schmeisser; I gave a Mathematica implementation here, and it should not be too hard to implement in a more traditional language.

— জেএম

2. With respect to the "theta function" approach for polynomial roots (which I agree is a bit too unwieldy for practical use), Umemura outlines an approach using Riemann theta functions.

— J. M.

Answer 3

For what reason do you assume that this is impossible?

Any symmetric real matrix $S$ can be orthogonally diagonalized, i.e. $S = G D G^t$, where $G$ is orthogonal and $D$ is diagonal.

Any orthogonal matrix of size n×n can be constructed as a product of at most n such reflections.Wikipedia. Therefore you have this decomposition.

I am not sure about the last statement, I just cite it (and I think it is correct). As far as I understand your question, it boils down to whether any orthogonal matrix can be decomposed into a sequence of Householder transforms.

Comments

I should have been more specific. The first step to diagonalizing a symmetric matrix is applying Householder until it is tridiagonal. Next, QR iterations are performed. This process cannot be completed using only closed-form Householder transformations. Why? (other than Abel's theorem)

— Victor Liu

You can do it with Jacobi rotations. Golub and Van Loan write that Jacobi is the same as Givens. Householder is just another way of doing Givens. In practice, the "correct" way might be with QR if it is faster.

— power

Answer 4

If the eigenvalues are already known (from a preliminary calculation based on the usual approach), one can use them to triangulize a nonsymmetric matrix (or diagonalize a symmetric matrix) by a product on $n-1$ Householder reflections. In the $k$th step the $k$th column is brought to triangular form. (This also provides a simple inductive proof of the existence of the Schur factorization.)

It is actually useful for methods where one repeatedly needs the orthoginal matrix in a numerically stable factored form.