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From what I know the Lanczos algorithm takes a symmetric matrix $A$ and a vector $v$ of norm $1$ as inputs, then generates tridiagonal matrix $J$ and orthonormal matrix $V$ such that $J = V^TAV$. I read the paper A Survey of Matrix Inverse Eigenvalue Problems By Boley and Golub (1986) that makes me confused.

In this paper having constructed a bordered diagonal matrix $B$ (a matrix whose graph is a star graph) using values $\lambda=\{\lambda_1, \lambda_2, \cdots, \lambda_n\}$ and $\mu=\{\mu_1, \mu_2, \cdots, \mu_{n-1}\}$ eigen data such that $\sigma(B)=\lambda$ and $\sigma(B_1)=\mu$ where $B_1$ is the matrix $B$ after removing its first row and column and $$ B = \begin{bmatrix} a1&b^T\\b&M \end{bmatrix} =\begin{bmatrix} a1&b_1&b_2& \dots&b_{n-1}\\ b_1&\mu_1&0&\cdots&0\\ b_2&0&\mu_2&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\\ b_{n-1}&0&0&\cdots&\mu_{n-1} \end{bmatrix} $$ it constructs tridiagonal matrix $J$ such that $\sigma(J)=\lambda$ and $\sigma(J_1) = \mu$ in the following way:

Using the orthogonal tranformation (Householder) we have

$$ \begin{bmatrix} 1&0\\0&Q_1^T \end{bmatrix} \begin{bmatrix} a_1&b_1e_1^T\\b_1e_1&J_1 \end{bmatrix} \begin{bmatrix} 1&0\\0&Q_1 \end{bmatrix} = \begin{bmatrix} a_1&b^T\\b&M \end{bmatrix} $$

It is clear the above transformation keeps the eigenvalues of $B$ and $B_1$. On carrying out the multiplication, we find

$$ Q_1^TJ_1Q_1 = M, Q_1b_1e_1 = b $$ The first equation shows that the eigenvectors of $J_1$ are the columns of $Q_1$ and the second equation shows that, apart from the factor $b_1$, the vector $b$ is the vector of first components of the eigenvectors of $J_1$: $$ b = b_1\{q_{11}, q_{12}, \cdots, q_{1, n-1}\} $$

This is what I can't understand

Thus, apart from the factor $b_1$, $b$ is the vector $v$ needed for the construction of $J_1$ from the Lanczos algorithm.

It is clear the method first builds a bordered diagonal matrix in some way, then uses the $b$ as the starting vector. If the Lanczos algorithm works by any starting vector with norm $1$, then why it tries to build this using the $b$ vector?

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The starting vectors in most of the eigenvalue algorithms are initialized with a random vector with entries $X \sim N(0,1)$ If you look in Trefethen. You should note that they randomly initialize most of the algorithms in numerical linear algebra.

enter image description here

It says "arbitrary" but what it means it is random. Some code follows after an explanation. The Lanzcos algorithm is in a suite of algorithms called the Krylov methods. The Krylov methods are a specific type of iterative eigenvalue algorithm as opposed to direct methods. The Krylov methods are listed below.

enter image description here

To explain how Lanczos works I am going to explain how Arnoldi works. The basis of both of these eigenvalue methods is first to use a similarity transform reducing to Hessenberg form. That is the following

$$A = QHQ^{*} $$

or $$AQ = QH $$

Instead, we compute $Q_{n} $ letting it be a $m \times n$ matrix whose first $n$ columns are of $Q$

Additionally, we haveenter image description here

In the book it calls it Hessenberg matrix however my professor said it is "hessenbergish". We end up with

$$ A Q_{n} = Q_{n+1} \bar{H_{n+1}} $$

The idea is simple that we are actually building Krylov subspaces

$$ \mathbf{K_{n}} = \langle b, Ab, \cdots, A^{n-1}b \rangle = \langle q_{1} ,q_{2} , \cdots q_{n} \rangle $$

We form these throught the QR decomp

$$ K_{n} = Q_{n}R_{n}$$

Now the we had $H_{n+1}$ to remove this row we note the following. $ Q_{n}^{*}Q_{n+1}$ is the $n \times (n+1)$ identity then $Q_{n}^{*}Q_{n+1}\bar{H_{n}}$ is $ n \times n$ Hessenberg matrix obtained by removing the last row.

$$ H_{n} = Q_{n}^{*}AQ_{n} $$

Now if you see we are iteratively building our projections through these Krylov subspaces.

The Lanczos iteration is simply when the matrix is symmetric.

function [T varargout] = myLanczos(mat,varargin)
%--------------------------------------------------------------------------
% Syntax:       T = myLanczos(mat);
%               T = myLanczos(mat,k);
%               T = myLanczos(mat,v);
%               T = myLanczos(mat,k,v);
%               [T Q] = myLanczos(mat);
%               [T Q] = myLanczos(mat,k);
%               [T Q] = myLanczos(mat,v);
%               [T Q] = myLanczos(mat,k,v);
%               
% Inputs:       mat is a symmetric or Hermitian N x N matrix
%
%               k is the number of Lanczos iterations to perform. The
%               default value is N
%
%               v is an N x 1 vector (internally forced to have unit norm)
%               specifying a particular starting vector to use. The default
%               vector is chosen uniformly at random from the unit sphere
%
% Outputs:      T is a k x k symmetric tridiagonal matrix. When k < N, T is
%               approximately similar to mat up to rank k. When k = N, T is
%               similar (in the linear algebra sense) to mat. That is, when
%               k = N, T has identical eigenvalues to mat
%
%               Q is the N x k similarity transformation such that
%               mat = Q * T * Q'. Note that equality holds only when k = N.
%               For k < N, mat ~ Q * T * Q'
%
% Description:  This function uses the Lanczos algorithm with full
%               reorthogonalization to compute k x k symmetric tridiagonal
%               matrix T that approximates mat up to rank k with respect to
%               transformation Q. That is, mat = Q * T * Q'. Note that
%               equality holds only for k = N. For k < N, mat ~ Q * T * Q'
%
%               NOTE: In particular, when k = N, T has identical
%               eigenvalues to mat
%
% Author:       Brian Moore
%               brimoor@umich.edu
%
% Date:         July 5, 2013
%--------------------------------------------------------------------------

% Knobs
symTol = 1e-8;

% Check input matrix size
[m,n] = size(mat);
if (m ~= n)
    error('Input matrix must be square');
end

% Make sure input matrix is symmetric
if max(max(abs(mat - mat'))) > symTol
    error('Input matrix is not symmetric to working precision');
end

% Parse user inputs
if (nargin == 2)
    if (length(varargin{1}) == n)
        k = n;
        v = varargin{1};
    else
        k = varargin{1};
        v = randn(n,1);
    end
elseif (nargin == 3)
        k = varargin{1};
        v = varargin{2};
else
    k = n;
    v = randn(n,1);
end

% Initialize variables
Q = nan(n,k);
q = v / norm(v);
Q(:,1) = q;
d = nan(k,1);
od = nan(k-1,1);

% Perform Lanczos iterations
for i = 1:k
    z = mat * q;
    d(i) = q' * z;

    %----------------------------------------------
    % Uncomment only one of the following 3 options
    %----------------------------------------------
    % Option 1: Full re-orthogonalization
    %z = z - Q(:,1:i) * (Q(:,1:i)' * z);

    % Option 2: Full re-orthogonalization (x2)
    z = z - Q(:,1:i) * (Q(:,1:i)' * z);
    z = z - Q(:,1:i) * (Q(:,1:i)' * z);

    % Option 3: No re-orthogonalization
    %z = z - d(i) * q;
    %if (i > 1)
    %    z = z - od(i-1) * Q(:,i-1);
    %end
    %----------------------------------------------

    if (i ~= k)
        od(i) = norm(z);
        q = z / od(i);
        Q(:,i + 1) = q;
    end
end

% Construct T
T = diag(d) + diag(od,-1) + diag(od,1);

% Return user-requested information
if (nargout == 2)
    varargout{1} = Q;
end

if you note if v is not specified it is these lines

else
        k = n;
        v = randn(n,1);
    end

    % Initialize variables
    Q = nan(n,k);
    q = v / norm(v);

It is constructing an orthogonal eigenvector. $q_{1}$ and each one after that from it.

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  • $\begingroup$ Thank you for your reply. Then why the paper didn't do that? it made lot's of computations to get the initial vector! $\endgroup$ – M a m a D Jul 27 '18 at 5:39
  • $\begingroup$ One moment I'll amend this. $\endgroup$ – user3417 Jul 27 '18 at 14:40
  • $\begingroup$ @Drupalist some changes made $\endgroup$ – user3417 Jul 27 '18 at 20:46
  • $\begingroup$ Thanks you for your explanation. So from what I understand it did all the computations to get the orthogonal matrix $Q_1$, right? $\endgroup$ – M a m a D Jul 28 '18 at 6:49
  • $\begingroup$ Well, the initial vector $x_{1}$ is just some random unit normal vector. it isn't orthogonal. The preceding steps are attempting to create orthogonal vectors by subtracting off projections using the inner product. $\endgroup$ – user3417 Jul 28 '18 at 12:29

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