MATLAB: Is ‘eig’ slower for tridiagonalized matrices

eig; hess

Consider the list of commands below:

>> A = gallery('poisson', 60);

>> tic; [V,D] = eig(full(A)); toc

Elapsed time is 16.074685 seconds.

>> B = hess(full(A)); tic; [V,D] = eig(B); toc

Elapsed time is 28.709543 seconds.

Why does 'eig' take longer on the tridiagonalized version of A ?

This behaviour occurs with other matrices. I'm running MATLAB R2011b.

Best Answer

Hello Douglas,

The brief answer is that it isn't anymore. In R2013a, there was an improvement in the handling of tridiagonalized matrices in "eig". Testing on my computer in R2011b and R2015b, I get:

R2011b:
eig(full(A)) - ~9 seconds
eig(B) - ~14.5 seconds
R2015b:
eig(full(A)) - ~3.7 seconds
eig(B) - ~0.6 seconds

I would recommend upgrading to the latest version of MATLAB, if you are able to.

-Cam

Related Solutions

MATLAB: Diagonalising a Skew-Symmetric Matrix

When EIG is called with an exactly symmetric/hermitian matrix, MATLAB falls back to a specialized algorithm that guarantees that U is orthogonal/unitary and that D is real. There is no such special algorithm choice for skew-symmetric matrices, so there is no guarantee here, even though if the problem is nicely conditioned, the result will be close to that:

>> rng default; A = randn(10); A = A - A';
>> [U, D] = eig(A);
>> max(abs(real(diag(D))))
ans =
   2.1034e-16
>> norm(U'*U - eye(10))
ans =
   4.7239e-15

However, if matrix B is (exactly) skew-symmetric, it implies that matrix A = 1i*B is hermitian, and passing this matrix to EIG will result in unitary eigenvectors and all-real eigenvalues, which you can then transform back:

[U, D] = eig(1i*A);
D = -1i*D;
>> max(abs(real(diag(D))))
ans =
     0
>> norm(U'*U - eye(10))
ans =
   1.5001e-15

MATLAB: How can I solve the following equation ([K]-t[G])w=0

help eig

    E = eig(A,B) is a vector containing the generalized eigenvalues
    of square matrices A and B.
    [V,D] = eig(A,B) produces a diagonal matrix D of generalized
    eigenvalues and a full matrix V whose columns are the
    corresponding eigenvectors so that A*V = B*V*D.

Hope this helps.

Thank you for formally accepting m answer.

Greg

Best Answer

Related Solutions

MATLAB: Diagonalising a Skew-Symmetric Matrix

MATLAB: How can I solve the following equation ([K]-t[G])w=0

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