2
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Is there a faster way to check whether $A\in\mathbb{R}^{n\times n}$ is positive definite on $b^{\bot}:=\{x\in \mathbb{R}^{n}: x\cdot b=0\}$ than

function foo(A,b)
    N = null(b);
    e = eig(N'*A*N);
    return all(e>0);
end
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5
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Instead of e = eig(N'*A*N);, you can use [R,p]=chol(N'*A*N);, and test for p==0. Matlab function chol returns the cholesky factor R and a p which is zero if N'*A*N is positive definite. p is a positive integer if the matrix is not positive definite.

Usually chol is much faster than eig. For example, with dimension n=2000, I found that chol(N'*A*N) is about 10 times faster than eig(N'*A*N) (tested with Octave instead of Matlab).

The matrix product N'*A*N is relatively expensive to compute. It might be faster to use a sparse basis N for $b^{\bot}$: Matlab/Octave code:

function r = foo2(A,b)
    [dummy,p] = sort(abs(b),"descend");
    A = A(p,p);
    m = size(A,1);
    b = b(p);
    N = [b(2:m)/-b(1);speye(m-1,m-1)];   % now b*N = 0
    [L,p] = chol(N'*A*N);
    r = (p==0);
end
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  • $\begingroup$ That construction of N is brilliant. The sparsity is great of course, but I wasn't even aware about that explicit construction. $\endgroup$ – Bananach Dec 13 '18 at 9:31
  • $\begingroup$ (Now that I think of it, it's kind of a super simple version of Gram-Schmidt starting from $(b,e_2,\dots,e_m)$ without the need for an orthogonality nor normality. Still, awesome) $\endgroup$ – Bananach Dec 13 '18 at 9:34
  • $\begingroup$ Note that N = null(b); produces an orthogonal N (in contrast to the one in foo2), which should be preferred if your original matrix A is ill conditioned. I haven't looked into the details, but I guess that the matrix N = [b(2:m)/-b(1);speye(m-1,m-1)]; has a condition number $\kappa(N) \le \sqrt{m}$. $\endgroup$ – wim Dec 13 '18 at 13:49
  • $\begingroup$ For $b \neq 0$ you can easily check the condition number of the sparse N with cond_N = norm(N*b(2:m)'/norm(b(2:m))), and decide to use N = null(b); if cond_N is too large. $\endgroup$ – wim Dec 13 '18 at 14:21
  • $\begingroup$ (Note that if the division in cond_N = norm(N*b(2:m)'/norm(b(2:m))) fails due to norm(b(2:m)) == 0, then cond_N = 1.) $\endgroup$ – wim Dec 13 '18 at 14:47

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