3/8/2024 0 Comments Matrix minus vector matlabThere is a particular S there that I'll use as a test matrix. Any one of these tests is sufficient to make the matrix S positive definite. Can I show you them? I'm going to have five tests. We don't know that we need the number, we just want to know is it a positive number.Īnd there are several neat tests. So how can we tell that all the eigenvalues are positive? Well, we only want to know their sign. Now there are good ways to do it, but it's not for paper and pencil, and not for I. We didn't know how to do it a little while ago. For a large matrix, we take time on that. That's easy to see.īut how can we tell if its eigenvalues are positive? That's not fun because computing eigenvalues is a big job. How do we recognize a positive definite matrix? It has to be symmetric. More often we're in this good case where all the eigenvalues are above 0. Of course, that's not a case that we are really thinking about. If I took the 0 matrix, all 0's, as A, A transpose A would be the 0 matrix. So if I want to move from semidefinite to definite, then I must rule out A x equals 0 there. In that case, I would only learn lambda equals 0, and I'd be in this semidefinite case. If we were in a singular case, A x could be the 0 vector. So lambda is never negative.Ī x could be the 0 vector. Over here, y is x, so I'm getting the length of x squared.Īnd you see that number lambda is, in this equation, I have a number that can't be negative. Here y is A x, so I'm getting the length of A x squared. Any time I have y transpose y, I'm getting the length of y squared. Then I have x transpose times the left side, is x transpose times the right side. Take the inner product of both sides with x. What's special about A transpose A x equal lambda x? The good idea? Multiply both sides by x transpose. This is the simple step that is worth remembering. And it will be at least positive semidefinite. If I take any matrix A, could be rectangular. The matrix could be singular, but all the eigenvalues have to be greater or equal to 0.Īnd let me show you exactly where those matrices come from. Positive definite matrices- automatically symmetric, I'm only talking about symmetric matrices- and positive eigenvalues. Symmetric and positive definite, or positive semidefinite, which means the eigenvalues are not only real, they're real for symmetric matrices. So that is the best of the best matrices. And lambda, which was omega squared, is the eigenvalue.īut I didn't stop to point out that if we want lambda to be omega squared, we need to know lambda greater or equal to 0. We substituted, and we discovered x had to be an eigenvector of S, as usual. Second derivative, plus S times y, equals 0.Īnd you maybe remember how we solved it. Our application was the second-order equation with a symmetric matrix, S.
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