Problem with cholesky decomposition
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When I apply the chol function to A = [1 -1; 0 1], it correctly informs me that the matrix is not positive definite.
But when I run chol(A, 'lower'), the answer is the identity matrix [1 0; 0 1].
Can anyone replicate this? Any reasons why this should be so?
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David Goodmanson
2020-5-19
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Hi Peter,
when you use the 'lower' option, chol assumes that the upper triangle is the complex conjugate transpose of the lower triangle. In this case that means that chol assumes the matrix is [1 0; 0 1], the identity matrix. So of course the cholesky decomposition is also the identity matrix.
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Christine Tobler
2020-5-19
When chol(A) is called without the 'lower' or 'upper' option, this is treated as if the 'upper' option had been chosen: So in the first example, chol assumes the matrix is [1 -1; -1 1]. This is because the Cholesky decomposition only works for symmetric matrices.
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