A positive definite matrix $A$ can be factored in to $LDt(L)$form. Is the statement the eigenvalues of $A$ are the diagonals of $D$ true? If so , how to prove it?
2026-03-30 02:03:33.1774836213
LD$t(L)$ factorization and eigenvalues
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No, far from true. The product of the diagonals is the determinant of $A$, hence is equal to the product of the eigenvalues. What is true (Sylvester's law of inertia) is that — for any symmetric matrix $A$ — the signs of the entries of $D$ match up with the signs of the set of eigenvalues, including the number of zeroes (which is, of course, the dimension of $\mathbf N(A) = \ker(A)$).