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?
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Question: What is $LDt(L)$ form? Is this another notation for $LDL^{T}$? – user14717 Jun 28 '14 at 14:56
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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 $\mathbf N(A) = \ker(A)$).
Ted Shifrin
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