JonaGik
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 Jun 13 awarded Notable Question Apr 2 awarded Notable Question Jun 27 awarded Notable Question May 3 awarded Popular Question Jan 28 awarded Popular Question Sep 26 awarded Popular Question Aug 16 awarded Popular Question May 17 comment Positive-Definiteness of a Quadratic Form Matrix The stement to be proved is: "A matrix is positive-definite if, and only if, it has, and only has, positive eigenvalues." Is that what you're looking for? Thanks again for your help. May 8 comment Positive-Definiteness of a Quadratic Form Matrix The idea that we can choose a vector X to draw inference about the eigenvectors seems fallacious. Presumably we need to draw inferences about the eigenvectors for all X which isn't necessarily the same as for some particular X. May 8 comment Positive-Definiteness of a Quadratic Form Matrix Unless I'm missing something, this only proves that this is inconsistent when x is the negative eigenvalue's eigenvector - it isn't a proof for the general case. May 1 comment Positive-Definiteness of a Quadratic Form Matrix @WillOrrick no problem and fixed in the second gray box too :) May 1 revised Positive-Definiteness of a Quadratic Form Matrix added 11 characters in body Apr 30 comment Positive-Definiteness of a Quadratic Form Matrix My understanding is that we want to classify A (as positive-definite, negative-definite, etc) on the basis of its eigenvalues. In that case, we don't know that $\lambda_1X_1^2+\ldots+\lambda_nX_n^2$ is positive. We wish to determine this based on its eigenvalues. What you have given is a case where all the eigenvalues are positive and the matrix is positive definite, but not a general, proven rule for this. Apr 30 comment Positive-Definiteness of a Quadratic Form Matrix Thanks for your response. I don't understand how the latter is an "only if" proof - it seems to only prove that the expression is positive in the particular (not general) case where X is any vector with a single element equal to one and the others equal to zero. Consider the case where $X = [1, 1, 0]$ (which appears to be legal). Now we have $\lambda_1 + \lambda_2 > 0$ which isn't constrained to all $\lambda_i > 0$. Presumably this is a valid counter-example? Apr 30 comment Positive-Definiteness of a Quadratic Form Matrix Are you saying that $e_k$ is some arbitrary eigenvector of $D = C^TAC$? Because then I can see this working. However, overall, the proof seems to still only prove that if (but not iff) all the eigenvectors are positive, the matrix is positive-definite. Is this right? Apr 30 accepted Frequency Response of Circuits - Laplace Transforms Apr 30 revised Positive-Definiteness of a Quadratic Form Matrix added 12 characters in body Apr 30 accepted Ellipse in Quadratic Form: Finding Intercepts with Principal Axes Apr 30 comment Ellipse in Quadratic Form: Finding Intercepts with Principal Axes Thank you very much! Apr 30 awarded Editor