167 reputation
8
bio website
location
age
visits member for 2 years, 2 months
seen Jan 29 at 15:55

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
Apr
30
revised Ellipse in Quadratic Form: Finding Intercepts with Principal Axes
edited body
Apr
30
comment Ellipse in Quadratic Form: Finding Intercepts with Principal Axes
@BrettFrankel sorry - that was a typo. Thanks for letting me know!
Apr
30
comment Positive-Definiteness of a Quadratic Form Matrix
As far as I can tell, your proof assumes that A is a diagonal matrix so that the eigenvectors are [1, 0, ..., 0], [0, 1, 0, ..., 0], ..., [0, ..., 0, 1] and thus $x^TAx = \sum_{i=1}^n \lambda_i X_i = \lambda_k X_k = \lambda_k$ is true. It is possible for more than one $\lambda_iX_i^2$ combination to be non-zero in the general case. Is this right? If so, this doesn't produce a general proof. Also, does this proof prove only that if all eigenvectors are positive, the matrix is positive-definite, not if and only if all eigenvectors are positive the matrix is positive-definite?
Apr
30
comment Positive-Definiteness of a Quadratic Form Matrix
Thanks very much for your response. This seems like its definitely on the right track. For $\sum_{i-1}^n X_i^2 = \underline \lambda \sum_{i-1}^n x_i^2$, I'm having trouble proving that to myself. Do you think you could prove it? Thanks again!