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I have studied 4 tests to prove for positive definite and positive semi-definite matrices:

  1. Quadratic Form Test (Sum of Squares Test): Q(x) = (x^T)Ax > 0 and >=0 respectively
  2. Eigen Values Test: if all eigen values >0 and >=0 respectively
  3. Upper Left Submatrices Determinant Test: if all determinants are >0 and >=0 respectively
  4. Pivot Test: If all pivots of the Echelon Form of the Matrix are >0 and >=0 respectively

I have also learned how to apply tests 1 and 2 to find negative definite, negative semi-definite, and indefiniteness of matrices - <0, <=0, and both positive and negative values respecively.

I wish to find out if tests 3 and 4 are possible to apply to find the remaining classifications.

I assumed that finding all the pivots to be negative (<0) in the echelon form would cause it to be negative definite, and henceforth the Determinant test would have Di<0 for odd i, and Di>0 for even i.
And similarly having (<=0) for negative semi-definite.
But I have found only one reference so far to corroborate this statement, all of the others use the eigen value test. Link

Is my logic right, and is it possible to find the remaining classifcation using tests 3 and 4?

PraxVamp
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  • Criterion 3 for PSD matrices is false and I've never heard of 4. What is your source for this? It needs to be stated that you require your evidently real matrices to be symmetric otherwise e.g. 2 and 3 are false... definitions of PD and PSD are not consistent when working over $\mathbb R$ unfortunately. – user8675309 Apr 27 '23 at 15:22
  • @user8675309 I mean I did mention symmetric in the title question, maybe i should have clarified it in my explaination as well. And its been taught in my college courses, and the textbook they use mentions it as well, altho its possible that I misunderstood it, it actually mention for PSD that no principal submatrix has a negative determinant, I just assumed that meant Upper Left submatrix, since that's what I studied for the PD. – PraxVamp Apr 28 '23 at 02:51
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    re: symmetric fair enough. The reality is your (3) is wrong for PSD matrices on its own merits-- consider $\begin{bmatrix} 0 &0 \ 0& -1 \end{bmatrix}$. It also has a consistency problem as your proposed criterion for negative semi-definite would also capture all matrices with leading principal minors of zero $\implies$ a lot of matrices are both PSD and NSD which can only happen with the zero matrix (contradiction). I gave proof of the generalized Sylvester Criterion for PSD matrices here: https://math.stackexchange.com/questions/4145638/a-is-positive-semidefinite-iff-textdet-b-k-geq-0/ – user8675309 Apr 28 '23 at 04:48
  • @user8675309 Ah,I see. Thanks for correcting me, I'll look into it. – PraxVamp Apr 28 '23 at 09:58

1 Answers1

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Yes, you're right, and this follows simply from applying the original tests to the matrix $-A$.

Hans Lundmark
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  • Thanks for the answer. Just one last question - would a matrix just be indefinite if it is none of the above 4 classifications? – PraxVamp Apr 27 '23 at 05:34
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    Well, if it's neither positive (semi)definite nor negative (semi)definite, then the only remaining possibility is that it's indefinite. – Hans Lundmark Apr 27 '23 at 07:19