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Consider the following matrix:

$ A=\begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&0 \end{pmatrix} $

The eigenvalues are $\lambda_1=0$, $\lambda_2=1-\sqrt{3}≤0$ and $\lambda_3=1+\sqrt{3}≥0$.

But when computing the principal minors we obtain:

$A_1=1≥0$

$A_2=\begin{vmatrix} 1&1\\ 1&1 \end{vmatrix} = 0$

$A_2= \begin{vmatrix} 1&1&1\\ 1&1&1\\ 1&1&0 \end{vmatrix} = 0$

So all leading principal minors are ≥ 0, but we have two eigenvalues with different sign and the third one is zero... Im really confused, thank you!

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    You should prove to yourself that if a symmetric PSD matrix has zero on the $k$th diagonal entry then all elements in row/column k must be zero -- this is violated here. Checking leading principal minors is not enough for symmetric PSD -- you need to check all principal minors. I gave a proof here: https://math.stackexchange.com/questions/4145638/a-is-positive-semidefinite-iff-textdet-b-k-geq-0/ – user8675309 Sep 06 '22 at 21:37
  • @user8675309 I've also found this other post – user Sep 06 '22 at 21:49
  • The matrix is symmetric, and it has eigenvalues of different signs. So, it cannot be positive definite, or positive semi-definite. – Disintegrating By Parts Sep 06 '22 at 21:59

1 Answers1

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Note that for the minor at the bottom right

$$\begin{vmatrix} 1&1\\ 1&0 \end{vmatrix} = -1$$

therefore by Sylvester criterion the signature is $(1,1,1)$ according to the sign of eigenvalues you have determined and the matrix is then indefinite.

Refer also to the related

user
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