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I need to find a counter example that satisfies the following:
1- the matrix $A_{n*n}, n>2$ symmetric and positive semidefinite, and the main diagonal is positive $a_{ii}>0$
2- the matrix $A$ is singular
3- let $G$ be the generalized inverse of the matrix $A$. The $g_{11}=0$

is there a counter example of a matrix that satisfies all these conditions together?

Hanan
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  • Comments are not for extended discussion; this conversation has been moved to chat. – Xander Henderson Mar 11 '22 at 16:33
  • @BenGrossmann Thanks for all the replies. It seems that it will be only valid when $h=0$ because otherwise, the big matrix $A$ won't be singular, right? if true then can I say that $g_11^{+}\neq 0$ if at least $a_{ii}\neq 0$ – Hanan Mar 11 '22 at 16:53

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If $A$ is a positive semidefinite matrix and $G$ is its pseudoinverse, then $g_{11} = 0 \iff a_{11} = 0$.

To see that this is the case, note the following.

  • As is proved here, $x^TAx = 0 \iff Ax = 0$.
  • For a symmetric matrix, $Ax = 0 \iff A^+x = 0$, where $A^+$ denotes the MP-pseudoinverse of $A$

Let $e_1$ denote the column vector $e_1 = (1,0,\dots,0)$. We have $$ a_{11} = 0 \iff e_1^TAe_1 = 0 \iff Ae_1 = 0 \\ \iff A^+e_1 = 0 \iff e_1^TGe_1 = 0 \iff g_{11} = 0. $$

Ben Grossmann
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  • Thanks for the answer. Regarding the continuity point, I'm asking if the generalized inverse MP for this matrix is necessary continuous ($\hat{G}^+ \rightarrow {G}^+$) or not? can the condition that the main diagonal elements are all positive guarantee the continuity of the MP inverse matrix – Hanan Mar 09 '22 at 20:31
  • The pseudoinverse is discontinuous whenever there is a change in matrix rank. It is continuous over sets of matrices that have constant rank. – Ben Grossmann Mar 09 '22 at 21:55
  • do you mean a change in the rank for the sequence of A?.. my problem is that I have a covariance matrix that has the same structure as mentioned in the original problem (additionally $0<a_{ii}<1$), and I use the generalized inverse to get MLE for a certain parameter. To prove the asymptotic results of the estimator I need to guarantee g_{11}≠0 (this one is true now) and $g^{+}{ij} \xrightarrow{P} g^{+}{ij}$ does the continuity follows based on all these input or I have to assume seq of matrices (A or G?) have a constant rank? or something additional required to be proved – Hanan Mar 10 '22 at 12:28
  • A sufficient condition would be that the entire sequence and its limit has the same rank. That said, I suspect that if the generalized inverses converge, then they must converge to the generalized inverse of the limit – Ben Grossmann Mar 10 '22 at 13:48
  • Can you elaborate more the if part you have mentioned?.. Based on jstor.org/stable/2099241 ,, I see the sufficient condition about that rank(A+En)=rank(A) which is the same as you mentioned implies that $G^{+}_{n}→G^{+}$.. and then sciencedirect.com/science/article/pii/0024379577900799 shows that $\hat{G}^{+}→G^{+}$ for the MP inverse(see theorem 1) that for n x m matrices if it applies to n x n matrices, would that guarantee that the generalized inverses converge to the generalized inverse of the limit? – Hanan Mar 10 '22 at 18:58
  • I don’t understand what you said after “MP inverse” – Ben Grossmann Mar 10 '22 at 19:35
  • I'm asking if what you wrote " the generalized inverses converge, then they must converge to the generalized inverse of the limit" means $\hat{G}^+\rightarrow G^+$? if so the paper https://www.sciencedirect.com/science/article/pii/0024379577900799 proved in theorem 1 that it holds for matrices of size $n*m$.. if this is not what you mean can you elaborate more or give example why in case the entire sequence and its limit has the same rank, $\hat{g}{ij}^{+}\rightarrow g{ij}^{+}$ will not hold – Hanan Mar 10 '22 at 21:36
  • Thanks for clarifying. That's close to what I meant. What I really mean is that if $\hat G^+ \to M$, then $M = G^+$. – Ben Grossmann Mar 11 '22 at 00:03
  • I think $\hat{G}^{+}\rightarrow M,$ then $M=G^+$ is granted if \hat{G}^{+} satisfices the assumptions as in corollary 1 of this paper https://www.sciencedirect.com/science/article/pii/0024379577900799, which should consequently imply that $\hat{g}+{ij}\rightarrow g^+{ij}$, right? ...2. another case if $a_{ij}$ is uniformly consistent and continuous ($\hat{a}{ij}\rightarrow a{ij}$, would it imply that $\hat{g}+{ij}\rightarrow g^+{ij}$ ? (bec I know that it is valid if we have the standard inverse, $\hat{a}{ij}\rightarrow a{ij}$ then $\hat{a}^{-1}{ij}\rightarrow a^{-1}{ij}$)
    • – Hanan Mar 11 '22 at 12:14
  • It is true that ${\hat G}^+ \to \hat G$ implies that ${\hat g}^+{ij} \to \hat g{ij}$. I'm not sure what you mean when you say that $a_{ij}$ is "uniformly consistent and continuous". It is not generally true that $\hat a_{ij} \to a_{ij}$ implies that $\hat g_{ij} \to g^+_{ij}$; this is exactly what I mean when I say that the pseudoinverse is discontinuous. – Ben Grossmann Mar 11 '22 at 14:36
  • yes by "uniformly consistent and continuous" I mean $\hat{a}{ij}\rightarrow a{ij}$.. but do you mean $\hat{g}{ij}^{+}\rightarrow {g}{ij}^{+}$ or it wasn't a typo? – Hanan Mar 11 '22 at 14:42