How to extract a positive definite submatrix from a PSD matrix?












1












$begingroup$


Let $M$ be a real symmetric positive semi-definite matrix s.t. there is only one zero eigenvalue.



Question: Is it true that there is a unique principal submatrix that is positive definite? If so, how to determine this submatrix?



First trial: Let the eigenvector corresponding to the zero eigenvalue be $v$. We can represent $M$ as
begin{equation}M=Q'DQ,tag{*}end{equation}
where $Q'Q=I$ and $D$ is a diagonal matrix. Suppose for certainty that the last element of D is equal to $0$, i.e., $d_{nn}=0$. This means that the last column of $Q$ is equal to $v$.



Now let's write $D=begin{bmatrix}D_1& 0\0& 0end{bmatrix}$ and $Q=begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}$, where $D_1$ is a PD diagonal matrix, $tilde Q$ is an $[n-1,n-1]$ matrix, $q_1$ and $q_2$ are vectors, and $x$ is a scalar.



With this, we rewrite (*) as
$$M=begin{bmatrix}tilde Q'& q_2\q_1'& xend{bmatrix}begin{bmatrix}D_1& 0\0& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1& 0\q_1'D_1& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1tilde Q& tilde Q'D_1q_1\q_1'D_1tilde Q& q_1'D_1q_1end{bmatrix}$$



Here, $tilde Q'D_1tilde Q$ is the $(n,n)$-principal submatrix of $M$, which is positive definite if $tilde Q$ is non-singular. Thus the problem boils down to deternining non-singular principal submatrices of $Q$. However, I'm not sure is this makes the problem any simpler..



Solution (?): The number of PD principal submatrices id equal to the number of non-zero elements in the eigenvector $v$. If $v_ineq 0$, $M[i]$ is PD.



I will post the proof in a day or so.






matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can extract principal submatrices using "elimination matrices"
    $endgroup$
    – Bertrand
    Jan 11 at 14:54










  • $begingroup$
    @Bertrand, I know how to extract submatrices. The question is rather which one is to extract?
    $endgroup$
    – Dmitry
    Jan 11 at 15:02
















1












$begingroup$


Let $M$ be a real symmetric positive semi-definite matrix s.t. there is only one zero eigenvalue.



Question: Is it true that there is a unique principal submatrix that is positive definite? If so, how to determine this submatrix?



First trial: Let the eigenvector corresponding to the zero eigenvalue be $v$. We can represent $M$ as
begin{equation}M=Q'DQ,tag{*}end{equation}
where $Q'Q=I$ and $D$ is a diagonal matrix. Suppose for certainty that the last element of D is equal to $0$, i.e., $d_{nn}=0$. This means that the last column of $Q$ is equal to $v$.



Now let's write $D=begin{bmatrix}D_1& 0\0& 0end{bmatrix}$ and $Q=begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}$, where $D_1$ is a PD diagonal matrix, $tilde Q$ is an $[n-1,n-1]$ matrix, $q_1$ and $q_2$ are vectors, and $x$ is a scalar.



With this, we rewrite (*) as
$$M=begin{bmatrix}tilde Q'& q_2\q_1'& xend{bmatrix}begin{bmatrix}D_1& 0\0& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1& 0\q_1'D_1& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1tilde Q& tilde Q'D_1q_1\q_1'D_1tilde Q& q_1'D_1q_1end{bmatrix}$$



Here, $tilde Q'D_1tilde Q$ is the $(n,n)$-principal submatrix of $M$, which is positive definite if $tilde Q$ is non-singular. Thus the problem boils down to deternining non-singular principal submatrices of $Q$. However, I'm not sure is this makes the problem any simpler..



Solution (?): The number of PD principal submatrices id equal to the number of non-zero elements in the eigenvector $v$. If $v_ineq 0$, $M[i]$ is PD.



I will post the proof in a day or so.






matrices







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can extract principal submatrices using "elimination matrices"
    $endgroup$
    – Bertrand
    Jan 11 at 14:54










  • $begingroup$
    @Bertrand, I know how to extract submatrices. The question is rather which one is to extract?
    $endgroup$
    – Dmitry
    Jan 11 at 15:02














1












1








1





$begingroup$


Let $M$ be a real symmetric positive semi-definite matrix s.t. there is only one zero eigenvalue.



Question: Is it true that there is a unique principal submatrix that is positive definite? If so, how to determine this submatrix?



First trial: Let the eigenvector corresponding to the zero eigenvalue be $v$. We can represent $M$ as
begin{equation}M=Q'DQ,tag{*}end{equation}
where $Q'Q=I$ and $D$ is a diagonal matrix. Suppose for certainty that the last element of D is equal to $0$, i.e., $d_{nn}=0$. This means that the last column of $Q$ is equal to $v$.



Now let's write $D=begin{bmatrix}D_1& 0\0& 0end{bmatrix}$ and $Q=begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}$, where $D_1$ is a PD diagonal matrix, $tilde Q$ is an $[n-1,n-1]$ matrix, $q_1$ and $q_2$ are vectors, and $x$ is a scalar.



With this, we rewrite (*) as
$$M=begin{bmatrix}tilde Q'& q_2\q_1'& xend{bmatrix}begin{bmatrix}D_1& 0\0& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1& 0\q_1'D_1& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1tilde Q& tilde Q'D_1q_1\q_1'D_1tilde Q& q_1'D_1q_1end{bmatrix}$$



Here, $tilde Q'D_1tilde Q$ is the $(n,n)$-principal submatrix of $M$, which is positive definite if $tilde Q$ is non-singular. Thus the problem boils down to deternining non-singular principal submatrices of $Q$. However, I'm not sure is this makes the problem any simpler..



Solution (?): The number of PD principal submatrices id equal to the number of non-zero elements in the eigenvector $v$. If $v_ineq 0$, $M[i]$ is PD.



I will post the proof in a day or so.






matrices







share|cite|improve this question











$endgroup$




Let $M$ be a real symmetric positive semi-definite matrix s.t. there is only one zero eigenvalue.



Question: Is it true that there is a unique principal submatrix that is positive definite? If so, how to determine this submatrix?



First trial: Let the eigenvector corresponding to the zero eigenvalue be $v$. We can represent $M$ as
begin{equation}M=Q'DQ,tag{*}end{equation}
where $Q'Q=I$ and $D$ is a diagonal matrix. Suppose for certainty that the last element of D is equal to $0$, i.e., $d_{nn}=0$. This means that the last column of $Q$ is equal to $v$.



Now let's write $D=begin{bmatrix}D_1& 0\0& 0end{bmatrix}$ and $Q=begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}$, where $D_1$ is a PD diagonal matrix, $tilde Q$ is an $[n-1,n-1]$ matrix, $q_1$ and $q_2$ are vectors, and $x$ is a scalar.



With this, we rewrite (*) as
$$M=begin{bmatrix}tilde Q'& q_2\q_1'& xend{bmatrix}begin{bmatrix}D_1& 0\0& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1& 0\q_1'D_1& 0end{bmatrix}begin{bmatrix}tilde Q& q_1\q_2'& xend{bmatrix}=
begin{bmatrix}tilde Q'D_1tilde Q& tilde Q'D_1q_1\q_1'D_1tilde Q& q_1'D_1q_1end{bmatrix}$$



Here, $tilde Q'D_1tilde Q$ is the $(n,n)$-principal submatrix of $M$, which is positive definite if $tilde Q$ is non-singular. Thus the problem boils down to deternining non-singular principal submatrices of $Q$. However, I'm not sure is this makes the problem any simpler..



Solution (?): The number of PD principal submatrices id equal to the number of non-zero elements in the eigenvector $v$. If $v_ineq 0$, $M[i]$ is PD.



I will post the proof in a day or so.






matrices




matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 10:04







Dmitry

















asked Jan 11 at 14:45









DmitryDmitry

657518




657518








  • 1




    $begingroup$
    You can extract principal submatrices using "elimination matrices"
    $endgroup$
    – Bertrand
    Jan 11 at 14:54










  • $begingroup$
    @Bertrand, I know how to extract submatrices. The question is rather which one is to extract?
    $endgroup$
    – Dmitry
    Jan 11 at 15:02














  • 1




    $begingroup$
    You can extract principal submatrices using "elimination matrices"
    $endgroup$
    – Bertrand
    Jan 11 at 14:54










  • $begingroup$
    @Bertrand, I know how to extract submatrices. The question is rather which one is to extract?
    $endgroup$
    – Dmitry
    Jan 11 at 15:02








1




1




$begingroup$
You can extract principal submatrices using "elimination matrices"
$endgroup$
– Bertrand
Jan 11 at 14:54




$begingroup$
You can extract principal submatrices using "elimination matrices"
$endgroup$
– Bertrand
Jan 11 at 14:54












$begingroup$
@Bertrand, I know how to extract submatrices. The question is rather which one is to extract?
$endgroup$
– Dmitry
Jan 11 at 15:02




$begingroup$
@Bertrand, I know how to extract submatrices. The question is rather which one is to extract?
$endgroup$
– Dmitry
Jan 11 at 15:02










1 Answer
1






active

oldest

votes


















0












$begingroup$

This result stated for instance by Moschini may be useful:



Lemma. Let $S$ be an $n times n$ symmetric matrix that satisfies $Sp = 0$, where $p ne 0$, and let $tilde{S}$ be an $(n-1) times (n-1)$ matrix obtained from $S$ by deleting any one row and the corresponding column. Then a necessary and sufficient condition
for $S$ to be negative semidefinite is that $tilde{S}$ is negative semidefinite.



Moschini, G., 1999, "Imposing Local Curvature Conditions in Flexible Demand Systems," Journal of Business & Economic Statistics, 17, 487-490.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Shouldn't it read: "... is that $tilde S$ is negative definite"? Otherwise I'm not sure that this statement is true. Anyway, this does not solve my problem.
    $endgroup$
    – Dmitry
    Jan 11 at 15:17












  • $begingroup$
    It is negative semidefinite, and not negative definite. The reason is clear in the proof. The lemma does not apply to diagonal matrices, as diagonal matrices do not satisfy $Sp = 0$ for $p ne 0$.
    $endgroup$
    – Bertrand
    Jan 11 at 16:38











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

This result stated for instance by Moschini may be useful:



Lemma. Let $S$ be an $n times n$ symmetric matrix that satisfies $Sp = 0$, where $p ne 0$, and let $tilde{S}$ be an $(n-1) times (n-1)$ matrix obtained from $S$ by deleting any one row and the corresponding column. Then a necessary and sufficient condition
for $S$ to be negative semidefinite is that $tilde{S}$ is negative semidefinite.



Moschini, G., 1999, "Imposing Local Curvature Conditions in Flexible Demand Systems," Journal of Business & Economic Statistics, 17, 487-490.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Shouldn't it read: "... is that $tilde S$ is negative definite"? Otherwise I'm not sure that this statement is true. Anyway, this does not solve my problem.
    $endgroup$
    – Dmitry
    Jan 11 at 15:17












  • $begingroup$
    It is negative semidefinite, and not negative definite. The reason is clear in the proof. The lemma does not apply to diagonal matrices, as diagonal matrices do not satisfy $Sp = 0$ for $p ne 0$.
    $endgroup$
    – Bertrand
    Jan 11 at 16:38
















0












$begingroup$

This result stated for instance by Moschini may be useful:



Lemma. Let $S$ be an $n times n$ symmetric matrix that satisfies $Sp = 0$, where $p ne 0$, and let $tilde{S}$ be an $(n-1) times (n-1)$ matrix obtained from $S$ by deleting any one row and the corresponding column. Then a necessary and sufficient condition
for $S$ to be negative semidefinite is that $tilde{S}$ is negative semidefinite.



Moschini, G., 1999, "Imposing Local Curvature Conditions in Flexible Demand Systems," Journal of Business & Economic Statistics, 17, 487-490.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Shouldn't it read: "... is that $tilde S$ is negative definite"? Otherwise I'm not sure that this statement is true. Anyway, this does not solve my problem.
    $endgroup$
    – Dmitry
    Jan 11 at 15:17












  • $begingroup$
    It is negative semidefinite, and not negative definite. The reason is clear in the proof. The lemma does not apply to diagonal matrices, as diagonal matrices do not satisfy $Sp = 0$ for $p ne 0$.
    $endgroup$
    – Bertrand
    Jan 11 at 16:38














0












0








0





$begingroup$

This result stated for instance by Moschini may be useful:



Lemma. Let $S$ be an $n times n$ symmetric matrix that satisfies $Sp = 0$, where $p ne 0$, and let $tilde{S}$ be an $(n-1) times (n-1)$ matrix obtained from $S$ by deleting any one row and the corresponding column. Then a necessary and sufficient condition
for $S$ to be negative semidefinite is that $tilde{S}$ is negative semidefinite.



Moschini, G., 1999, "Imposing Local Curvature Conditions in Flexible Demand Systems," Journal of Business & Economic Statistics, 17, 487-490.






share|cite|improve this answer









$endgroup$



This result stated for instance by Moschini may be useful:



Lemma. Let $S$ be an $n times n$ symmetric matrix that satisfies $Sp = 0$, where $p ne 0$, and let $tilde{S}$ be an $(n-1) times (n-1)$ matrix obtained from $S$ by deleting any one row and the corresponding column. Then a necessary and sufficient condition
for $S$ to be negative semidefinite is that $tilde{S}$ is negative semidefinite.



Moschini, G., 1999, "Imposing Local Curvature Conditions in Flexible Demand Systems," Journal of Business & Economic Statistics, 17, 487-490.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 11 at 15:04









BertrandBertrand

964




964












  • $begingroup$
    Shouldn't it read: "... is that $tilde S$ is negative definite"? Otherwise I'm not sure that this statement is true. Anyway, this does not solve my problem.
    $endgroup$
    – Dmitry
    Jan 11 at 15:17












  • $begingroup$
    It is negative semidefinite, and not negative definite. The reason is clear in the proof. The lemma does not apply to diagonal matrices, as diagonal matrices do not satisfy $Sp = 0$ for $p ne 0$.
    $endgroup$
    – Bertrand
    Jan 11 at 16:38


















  • $begingroup$
    Shouldn't it read: "... is that $tilde S$ is negative definite"? Otherwise I'm not sure that this statement is true. Anyway, this does not solve my problem.
    $endgroup$
    – Dmitry
    Jan 11 at 15:17












  • $begingroup$
    It is negative semidefinite, and not negative definite. The reason is clear in the proof. The lemma does not apply to diagonal matrices, as diagonal matrices do not satisfy $Sp = 0$ for $p ne 0$.
    $endgroup$
    – Bertrand
    Jan 11 at 16:38
















$begingroup$
Shouldn't it read: "... is that $tilde S$ is negative definite"? Otherwise I'm not sure that this statement is true. Anyway, this does not solve my problem.
$endgroup$
– Dmitry
Jan 11 at 15:17






$begingroup$
Shouldn't it read: "... is that $tilde S$ is negative definite"? Otherwise I'm not sure that this statement is true. Anyway, this does not solve my problem.
$endgroup$
– Dmitry
Jan 11 at 15:17














$begingroup$
It is negative semidefinite, and not negative definite. The reason is clear in the proof. The lemma does not apply to diagonal matrices, as diagonal matrices do not satisfy $Sp = 0$ for $p ne 0$.
$endgroup$
– Bertrand
Jan 11 at 16:38




$begingroup$
It is negative semidefinite, and not negative definite. The reason is clear in the proof. The lemma does not apply to diagonal matrices, as diagonal matrices do not satisfy $Sp = 0$ for $p ne 0$.
$endgroup$
– Bertrand
Jan 11 at 16:38


















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