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 $D=\begin{bmatrix}D_1& 0\\0& 0\end{bmatrix}$ and $Q=\begin{bmatrix}\tilde Q& q_1\\q_2'& x\end{bmatrix}$, 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 $$M=\begin{bmatrix}\tilde Q'& q_2\\q_1'& x\end{bmatrix}\begin{bmatrix}D_1& 0\\0& 0\end{bmatrix}\begin{bmatrix}\tilde Q& q_1\\q_2'& x\end{bmatrix}= \begin{bmatrix}\tilde Q'D_1& 0\\q_1'D_1& 0\end{bmatrix}\begin{bmatrix}\tilde Q& q_1\\q_2'& x\end{bmatrix}= \begin{bmatrix}\tilde Q'D_1\tilde Q& \tilde Q'D_1q_1\\q_1'D_1\tilde Q& q_1'D_1q_1\end{bmatrix}$$

Here, $\tilde Q'D_1\tilde Q$ 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 $v_i\neq 0$, $M[i]$ is PD.

I will post the proof in a day or so.

$\endgroup$
  • 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
$\begingroup$

Here is a counter-example to your conjecture: the matrix $$ M=\left( \begin{array}{rrr} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{array} \right) $$ is positive semi-definite and has rank $2$, but any matrix extracted by eliminating the $i^{th}$ row and column is positive definite.

$\endgroup$
  • $\begingroup$ You are right. I've already recognized that. Please see the last lines in my answer. There, $v$ is the eigenvector corresponding to the zero eigenvalue. That result agrees with your observation: $v=[1,1,1]$ hence, all principal submatrices will be PD. $\endgroup$ – Dmitry Jan 28 at 11:46
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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.