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.

  • 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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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