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I have the following matrix strict inequality: where $X,Y,A,B \in \mathbb{R}^{n\times n}$, $X,Y$ are symmetric matrices $(X=X^T, Y=Y^T)$, there are no conditions imposed on $A$, nor on $B$

$\begin{bmatrix} I & A^T \end{bmatrix} \begin{bmatrix} X & B^T \\ B &Y \end{bmatrix} \begin{bmatrix} I \\ A \end{bmatrix} <0$

It is easy to see that $\begin{bmatrix} I & A \end{bmatrix}_{n \times 2n}$ has full row rank:

  • Since there are $n$ rows, $rank[I \ A]\leq n$
  • Since, due to the identity matrix, there are at least $n$ linearly independent columns, then $rank [I \ A]\geq n$

Notice that this implies $null([I \ A])=n$

Now, what I have trouble understanding is this step: the paper I am reading says that since $\begin{bmatrix} I & A \end{bmatrix}$ has full row rank the last inequality $\textit{becomes}$ (I don't know if the author means that now we have a sufficient, necessary or equivalent condition) :

$ \begin{bmatrix} X & B^T \\ B &Y \end{bmatrix}<0$

Finally, (not immediately) before what I wrote, the paper states this Lemma, so it may be useful:

Lemma: let $P,Q$ two symmetric matrices (possibly of different orders), and S a matrix with appropiate dimensions. All matrices have real entries. Then the following are equivalent:

  1. $\begin{bmatrix}P &S^T \\ S & Q\end{bmatrix}>0$
  2. $Q>0$ and $P-S^T Q^{-1}S>0 $
  3. $P>0$ and $Q-SP^{-1}S^T >0$
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  • $\begingroup$ Is the inequality entry-wise or in the sense of "positive-definiteness"? $\endgroup$ – bartgol Sep 24 '15 at 3:45
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This is definitely not true in general. Definiteness on a subspace does not entail definiteness on the whole space. E.g. we have $$ \pmatrix{1&0}\pmatrix{-1&1\\ 1&1}\pmatrix{1\\ 0}=-1<0 $$ but $$ \pmatrix{-1&1\\ 1&1} $$ is indefinite. Either the authors of the paper you read or you have probably left out some conditions, or you have misunderstood the paper.

By the way, the converse of what you mentioned is true. If $\begin{bmatrix} X & B^T \\ B &Y\end{bmatrix}<0$, then $\begin{bmatrix} I & A^T \end{bmatrix} \begin{bmatrix} X & B^T \\ B &Y \end{bmatrix} \begin{bmatrix} I \\ A \end{bmatrix} <0$ too.

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  • $\begingroup$ user1551. Could you prove the converse please? does it use something about the rank of [I A] ? The focus of the paper is on sufficient conditions, rather than necessary conditions. So, proving the converse would be great help $\endgroup$ – Luis Vásquez Sep 24 '15 at 5:43
  • $\begingroup$ @LuisVásquez You may simply show that $v^T[I,A^T][...][...]v<0$ for every nonzero vector $v$. Note that, since $[I,A^T]$ has full row rank, $v^T[I,A^T]$ is nonzero. Then apply the negative definiteness of the square block matrix to conclude. $\endgroup$ – user1551 Sep 24 '15 at 5:46
  • $\begingroup$ but what if $v\in ker([I \ A]_{n \times 2n})$ ? Remember that the kernel of $[I \ A]$ has dimension $n$ $\endgroup$ – Luis Vásquez Sep 24 '15 at 18:09
  • $\begingroup$ never mind, I found out, just had to do the multiplication $\endgroup$ – Luis Vásquez Sep 24 '15 at 18:18
  • $\begingroup$ yet, I have not used the condition that the kernel of $[I \ A]$ has dimension $n$. Any ideas? $\endgroup$ – Luis Vásquez Sep 25 '15 at 2:23

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