# Symmetric Matrix Sign equivalence and rank of extenden matrix

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 $Q-SP^{-1}S^T >0$
• Is the inequality entry-wise or in the sense of "positive-definiteness"? Sep 24 '15 at 3:45

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.
• @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. Sep 24 '15 at 5:46
• but what if $v\in ker([I \ A]_{n \times 2n})$ ? Remember that the kernel of $[I \ A]$ has dimension $n$ Sep 24 '15 at 18:09
• yet, I have not used the condition that the kernel of $[I \ A]$ has dimension $n$. Any ideas? Sep 25 '15 at 2:23