,2}
\end{bmatrix} \text{ and } B=\begin{bmatrix}
B_{1,1}&B_{1,2}\\
B_{2,1}&B_{2,2}
\end{bmatrix}
\end{equation*}
be $m\times n$ and $r\times s$ matrices with real entries, respectively, where $A_{2,2}$ and $B_{1,1}$ are $k\times k$. In \cite{FalJoh1999a}, the \emph{$k-$subdirect sum} of $A$ and $B$, denoted by $A\oplus_k B$, was introduced; this is the matrix 
\begin{equation*}A \oplus_{k} B=
\begin{bmatrix}
A_{1,1}&A_{1,2}&0\\
A_{2,1}&A_{2,2}+B_{1,1} & B_{1,2}\\
0 & B_{2,1} & B_{2,2}
\end{bmatrix}
\end{equation*}

\begin{lemma}\cite{MR3010007}\label{lem:sum}
Let 
\begin{equation*}
C = \begin{bmatrix}
C_{1,1} & C_{1,2}\\
C_{2,1} & C_{2,2}
\end{bmatrix}
\quad\text{and}\quad
D = \begin{bmatrix}
D_{1,1} & D_{1,2}\\
D_{2,1} & D_{2,2}
\end{bmatrix},
\end{equation*}
where $C_{2,2}$ and $D_{1,1}$ are $k\times k$ matrices.
Then $\rank(C\oplus_k D)\leq \rank(C\oplus D)$. 
\end{lemma}

In the next theorem, we show that each term in the minimum of Formula~(\ref{mainformula}) is at least $\mbox{mr}(M)$.

\begin{thm}
Let 
\begin{equation*}
M=\begin{bmatrix}
A_{1,1}&A_{1,2}&0\\
A_{2,1}& a_{2,2} & A_{2,3}\\
0 & A_{3,2} & A_{3,3}\\
\end{bmatrix}
\end{equation*} 
be a sign pattern matrix, where $A_{1,1}$ is $m_1\times n_1$, $A_{1,2}$ is $m_1\times 1$, $A_{2,1}$ is $1\times n_1$, $a_{2,2}$ is $1\times 1$, $A_{2,3}$ is $1\times n_2$, $A_{3, 2}$ is $m_2\times 1$ and $A_{3,3}$ is $m_2\times n_2$. Then each of the following inequalities hold:
\begin{enumerate}[(i)]
\item $\mbox{mr}(A_{1,1})+\mbox{mr}(A_{3,3})+2\geq \mbox{mr}(M)$,
\item $\mbox{mr}([A_{1,1}\;A_{1,2}])+\mbox{mr}([A_{3,2}\;A_{3,3}])+1\geq \mbox{mr}(M)$,
\item $\mbox{mr}(\begin{bmatrix}
A_{1,1}\\
A_{2,1}\\
\end{bmatrix})+\mbox{mr}(\begin{bmatrix}
A_{2,3}\\
A_{3,3}\\
\end{bmatrix})+1\geq \mbox{mr}(M)$, and
\item 
for each $p\in \{+,-,0\}$, 
\begin{equation*}
\mbox{mr}(\begin{bmatrix}
A_{1,1} & A_{1,2}\\
A_{2,1} & p
\end{bmatrix}
)+\mbox{mr}(
\begin{bmatrix}
a_{2,2}-p & A_{2,3}\\
A_{3,2} & A_{3,3}
\end{bmatrix}
)\geq \mbox{mr}(M).
\end{equation*} 
\end{enumerate}
\end{thm}


\begin{proof}
To see that $\mbox{mr}(A_{1,1})+\mbox{mr}(A_{3,3})+2\geq \mbox{mr}(M)$, let $C_{1,1}\in Q(A_{1,1})$ and $C_{3,3} \in Q(A_{3,3})$ such that $\rank(C_{1,1}) = \mbox{mr}(A_{1,1})$ and $\rank(C_{3,3})=\mbox{mr}(A_{3,3})$.
Let $C_{1,2}\in Q(A_{1,2})$, $C_{2,1}\in Q(A_{2,1})$, $C_{2,3} \in Q(A_{2,3})$, $C_{3,2}\in Q(A_{3,2})$, and $\mbox{sgn}(c_{2,2}) = a_{2,2}$. Then
\begin{equation*}
\begin{split}
\mbox{mr}(A_{1,1}) + \mbox{mr}(A_{3,3})+2 & = \rank(\begin{bmatrix}
C_{1,1} & 0\\
0 & C_{3,3}
\end{bmatrix}) + 2\\
&\geq \rank(\begin{bmatrix}
C_{1,1} & C_{1,2} & 0\\
C_{2,1} & c_{2,2} & C_{2,3}\\
0 & C_{3,2} & C_{3,3}
\end{bmatrix})\\
&\geq \mbox{mr}(M),
\end{split}
\end{equation*}

To see that $\mbox{mr}([A_{1,1}\;A_{1,2}])+\mbox{mr}([A_{3,2}\;A_{3,3}])+1\geq \mbox{mr}(M)$, let $[C_{1,1}\;C_{1,2}]\in Q([A_{1,1}\;A_{1,2}])$ and $[C_{3,2}\;C_{3,3}]\in Q([A_{3,2}\;B_{3,3}])$ be such that $\rank([C_{1,1}\;C_{1,2}])=\mbox{mr}([A_{1,1}\;A_{1,2}])$ and $\rank([C_{3,2}\;C_{3,3}])=\mbox{mr}([A_{3,2}\;A_{3,3}])$. Clearly,
\begin{equation*}
\rank([C_{1,1}\;C_{1,2}]) + \rank([C_{3,2}\;C_{3,3}])\geq \rank(\begin{bmatrix}
C_{1,1} & C_{1,2}  & 0\\
0 & C_{3,2} & C_{3,3}
\end{bmatrix}).
\end{equation*}
Since
\begin{equation*}
\rank(\begin{bmatrix}
C_{1,1} & C_{1,2}  & 0\\
0 & C_{3,2} & C_{3,3}
\end{bmatrix}) + 1\geq \rank(
\begin{bmatrix}
C_{1,1} & C_{1,2} & 0\\
C_{2,1} & c_{2,2} & C_{2,3}\\
0 & C_{3,2} & C_{3,3}
\end{bmatrix}),
\end{equation*}
we obtain $\mbox{mr}([A_{1,1}\;A_{1,2}])+\mbox{mr}([A_{3,2}\;A_{3,3}])+1\geq \mbox{mr}(M)$.

The proof that 
$\mbox{mr}(\begin{bmatrix}
A_{1,1}\\
A_{2,1}\\
\end{bmatrix})+\mbox{mr}(\begin{bmatrix}
B_{1,2}\\
B_{2,2}\\
\end{bmatrix})+1\geq \mbox{mr}(M)$ is similar to the proof of the previous case.

Let $p\in \{+,-,0\}$. To shorten notation, let
\begin{equation*}
R_p = \begin{bmatrix}
A_{1,1} & A_{1,2}\\
A_{2,1} & p
\end{bmatrix}
\text{ and }
S_p = \begin{bmatrix}
a_{2,2}-p & A_{2,3}\\
A_{3,2} & A_{3,3}
\end{bmatrix}.
\end{equation*}
To see that $\mbox{mr}(R_p) +