Let $
\begin{bmatrix}
A & -B\\
B & A
\end{bmatrix}
\quad
$ be a full rank $2m\times 2n$ real matrix s.t. $A,B$ are both $m\times n$ real matrices and $n\ge m$. Then can we say the associted complex matrix $A+iB$ is also of full rank?
My attempt: Since elementary matrix operation doesn't change the rank and we can see the real matrix as a complex matrix, we do the following:
$
\begin{bmatrix}
A & -B\\
B & A
\end{bmatrix}
\quad
\to$
$
\begin{bmatrix}
A +iB & -B+iA\\
B & A
\end{bmatrix}
\quad
\to$
$
\begin{bmatrix}
A +iB & 0\\
B & A-iB
\end{bmatrix}
\quad
$
Now suppose the rank of $A+iB$ is $r<m$, then we may find matrices $X,Y$, s.t. $X(A+iB)Y=\begin{bmatrix}
I_r & 0\\
0 & 0
\end{bmatrix} :=C
\quad
,$ where $I_r$ is the $r\times r$ identity matrix. Now we get
$\begin{bmatrix}
X & 0\\
0 & I
\end{bmatrix}
\begin{bmatrix}
A +iB & 0\\
B & A-iB
\end{bmatrix}
\begin{bmatrix}
Y & 0\\
0 & I
\end{bmatrix}
=\begin{bmatrix}
C & 0\\
B & A-iB
\end{bmatrix} :=D.
\quad$
Now we see $D$ has rank at most $r+m<2m$, thus $D$ is not of full rank thus the initial matrix $
\begin{bmatrix}
A & -B\\
B & A
\end{bmatrix}
\quad
$ is not of full rank, contradiction. Am I right?
Best Answer
Yes, the statement is true and your proof is correct.
Here is a faster approach. Because $m \leq n$, we see that the matrices have full rank if and only if they have a trivial left kernel. Now, we note that for real vectors $x,y \in \Bbb R^{m}$, we have $$ (x +iy)^T(A + iB) = 0 \iff\\ x^TA - y^TB = 0, \quad y^TA + x^TB = 0 \iff\\ \pmatrix{x^T & y^T} \pmatrix{A & -B\\ B & A} = 0. $$ Thus, $A + iB$ has full rank if and only if $M = \pmatrix{A & -B\\B & A}$ has full rank.
In fact, because the above defines an $\Bbb R$-linear bijection between the left kernel of $A + iB$ (a subspace of $\Bbb C^m$) and the left kernel of $M$ (a subspace $\Bbb R^{2m}$), we can deduce that $\operatorname{rank}(M) = 2 \operatorname{rank}(A + iB)$.
An alternative approach: note that $M$ has the same rank as $$ \pmatrix{-i I_m & I_m\\ I_m & -iI_m} \pmatrix{A & - B\\ B & A}\pmatrix{i I_n & I_n\\ I_n & iI_n} = \\ 2\pmatrix{A + iB & 0\\ 0 & A - iB}. $$ Thus, the rank of $M$ is the sum of the ranks of $A + iB$ and $A - iB$. Now, I claim that $A + iB$ and $A - iB$ necessarily have the same rank. To see that this holds, it suffices to show that $v \in \Bbb C^m$ is in the left kernel of $A + iB$ if and only if the conjugate $\bar v \in \Bbb C^m$ is an element of the left kernel of $A - iB$.
With that, the rank of $M$ is always $2\operatorname{rank}(A + iB)$.