Full rank real matrix and the rank of its associated complex matrix

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 $$r, then we may find matrices $$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?

• Please give a reason if you want my question to be closed
– 6666
Jul 28 '20 at 1:53
• Presumably, the close votes were cast before you added the context of your attempt. That said, I agree that they should have indicated what it was that they found problematic. Jul 28 '20 at 10:59
• @BenGrossmann thanks for the explanation. But the second close vote was after I added my attempt. And at the very beginning, I had no idea about my question. I wondered what I should add at that time
– 6666
Jul 28 '20 at 18:33
• For future reference, some context such as where you encountered the problem (i.e. if it's from a course or textbook) or the fact that you were not sure where to start would probably have helped Jul 28 '20 at 18:57

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 $$2\operatorname{rank}(A + iB)$$.