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?

  • 2
    $\begingroup$ Please give a reason if you want my question to be closed $\endgroup$
    – 6666
    Jul 28 '20 at 1:53
  • $\begingroup$ 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. $\endgroup$ Jul 28 '20 at 10:59
  • $\begingroup$ @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 $\endgroup$
    – 6666
    Jul 28 '20 at 18:33
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    $\begingroup$ 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 $\endgroup$ Jul 28 '20 at 18:57

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


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