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Wikipedia provides two methods to prove row rank of a matrix is equal to its column rank. My doubt is regarding the second method. But the wikipedia page mentions that this proof is valid only for real matrices. What part of this proof breaks down if the matrix is complex? I've added the proof here for easy reference.

Proof:

Let $r$ be the row rank of a matrix $A$. Let $x_1, x_2, \ldots x_r$ be a basis for row space of $A$.
Claim: $Ax_1, Ax_2, \ldots Ax_r$ are linearly independent (LI).
To prove, I need to show that $$\sum_{i=1}^{r}c_iAx_i = 0 \iff c_i=0, \forall i=1,2,\dots r$$ $$\sum_{i=1}^{r}c_iAx_i = A\sum_{i=1}^{r}c_ix_i$$ Let $$v=\sum_{i=1}^{r}c_ix_i \implies v \in C(A')$$ $$\sum_{i=1}^{r}c_iAx_i = Av = 0 \implies v \in N(A) $$ $$ \therefore v \in C(A') \bigcap N(A) \implies v = \vec{0}$$ $$ v = \sum_{i=1}^{r}c_ix_i = 0 \implies c_i=0 \forall i $$ Because $x_i$ are LI. This proves $Ax_i$ are LI.

We know that $Ax_i \in C(A)$ and $Ax_i$ are LI, dim($C(A)$)$\ge r$ i.e. column rank is greater than or equal to row rank. By doing the same steps with $A'$, we get that row rank of $A$ is greater than or equal to its column rank. This proves that row rank is equal to column rank.

Nagabhushan S N
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  • Where does Wikipedia state that the proof is only valid for real matrices? – Ben Grossmann Jun 04 '20 at 15:10
  • Note also that your attempt at rewriting the proof symbolically has made it less clear than it was in its original form – Ben Grossmann Jun 04 '20 at 15:14
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    For real matrices, the row space is orthogonal to the null space, so any element in the intersection has to be $0$. However, this is not necessarily true for complex matrices; you can see here for an example. – twosigma Jun 04 '20 at 15:26
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    Note that there are two analogous statements for complex matrices. One is that the rank of $A$ is equal to that of its entrywise transpose $A^T$, and the other is that the rank of $A$ is equal to that of its conjugate transpose $A^$ (or $A^H$, or $A'$, per your preference). Only the fact that $\operatorname{rank}(A) = \operatorname{rank}(A^)$ can be shown by generalizing this proof. – Ben Grossmann Jun 04 '20 at 15:33
  • @twosigma thanks a ton! That gave a lot of insights into subspaces of complex matrices. So, for complex matrices, if a vector is in both row space and null space, the vector need not necessarily be a zero vector. Can you suggest some articles/webpages to understand more about the linear algebra of complex matrices? – Nagabhushan S N Jun 04 '20 at 17:23
  • @Omnomnomnom, thanks for the feedback. I'll try to improve myself in future posts. Do you mean, by generalizing the above proof for complex matrices, one can show $rank(A) = rank(A^)$ but not $rank(A)=rank(A')$? Usually how is row rank defined for complex matrices? $rank(A')$ or $rank(A^)$? – Nagabhushan S N Jun 04 '20 at 17:27
  • @NagabhushanSN For me, $A'$ means the same thing as $A^*$. I use $A^T$ to indicate the entrywise transpose – Ben Grossmann Jun 04 '20 at 18:08
  • See. Also. – PinkyWay Jun 04 '20 at 19:38
  • @Cheesecake thanks for the suggestions, but none of them solve my problem. In particular, Proof verification: column rank = row rank asks if the OP's proof is valid or not. It doesn't concern with complex numbers and the proof is different than the one I've posted here. Again, thanks for your time :) – Nagabhushan S N Jun 05 '20 at 03:20

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Consider the following statement in the second proof on the Wikipedia page:

(b) since $Av=0$, the vector $v$ is orthogonal to every row vector of $A$.

This is true if we are working over the real numbers. On the other hand, over the complex numbers, $Av=0$ implies that $\overline{v}$ (taking the complex conjugate of each entry in $v$) is orthogonal to every row vector of $A$.

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    This ultimately depends, however, on how one defines the row-space of a matrix $A$ that has complex entries. In some contexts, it is common to define the row-space as the column-space of the conjugate-transpose $A^*$, which means that the fact that the null space forms the orthogonal complement to the row space still holds. – Ben Grossmann Jun 04 '20 at 15:36