Proof that row rank=column rank using orthogonality

Question

I wanted to find a proof that row rank = column rank using orthogonality, and I came across the following proof:

Let $A \in \Bbb R^{m \times n}$ be a matrix with row rank $r(A)=r$. Let $\{ x_1,x_2,...,x_r \}$ be a basis of the row space. We shall prove that $\{ Ax_1,Ax_2,...,Ax_r \}$ is a linearly independent set. Let $a_1,a_2,...,a_r \in \Bbb R$ be scalars such that $a_1Ax_1+a_2Ax_2+...+a_rAx_r=0$. It follows that $A(a_1x_1+a_2x_2+...+a_rx_r)=0$. Denote $v=a_1x_1+a_2x_2+...+a_rx_r$. We see that $Av=0$ and that $v$ is a linear combination of basis vectors of the row space. Hence $v \in KerA \cap RowA$. Since the kernel and the row space are orthogonal, their intersection is trivial, hence $v=0$, and $a_1x_1+a_2x_2+...+a_rx_r=0$ and since $\{ x_1,x_2,...,x_r \}$ is linearly independent we obtain the desired result $a_1=a_2=...=a_r=0$.

We have proven $\{ Ax_1,Ax_2,...,Ax_r \}$ is a linearly independent set, and every vector in that set is a linear combination of the columns of $A$, hence the column space of $A$ has at the least dimension $r$, or $r(A^T) \geq r(A)$. Now, repeat the whole argument for the matrix $A^T$, and we'll get $r((A^T)^T)=r(A) \geq r(A^T)$. Finally, $r(A)=r(A^T)$.

This proof is great for real matrices, but obviously it won't work for complex matrices, since the standard dot product is not a valid inner product in complex spaces (for example, the norm of the vector $(1,i)$ in the standard inner product space is $0$, but that is impossible) and subsequently, the kernel of $A$ is not orthogonal to the column space of the transpose of $A$, but rather, the kernel is orthogonal to the column space of the Hermitian transpose of $A$. My question is, is there a way to prove nevertheless that the column rank is equal to the row rank using orthogonality in complex spaces?

Answer 1

Here is a proof that will work for complex spaces over $ \Bbb C$, where the inner product is defined as $\langle x,y \rangle=x \cdot \overline y$ (so $x$ is dotted with the complex conjugate of $y$)

Let $A \in \Bbb C^{m \times n}$ be a matrix with row rank $r(A)=r$. Let $B_1=\{ x_1,x_2,...,x_r \}$ be a basis of the row space. We shall prove that $\{ A\overline{x_1},A\overline{x_2},...,A\overline{x_r} \}$ is a linearly independent set. Let $a_1,a_2,...,a_r \in \Bbb C$ be scalars such that $a_1A\overline{x_1}+a_2A\overline{x_2}+...+a_rA\overline{x_r}=0$. Denote $\overline{b_1}=a_1,\overline{b_2}=a_2,...,\overline{b_r}=a_r$. It follows that $A(\overline{b_1x_1}+\overline{b_2x_2}+...+\overline{b_rx_r})=0$. And therefore $A(\overline{b_1x_1+b_2x_2+...+b_rx_r})=0$. Denote $v=b_1x_1+b_2x_2+...+b_rx_r$. We see that $A\overline v=0$. By definition of matrix multiplication, and by the definition of the inner product, every row of $A$ is orthogonal to $v$, and every linear combination of the rows is also orthogonal to $v$, so $v$ is orthogonal to itself (because by its definition we get that $v$ is a linear combination of basis vectors of the row space), and hence $v=0$. By definition of basis we get that $b_1=b_2=...=b_r=0$, and hence $a_1=a_2=...=a_r=0$. We've proven $\{ A\overline{x_1},A\overline{x_2},...,A\overline{x_r} \}$ is a linearly independent set, and it's not hard to see that it consists of linear combinations of the columns of $A$.

The continuation of the proof is the same as in the question. (This proof might have some redundant claims, but the idea is coherent)