# Relationship between row space and orthogonal component of kernel of complex vector space

When we consider the real vector space, row space is equal to orthogonal complement of the null space (kernel). This fact can be proved as follows.

Let's consider linear map $A : \mathbb{R}^n \rightarrow \mathbb{R}^m$.

Row space of the matrix is $\mathrm{Row}(A) := \mathrm{span}(\{\mathbf{r}_i\}_{i=1,2,...,m})$. Where, $\{\mathbf{r}_i\}_{i=1,2,...,n}$ is row vector of matrix A.

On the other hand, the kernel of the matrix is $\mathrm{Kernel}(A) := \{ \mathbf{v}\in\mathbb{R}^n | A\mathbf{v}=0 \} = \{ \mathbf{v}\in\mathbb{R}^n | \mathbf{r}_i\cdot\mathbf{v}=0 \mathrm{\ for\ all\ i }\}$

Then, orthogonal subspace of the kernel is nothing but row space.

However, when we consider complex vector space $A : \mathbb{C}^n \rightarrow \mathbb{C}^m$, inner product of row vector is $\mathbf{r}_i^{*}\cdot\mathbf{v}$. So, in this case, the condition $\mathbf{r}_i^{*}\cdot\mathbf{v}=0$ is different from the definition of the kernel.

Is there some relationship between row space and orthogonal space of the kernel of complex vector space?

• This question have been answered here. Nov 3, 2019 at 5:22

Let $$A \in M_{\text{m}\times \text{n}}(\mathbb C)$$
Let $$v \in \mathbb C^n$$ and $$w \in \mathbb C^m$$
$$\text{}v \in \mathrm{Kernel}(A)$$
$$\iff Av = \vec 0$$
$$\iff \langle Av,w\rangle = 0 \text{ for all } w \in \mathbb C^m$$
$$\iff w^\dagger Av = 0 \text{ for all } w \in \mathbb C^m$$
$$\iff (A^\dagger w)^\dagger v = 0 \text{ for all } w \in \mathbb C^m$$
$$\iff \langle v, A^\dagger w\rangle = 0 \text{ for all } w \in \mathbb C^m$$
$$\text{}v \in \mathrm{Column}(A^\dagger)^\perp$$
Hence, $$\mathrm{Kernel}(A)=\mathrm{Column}(A^\dagger)^\perp \text{, where } A^\dagger \text{ is the conjugate transpose of }A$$
Therefore, $$\mathrm{Kernel}(A)^\perp=\mathrm{Column}(A^\dagger)$$