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?

Ref: https://en.wikipedia.org/wiki/Row_and_column_spaces#Relation_to_the_null_space

  • 2
    $\begingroup$ This question have been answered here. $\endgroup$ – Learn_and_Share Nov 3 '19 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) $


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