Orthogonal complement of the column space of a matrix Let $H =\operatorname{Col}(A)$, where $$A =\begin{pmatrix}
1&2\\
2&4\\
3& 1\end{pmatrix}$$
Find $H^\perp$, the orthogonal complement
of $H$.
$H$ is the same thing as $A$, and as I understand it, Orthogonal complement means the span of vectors that are orthogonal to the matrix, but I don't understand how to solve for this. Nor am I completely clear on how a vector can be orthogonal to a matrix, as I only no how to find the dot product between vectors with only $1$ row.
Thanks.

In general, for any matrix $A \in \mathbb{C}^{m\times n}$, the answer may be obtained, using those relations: 
$$ \text{im}(A^*) =\ker(A)^{\perp} ~~\text{and}~ ~\ker(A^*) = \text{im}(A)^{\perp}$$ furthermore
$$\mathbb{C}^{m} = \ker(A^*) \oplus \text{im}(A) ~~\text{and}~ ~ \mathbb{C}^{n} = \ker(A) \oplus \text{im}(A^*)$$ (this is sometimes called Fredholm alternative)
Where 


*

*$\ker(\cdot)$ is the kernel of a matrix.

*$\text{im}(\cdot)$ is the image of a matrix. 

*$A^*$ is the conjugate transpose. When dealing with real matrices only, this becomes the usual transpose.  

*$\oplus$ - direct sum.

