# Combining Column- and Row-wise meanings of a matrix

A matrix can be thought of in terms of columns (then it represents the basis vectors of an coordinate system) or in terms of rows (then it represents a set of linear equations).

How can we combine those two views?

Example: Let the matrix $A$ represent a set of linear equations. Then each row is a linear equation. But we can also see the matrix as the basis vectors of a coordinate system. What coordinate system does matrix $A$ represent? What is the link between the column-wise and row-wise viewpoints on matrix?

If $V$ is a vector space over $\mathbb K$, then its dual space $V^*$ consists of all linear functionals $\mathbf\alpha:V\to\mathbb K$. It’s easy to verify that $V^*$ is also a vector space and that for finite-dimensional spaces, $\dim V=\dim V^*$. If $(\mathbf{e}_1,\dots,\mathbf{e}_n)$ is a basis for $V$, the dual basis of $V^*$ is a basis $(\mathbf{\epsilon}^1,\dots\mathbf{\epsilon}^n)$ such that $\mathbf{\epsilon}^i[\mathbf{e}_j]=\delta_{ij}$. If we represent elements $\mathbf v$ of $V$ as column vectors relative to some basis, then we can use row vectors relative to the dual basis to represent elements $\mathbf\alpha$ of $V^*$, and $\mathbf\alpha[\mathbf v]$ can be computed via matrix multiplication. (Since $(\mathbb R^n)^*$ is isomorphic to $\mathbb R^n$, the distinction between them is often glossed over.) For the rest if this answer, I’ll take $\mathbb K=\mathbb R$, since that’s likely the most familiar to you.
Let $V=\mathbb R^m$ and $W=\mathbb R^n$ and $L:V\to W$ a linear map. We know that, relative to bases $(\mathbf{v}_1,\dots,\mathbf{v}_m)$ and $(\mathbf{w}_1,\dots,\mathbf{w}_n)$ of $V$ and $W$, respectively, $L$ can be represented by some $m\times n$ matrix $A$. If $\mathbf v=\sum_{i=1}^mc_i\mathbf{v}_i$, then $A\mathbf v=\sum_{i=1}^mc_iA\mathbf{v}_i$. This means that the columns of $A$ are the images of the basis vectors $\mathbf{v}_i$ and that $\operatorname{im}L$ is spanned by the columns of $A$.
Looking at the rows of $A$ instead, with the above interpretation of the product of a row and column vector, we can treat each row of $A$ as an element $\mathbf{\alpha}^i$ of $V^*$. If $L\mathbf v=\sum_{i=1}^nd_i\mathbf{w}_i$, then the rules of matrix multiplication tell us that $d_i=\mathbf{\alpha}^i[\mathbf v]$. For a vector $\mathbf v\in\ker L$, this gives us a system of equations \begin{align}\mathbf{\alpha}^1[\mathbf v]&=0\\\vdots\\\mathbf{\alpha}^m[\mathbf v]&=0\end{align} satisfied by $\mathbf v$. By linearity, for all $\mathbf{\alpha}\in\operatorname{span}{\{\mathbf{\alpha}^1,\dots,\mathbf{\alpha}^m\}}$, we also have $\mathbf\alpha[\mathbf v]=0$. A functional that maps every element of some space $V$ to zero is said to annihilate that space, and the space of linear functionals that annihilate $V$ is called the annihilator of $V$, commonly denoted by either $V^0$ or $V^\perp$. So, the row space of $A$ annihilates $\ker L$.
To complete the connection, we need the $adjoint$ of $L$. This is the linear operator $L^*:W^*\to V^*$ (note the reversal of direction) such that for $\mathbf\beta\in W^*$ and $v\in V$, $(L^*\mathbf\beta)[\mathbf v]=\mathbf\beta[L\mathbf v]$. If $A$ is the matrix of $L$ relative to some choice of bases for $V$ and $W$, then the matrix of $L^*$ relative to their dual bases is simply $A^T$. The columns of $A^T$ span the image of $L^*$, but the columns of $A^T$ are the rows of $A$, so we see that $\operatorname{im}{L^*}=(\ker L)^0$. It’s also fairly easy to show that $\ker{L^*}=(\operatorname{im}L)^0$.
If we set $V=W=\mathbb R^n$ and identify this space with its dual via the isomorphism mentioned earlier, we can interpret $\mathbf\alpha[\mathbf v]$ as an inner product. The above then becomes the well-known result for a square matrix $A$: its row space is the orthogonal complement of its null space and its column space is the orthogonal complement of the null space of its transpose.