Need help regarding intuition of rows in a coordinate/basis matrix, where the columns are vectors. Given a matrix M such that its columns are the vectors of a new basis with respect to another basis B.
To find the coordinates of v in the other basis, we can simply take $M[v]_M = [v]_B$.
Let me give an example of M
$$\begin{bmatrix}1&2\\ 4&3\end{bmatrix}$$
I believe they are linearly independent(i just pulled out some random number off my head and tested), but the numbers aren't that important.
What i am confused about is we know that the columns of M form a set of basis vectors but when doing $M[v]_m$ matrix multiplication, we iterate within each $row_i$ of M for each value in the corresponding row of the output vector instead.
Now, i learn that, in my school's materials convention, we represent linear functionals as row vectors instead, since column vectors are for things like coordinate vectors and this makes sense to me at least here, but above, i am using a basis matrix's rows like linear functionals?
So yeah, is it just "it is how it is because matrix multiplication rules", or is there some special property or something about rows in matrices.
 A: $$
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}=
\begin{bmatrix}
ax + by\\
cx + dy
\end{bmatrix}=
\begin{bmatrix}
ax\\
cx
\end{bmatrix} +
\begin{bmatrix}
by\\
dy
\end{bmatrix} =
x\begin{bmatrix}
a\\
c
\end{bmatrix}+
y\begin{bmatrix}
b\\
d
\end{bmatrix}
$$
When you do ordinary matrix multiplication, you usually iterate over the rows to do the calculation, which is the first equality above. But this can always be rearranged to show that the answer is a linear combination of the columns your matrix.
So if you have a linear transformation $T$ between finite dimensional vector spaces $V$ and $ W$, say $T:V\to W$, represented by a matrix $M$, so that for any $v\in V$ you have $T(v)=Mv$, then the $\textit{column space}$ of $M$ is the subspace of $W$ spanned by the linearly independent columns of $M$ and is the subspace in which all possible results $Mv$ reside.
On the other hand, the linearly independent rows of $M$ span a subspace of $V$ called the $\textit{row space}$ of $M$. The orthogonal complement of the row space of $M$ is the $\textit{null space}$ of $M$ and is the subspace of $V$ that contains all $v\in V$ such that $Mv=0$.
All of these spaces have more formal definitions that need to be understood, but hopefully this gives a bit of a roadmap to sort out these things.
A: First a note: as you correctly wrote, we have $M[v]_M=[v]_B$, but to obtain $[v]_M$ we need to calculate $M^{-1}[v]_B$, where $M^{-1}$ corresponds to the reversed basis transformation, i.e. its columns are just $[b_i]_M$.
Row vectors indeed act as linear functionals, and for any basis $m_1,\dots,m_n$, taking the $i$th coordinate with respect to this basis is a linear functional, and this is exactly what we get if only considering multiplication (from left) by the $i$th row of $M$.
