Why is a linear transformation expressed using its transpose? If $A$ is an invertible matrix with entries from $\mathbb{R}$, what is the reasoning behind defining an invertible linear transformation $f_A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ as $f_A=xA^t$, where $A^t$ represents the transpose of $A$? A more "natural" definition to me would be$f_A=Ax$.
 A: Let $f_A(x)=xA^t$ (where $x$ is a row vector) and $g_A(x)=Ax$ (where $x$ is a column vector).
Note that
$$
f_A(x)^t=(xA^t)^t=Ax^t=g_A(x^t).
$$
Hence, $f_A$ and $g_A$ are essentially the same transformation, except that $f_A$ uses row vectors and $g_A$ uses column vectors. If you begin with some $x\in\mathbb R^n$, treat it as a row vector, apply $f_A$ and then take the transpose to turn it into a column vector, you get the same result as you would if you apply $g_A$ to $x$ as a column vector.
A: If you consider $\mathbb{R}^n$ as the space of row vectors, that is, $1\times n$ matrices, the only reason for using the transpose is to make composition of linear maps corresponding to matrix multiplication.
Define $f_A(x)=xA^T$; if $B$ is another $n\times n$ matrix (invertibility is not of a concern), you also have $f_B(x)=xB^T$.
Then
$$
f_A\circ f_B(x)=f_A(xB^T)=xB^T\!A^T=x(AB)^T=f_{AB}(x)
$$
so
$$
f_A\circ f_B=f_{AB}
$$

Avoiding the transpose is the main reason why usually $\mathbb{R}^n$ is the space of column vectors and the map associated to a matrix is $g_A(x)=Ax$; in this case we have $g_A\circ g_B=g_{AB}$.
When using row vectors it's often customary to write maps on the right of the argument, so the transpose is not needed in the definition in order that map composition corresponds to matrix multiplication.
A: Left muliplication by $A$ is a linear map between spaces of column vectors. There is an obvious one-to-one correspondence between column vectors and row vectors. What is the linear map between spaces of row vectors which corresponds to left multiplication by $A\require{AMScd}$ on column vectors?
Stated with commutative diagrams, we want a map linear map $B$ for which
\begin{CD}
    \mathbb{R}^n_\mathrm{col} @>A>> \mathbb{R}^m_\mathrm{col}\\
    @V t V V @VV t V\\
    \mathbb{R}^n_\mathrm{row} @>>B> \mathbb{R}^m_\mathrm{row}
\end{CD}
Above, $A$ is an $m\times n$ matrix and $t$ stands for transpose, which is the way to convert between column vectors and row vectors. Since $t$ is invertible (just take the transpose twice), we can solve for the map $B$ by just chasing the diagram around the other way: first apply $t$ to a row vector to get a column vector, then left multiply by $A$, then apply $t$ again to get a row vector again. Algebraically the effect of this is $(Ax^t)^t=xA^t$, so the corresponding map is right-multiplication by $A^t$.
A: the notion of matrix product must respect how the matrix product
is done, and also how can it represent a vector in $K^n$, a matrix
$M$ that has $n$ rows and $m$ columns, it  is called  of type$(n,
m)$. Let matrix $M$ and $N$ of  type respectively $(n, m)$, $(l,
k)$, then the matrix  product $MN$ exists iff $m = l$ and in this
case the matrix is of the type$(n, k)$. a vector of $K^n$ as the
matrix vector is of type$(1, n)$.
If $f$ is a linear mapping from $ K^n $ to $k^m$, the matrix $ A = M_{(f, b, b ')} $ of $ f$  with respect
to a base $ b $ of $K^n$ and a  base $b '$ of $k^m$ is of type$(m,
n)$.
how can write $f (x)=v$ in the  matrix form?
 $xA^t$ or $Ax$ it is clear if $n\not=m$ we can not write
 $f(x)=Ax$, but always $f(x)=xA^t$, this the reason why we choose
 $xA^t$ and not $Ax$ even if $Ax$ is well.
