Matrix of Linear Transformation by right multiplication I am trying to solve the following problem: 
Let $A$ be an $n\times n$ matrix, and let $V$ denote the space of $n$-dimensional row vectors. What is the matrix of the linear operator ‘‘right multiplication by $A$’’ with respect to the standard basis of $V$?
I am not sure where to begin with this problem.
 A: Let $T:V \to V$ be the linear map defined by
$$T(v^T)=v^TA. $$
We want to think of $V$ as an abstract vector space, and express $T$ as a left matrix multiplication. In order to do this we use the canonical isomorphism $V \to F^n$ defined by $v \mapsto v^T$. Applying $T$ to the first basis vector $e_1^T$, gives $T(e_1^T)=e_1^TA=R_1(A)=(a_{11}, a_{12}, \dots, a_{1n})$, the first row of $A$, which can also be written as the linear combination $a_{11} e_1^T+a_{12} e_2^T+\dots+a_{1n} e_n^T$. Thus the first column of $[T]$ is the first row of $A$. Continuing in this manner, we get
$$[T]=A^T. $$
A: The matrix of the linear operator is the transpose of $A.$  Here is why:
Let $T$ be the linear operator you describe, $\alpha$ any vector in $V,$ $\mathscr B$ the standard ordered basis, and $[\alpha]_{\mathscr B}$ the coordinate matrix of $\alpha$ relative to $\mathscr B.$  Recall that $[\alpha]_{\mathscr B}$ is an $n \times 1$ column matrix.  The matrix of $T$ relative to $\mathscr B$ is the $n \times n$ matrix $[T]_{\mathscr B}$ such that
$$[T\alpha]_{\mathscr B} = [T]_{\mathscr B} [\alpha]_{\mathscr B}.$$
Now
$$\begin{align}
[T\alpha]_{\mathscr B} & = [\alpha A]_{\mathscr B}\\
                       & = \left[\left(A^t \alpha^t\right)^t\right]_{\mathscr B}\\
                       & = \left[A^t \alpha^t\right]_{\mathscr B}.\end{align}$$
We dispensed with the outer transpose because the $[ \cdot ]_{\mathscr B}$ notation forces the entries of the vector to be a column matrix, so it does not matter whether the vector in inside the $[ \cdot ]_{\mathscr B}$ symbol is a row or column vector. Continuing,
$$\begin{align}
[T\alpha]_{\mathscr B} & = \left[A^t \right]_{\mathscr B} \left[\alpha^t \right]_{\mathscr B}\\
                       & = A^t [\alpha]_{\mathscr B}\end{align}$$
where $\left[A^t \right]_{\mathscr B} = A^t$ because $\mathscr B$ is the standard ordered basis, and, once again, we can dispense with the transpose inside the $[ \cdot ]_{\mathscr B}$ notation.  We see, thus, that the matrix of $T$ relative to $\mathscr B$ is $A^t.$
A: It is the function $f:h^T\rightarrow h^TA$ where $h$ is a vector. We choose the basis $e_1^T,\cdots,e_n^T$ where $e_1\cdots,e_n$ is the canonical basis. EDIT: Then the matrix associated to $f$ is $A$ (for every $i$, calculate $f(e_i^T)$ and write the $n$ results row by row).
