Show that $\vec x^TA$ is a linear combination of the rows of the matrix $A$ Let $\vec x^T = (x_1,x_2,\ldots,x_n)$ be a row vector in $\mathbb R^n$, and let $A$ be an $n\times m$ matrix. Show that $\vec x^TA$ is a linear combination of the rows of the matrix $A$.
I think it should be something like $\vec x^TA = \operatorname{Row}(A) = \operatorname{span}\{\text{rows of }A\}$, but I'm not quite sure where to start with proving this. 
 A: Well, $A^tx$ is column vector that can be written as a linear combination of the columns of $A^t$, that is, the transpose of the rows of $A$. Hence, $x^tA$ is the transpose of the previous vector, which now is a row vector that can be written as a linear combination of the rows of $A$.
A: If you want to see this clearly, just write it out. We have 
\begin{equation} 
\begin{split} 
x^TA &= (x_1, x_2, \ldots, x_n)\left( \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1m} \\ a_{21} & a_{22} & \ldots & a_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nm} \end{array} \right) \\
&= \left( \begin{array}{c} a_{11}x_1+ \cdots + a_{n1}x_n \\ a_{12}x_1+ \cdots + a_{n2}x_n \\ \vdots \\ a_{1m}x_1+ \cdots + a_{nm}x_n \end{array} \right)^T
\end{split} 
\end{equation} 
After 'factorising' each $x_i$, you can see clearly we are left with a linear combination of the rows of the matrix $A$. 
A: Let $e_i$ be the $i$th row of the $n\times n$ identity matrix. The $j$th element of $e_i$ is then $\delta_{ij}$, the Kronecker delta, which is equal to one when the indices are equal and zero otherwise. By the definition of matrix multiplication, the $j$th element of $e_iA$ is equal to $\sum_{k=1}^n \delta_{ik}a_{kj} = a_{ij}$. In other words, $e_iA$ is equal to the $i$th row of $A$.  
Now, $x$ can be written as a linear combination of the $e_i$ (check this!), therefore by linearity $xA$ must be a linear combination of the rows of $A$. Indeed, you should be able to write down the exact linear combination in terms of the elements of $x$.
