Theorem:
Let Y be an $n \times$ 1 column vector and A be an $n \times K$ matrix. If $Y^{T}$ is a row vector then $Y^{T}A$ is a linear combination of rows of A with coefficient from $Y^{T}$
Proof:
$\left(Y^{T}A \right )_{ik}$
= $\sum_{j=1}^{n}y_{ij}A_{jk}$
but only one row exists for $Y^{T}$ so for sake of convenience, drop the row index
= $\sum_{j=1}^{m}y_{j}A_{jk}$
= $y_{1}A_{1k} + y_{2}A_{2k} + y_{3}A_{3k} + \cdot \cdot \cdot+y_{n}A_{nK} \in \mathbb{R}$
From here we observe that for a row vector of dimension $1 \times n$ and coefficient matrix A of dimension $n \times K$
$\left [ y_{1} \cdot \cdot \cdot y_{n} \right ] \cdot \begin{bmatrix} a_{11} &a_{12} &a_{13} &\cdot &\cdot &\cdot &a_{1K}\\ a_{21}&a_{22} &a_{23} &\cdot &\cdot &\cdot &a_{2K} \\ \cdot& & & & & &\\ \cdot& & & & & &\\ \cdot& & & & & &\\ a_{n1}&a_{n2} &a_{n3} &\cdot &\cdot &\cdot & a_{nK}\\ \end{bmatrix}$
which gives
$y_{1}a_{11}+y_{2}a_{21}+ \cdot \cdot \cdot + y_{n}a_{n1}$
$y_{1}a_{12}+y_{2}a_{22}+ \cdot \cdot \cdot + y_{n}a_{n2}$
$\cdot$
$\cdot$
$\cdot$
$y_{1}a_{1n}+y_{2}a_{2n}+ \cdot \cdot \cdot +y_{n}a_{nK}$
Clearly, it can be seen $Y^{T}A$ is a linear combination of columns from A rather than rows from A as is required.
Any explanation for this is highly appreciated. Thanks in advance.