I am wondering if there is a way to solve this equation for a:
$$(as^T ⊙ b)n = t.$$
where:
⊙ is element-wise multiplication
a is an unknown v x 1 vector
s is an i x 1 $\vec{1}$ vector
b is a known v x i matrix
n is a known i x 1 vector
t is a v x 1 $\vec{1}$ vector
Thank you in advance
Yes. Multiply it out as elements. You will have a system of $v$ uncoupled linear equations in the components of $a$. Then solve each, by itself.
Example with $v=3$, $i=2$: \begin{align*} \left( \begin{pmatrix}a_1 \\ a_2 \\ a_3\end{pmatrix} (s_1 \, s_2) \odot \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{pmatrix}\right)\begin{pmatrix}n_1 \\n_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\ \left( \begin{pmatrix}a_1s_1 & a_1s_2 \\ a_2s_1 & a_2s_2 \\ a_3s_1 & a_3s_2\end{pmatrix} \odot \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{pmatrix}\right)\begin{pmatrix}n_1 \\n_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\ \begin{pmatrix}a_1b_{11}s_1 & a_1b_{12}s_2 \\ a_2b_{21}s_1 & a_2b_{22}s_2 \\ a_3b_{31}s_1 & a_3b_{32}s_2\end{pmatrix} \begin{pmatrix}n_1 \\n_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\ \begin{pmatrix}a_1b_{11}n_1s_1 + a_1b_{12}n_2s_2 \\ a_2b_{21}n_1s_1 + a_2b_{22}n_2s_2 \\ a_3b_{31}n_1s_1 + a_3b_{32}n_2s_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\ \begin{cases} a_1 = \frac{t_1}{b_{11}n_1s_1 + b_{12}n_2s_2} \\ a_2 = \frac{t_2}{b_{21}n_1s_1 + b_{22}n_2s_2} \\ a_3 = \frac{t_3}{b_{31}n_1s_1 + b_{32}n_2s_2} \end{cases}\text{,} \end{align*} assuming those $v$ divisions are defined.