Let $j$ and $b$ be $k*1$ vectors and $y$ be an $n*1$ vector. I am bit stuck on how to solve for $b$ in the following equation: $$j^Tb=y^Ty$$ If $j$ were a matrix, we could, multiply both sides of the equation with $j$ and then with $(jj^T)^{-1}$. However, given that $j$ is a vector, $jj^T$ or $uj^T$ for any $k*1$ vector $u$ is a singular matrix and therefore cannot be inverted.
I suspect I cannot see something obvious here...Any ideas (other than applying singular value decomposition)?
Given the linear system $Mv=p$, all possible solutions (if they exist) are given by $$\eqalign{v &= M^+p + (I-M^+M)\,a}$$ where $M^+$ is the pseudoinverse of $M$, $I$ is the identity matrix, and $a$ is an arbitrary vector.
In the current problem $$\eqalign{ v &= b,\quad p = y^Ty,\quad M = j^T,\quad M^+ = \frac{j}{j^Tj} \\ }$$ yielding $$\eqalign{ b &= \Bigg(\frac{y^Ty}{j^Tj}\bigg)\,j + \Bigg(I-\frac{jj^T}{j^Tj}\Bigg)\,a \\ &= \frac{1}{j^Tj}\bigg(\big(y^Ty-j^Ta\big)j + \big(j^Tj\big)a\bigg) \\ }$$ The choice of the $a$-vector completely determines the solution vector.