Formulate $a_i-a_j\approx d_{ij}$ as least squares Given $d_{ij}\in\mathbb{R}$, $1\leq i<j\leq n$, I'm considering  the problem of finding $a\in\mathbb{R}^n$ such that $a_1=0$ and
$$
a_i-a_j\approx d_{ij}.
$$
I need to formulate this problem as a least square problem, in the context of linear systems.
The way i'm thinking is like that:
Defining $E_{ij}$ as the column matrix with all entrys $0$, except $1$ at the row $i$ and $-1$ at the row $j$, we have that
$$
E_{ij}^Ta=d_{ij}.
$$
So, we have
$$
\begin{bmatrix}
E_{12}^T \\
E_{13}^T \\
\vdots \\
E_{1n}^T
\end{bmatrix}
a=
\begin{bmatrix}
d_{12} \\
d_{13} \\
\vdots \\
d_{1n}
\end{bmatrix}.
$$
In general, for every $1\leq k < n$, we have a system of $n-k$ equations:
$$
\begin{bmatrix}
E_{k\,k+1}^T \\
E_{k\,k+2}^T \\
\vdots \\
E_{kn}^T
\end{bmatrix}
a=
\begin{bmatrix}
d_{k\,k+1} \\
d_{k\,k+2} \\
\vdots \\
d_{kn}
\end{bmatrix}.
$$
So, there are $n-1$ systems of equations for $a$. How exactly can I use least squares to find the best estimative for $a$?
 A: There's a nice way to formulate this problem using vectorization and the Kronecker product. If $D$ denotes the matrix with entries $d_{ij}$ and $e$ denotes the vector $(1,\dots,1) \in \Bbb R^n$, then the desired outcome can be written as
$$
ae^T - ea^T \approx D.
$$
Using the properties of vectorization, we can write
$$
\operatorname{vec}(ae^T - ea^T) = (e\otimes I_n  - I_n \otimes e)a,
$$
where vec denotes column-major vectorization (as in the wiki page), $\otimes$ denotes a Kronecker product, and $I_n$ denotes the identity matrix. With that, you are looking for a least squares solution to the equation $Ma = \operatorname{vec}(D)$
, where $M = e \otimes I_n - I_n \otimes e$.
Notably, $M$ does not have linearly independent columns. One convenient approach is to use the Moore Penrose pseudoinverse to get the least squares solution $a = M^+\operatorname{vec}(D)$.

You have found that
$$
M^TM = (2n I_n - ee^T).
$$
This matrix is positive semidefinite with eigenvalue $0$ of multiplicity $1$ and $2n$ of multiplicity $n-1$. It follows that its pseudoinverse is given by $(M^TM)^+ = \frac 1{(2n)^2}M^TM$. Thus, we have
$$
M^+ = (M^TM)^+M^T = \frac{M^TMM^T}{4n^2},
$$
which we can plug into $a = M^+ \operatorname{vec}(D)$.
