Equivalent Graph Resistance and the Moore-Penrose Inverse I have read that the ohmic resistace between two nodes $u$ and $v$ in a graph is given by the formula
$$\Omega = (e_u - e_v)^T * L^\dagger * (e_u - e_v)$$
where $L^\dagger$ is the Moore-Penrose pseudoinverse of the Graph-Laplacian matrix and $e_i$ is the indicator vector of the node $i$ on the graph.
Now, if I get it correctly, the pseudoinverse of any square matrix is not unique. If that is the case, how can the resistance be unique?
 A: I hadn't fully read the properties like the existence and uniqueness of the pseudoinverse prior to posting this question.
By definition, $A^\dagger(n * m)$ is a pseudoinverse of $A(m * n)$ if it satisfies the following properties.
\begin{align}
A A^\dagger A & = A\tag1\label1\\
A^\dagger A A^\dagger & = A^\dagger\tag2\label2\\
(A A^\dagger)^* & = A A^\dagger\tag3\label3\\
(A^\dagger A)^* & = A^\dagger A\tag4\label4\\
\end{align}
It turns out the Moore-Penrose inverse of any matrix $A$ is in fact unique.
Proof
Suppose $A_1^\dagger$ and $A_2^\dagger$ are two pseudoinverses of $A$. Then
\begin{align}
A A_1^\dagger & = (A A_2^\dagger A) A_1^\dagger \\
& = (A A_2^\dagger) (A A_1^\dagger) \\
& = (A A_2^\dagger)^* (A A_1^\dagger)^* \\
& = (A_2^\dagger)^* A^* (A A_1^\dagger)^* \\
& = (A_2^\dagger)^* (A A_1^\dagger A)^* \\
& = (A_2^\dagger)^* A^* \\
& = A A_2^\dagger\tag5\label5\\
A_1^\dagger A & = A_2^\dagger A \tag6\label6\\
A_1^\dagger & = A_1^\dagger A A_1^\dagger \\
& = A_1^\dagger A A_2^\dagger \\
& = A_2^\dagger A A_2^\dagger \\
& = A_2^\dagger \\
\end{align}
So, the pseudoinverse is unique.
