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?
Best Answer
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.