# 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?

• The pseudoinverse is unique. Oct 27, 2019 at 5:33
• @MishaLavrov, can you point me to a source? Oct 27, 2019 at 7:01
• My source is Wikipedia, so I don't have enough to go on to write an answer. Oct 27, 2019 at 14:00

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.
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.