pseudoinverse of vec-transpose operator

Question

I'm struggling to find closed form solution for the Moore-Penrose pseudoinverse of the following singular matrix:

$$ P + I $$

where P is a vec-transpose operator matrix, defined by:

$$P=\sum_{ij} E_{ij} \otimes E_{ij}^T$$

where $E_{ij} \in \mathbb{R}^{n \times n}$ is a single-entry matrix. $P\in \mathbb{R}^{n^2\times n^2}$ is a permutation matrix and has following properties (let $A\in\mathbb{R}^{n\times n}$ be some matrix):

• $vec(A^T) = P \cdot vec(A)$
• $P = P^T = P^{-1}$
• $P^2 = I$

What I've got so far is that the problem boils down to finding eigenvectors of $P$, because they are the same as of $P + I$, so I can use SVD formula to obtain $(P+I)^+$.

Answer 1

Figured it out. Define $M=(P+I)$. By definition, a generalized inverse matrix $A^-$ of a matrix $A$, is some matrix that satisfies $AA^-A = A$. By noting that $M^n=2^{n-1}M$ and hence $M^3=4M$, it's easy to see that the matrix $$ M^+=\frac{1}{4}M $$ is indeed a generalized inverse of M. Moreover, one can show that it also satisfies all the necessary conditions to be a Moore-Penrose pseudoinverse.

Answer 2

Here's another way to think about it.

You are basically adding a matrix to it's transpose, $$f(A) = A + A^T,$$ except in vectorized format. Vectorization is linear so you can think of what's going on equally in vectorized or unvectorized format.

The null space of this map are anti-symmetric matrices, whereas for symmetric matrices the map just adds the matrix to itself: $$f(A) = \begin{cases} 2A, & A \text{ is symmetric} \\ 0, & A \text{ is anti-symmetric}. \end{cases}$$

The pseudoinverse is the map that is the exact inverse on the space where the matrix is invertible, and zero on the null space. This is, $$f^+(A) = \begin{cases} \frac{1}{2}A, & A \text{ is symmetric} \\ 0, & A \text{ is anti-symmetric}. \end{cases}$$

Using the expansion of any matrix as a sum of it's symmetric and nonsymmetric parts as well as the linearity of $f^+$, we have $$ \begin{align} f^+(A) &= f^+\left(\frac{A + A^T}{2} + \frac{A - A^T}{2}\right) \\ &= f^+\left(\frac{A + A^T}{2}\right) \\ &= \frac{A + A^T}{4} \end{align} $$

Now it is clear why the vectorized version of this map is "$\text{vec}(f^+)$" $ = \frac{I + P}{4}$.