pseudoinverse of vec-transpose operator 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 $(P+I)^+$.
 A: 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.
A: 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}$.
