1
$\begingroup$

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 $(P+I)^+$.

$\endgroup$
1
$\begingroup$

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}$.

$\endgroup$
  • $\begingroup$ neat, very intuitive. cheers! $\endgroup$ – dmytro Jun 28 '14 at 11:29
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.