1
$\begingroup$

I'm trying to understand a proof that uses a Moore-Penrose (pseudo)inverse. I don't have a very deep understanding of the M-P inverse, but I understand the four-property definition and basic properties that one can find in the Wikipedia page and other introductory sources. However, I can't see how to derive the expression below from them. No doubt the derivation has to do with particular properties involved in the proof (perhaps the role of diagonal matrices or stochastic vectors and matrices in it).

Here are the assumptions for the proof of a lemma on pp. 38-39 of Hartfiel's Markov Set-Chains, Springer 1998. I'm probably giving more detail than necessary, but I don't know what matters. (I am simplifying Hartfiel's notation slightly. His $K$ is my $K$ plus another matrix $F$.)

$X$ is the diagonal matrix for stochastic row vector $x$; $Y$ is the same for stochastic row vector $y$. $\alpha$ and $\beta$ are nonnegative numbers such that $\alpha + \beta = 1$. $A$ and $B$ are stochastic matrices (each row sums to 1) with elements $a_{ij}$ and $b_{ij}$, respectively.

Define matrix $K$ by

$$K = (\alpha X + \beta Y)^+ \,\, (\alpha XA + \beta YB)$$

where $^+$ indicates the Moore-Penrose inverse. Hartfiel claims that

$$k_{ij} = \frac{\alpha x_i a_{ij} + \beta y_i b_{ij}}{\alpha x_i + \beta y_i}$$

if $\alpha x_i + \beta y_i > 0$, or

$$k_{ij} = 0$$

if $\alpha x_i + \beta y_i = 0$.

This is a (to me) surprisingly simple expression. The numerator in the first case is just element $i,j$ of $(\alpha XA + \beta YB)$.

So in this case the effect of multiplying by the M-P inverse of $\alpha X + \beta Y$ is that each element of the resulting matrix is either zero is or simply the result of dividing each element of the right-hand matrix by the weighted sum of the corresponding vector elements. Why is that?

Is there an easy way to see how this follows from one of the Moore-Penrose inverse definitions? (Or do I need a lot more study to understand this expression?!?)

$\endgroup$
2
$\begingroup$

In general, if $D$ is a diagonal matrix and $C=D^+$, then $C$ is also a diagonal matrix with $$ c_{ii}=\begin{cases} d_{ii}^{-1}&\text{ when }d_{ii}\ne0,\\ 0&\text{ when }d_{ii}=0. \end{cases} $$ It is easy to verify that the $D^+$ defined this way satisfies the four defining properties of Moore-Penrose pseudo-inverse. If you permute the standard basis vectors of $\mathbb R^n$ (or $\mathbb C^n$), so that $D=D_1\oplus0$, then the result above is just saying that $D^+=D_1^{-1}\oplus0$. Essentially, it means the Moore-Penrose pseudo-inverse of an invertible matrix $D_1$ is just the usual inverse $D_1^{-1}$, and the M-P pseudo-inverse of the zero matrix is the zero matrix itself.

$\endgroup$
4
  • $\begingroup$ Thanks user1551. You're right--that is a reasonable inference from what I wrote, but looking at the book again, it's not correct as it happens. I'll edit the question. I am curious why $\alpha X + \beta Y$ being invertible follows from that assumption, though. Perhaps it's not simple to explain though. I confess that although I have a basic understanding of how to invert a nonsingular matrix, insights about inverse matrices don't easily flow out of me. :-) $\endgroup$ – Mars Jul 25 '17 at 17:22
  • $\begingroup$ Thanks user1551. This is very helpful. What is $\oplus$, though? This page lists it as symmetric set difference (clearly not what you intend). Answers here and here, and here say that's it's the direct sum. $\endgroup$ – Mars Jul 26 '17 at 18:39
  • $\begingroup$ If direct sum is what you meant, I don't know how to apply it to matrices. Some of the answers also mention as exclusive or, i.e. symmetric set difference again. I see how the direct sum concept applies to vector spaces here, but you're applying it to individual vectors. Maybe I'm just being dense. $\endgroup$ – Mars Jul 26 '17 at 18:39
  • $\begingroup$ In any event, the important point for me is that the M-P inverse is also diagonal with reciprocals or zeros depending on whether the element of the original matrix was nonzero. That's exactly the information I needed, and I will think through the rest. Thank you. $\endgroup$ – Mars Jul 27 '17 at 1:31

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.