Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a Laplacian matrix $A\in\mathbb{R}^{n\times n}$ (symmetric, positive semi-definite matrix with positive diagonal elements and non-positive off-diagonal), and its Moore-Penrose pseudoinverse $A^+$, why is $$A^+Ab=b,$$ for some vector $b\in\mathbb{R}^n$ whose elements sum to zero. Note that the null-space of $A$ is spanned by vector of all ones, $1_n=(1,1,...1)\in\mathbb{R}^n$.

I find that $A^+=(A+1_n1_n^T)^{-1}-n^{-2}1_n1_n^T$, but the second term will vanish in multiplication with $A$.

share|improve this question
1  
What is $A^+$? $ $ –  Martin Argerami Nov 26 '12 at 17:02
    
@MartinArgerami see the edit. –  user506901 Nov 26 '12 at 17:09
    
Thanks. $\ \ \ \ \ $ –  Martin Argerami Nov 26 '12 at 17:11

1 Answer 1

up vote 2 down vote accepted

Since $A$ is real symmetric, it can be orthogonally diagonalized. Let's say $A=Q\ \mathrm{diag}(\lambda_1,\ldots,\lambda_r,0,\ldots,0)\,Q^T$, where $Q$ is orthogonal and $\lambda_1,\ldots,\lambda_r$ are its nonzero eigenvalues. As the nullspace of $A$ is one-dimensional and spanned by $1_n$, we have $r=n-1$ and the last column of $Q$ is $\pm\frac1{\sqrt{n}}1_n$. Furthermore, since $A$ is positive semidefinite, its eigenvalue decomposition is automatically a singular value decomposition. Hence $A^+A=Q(I_{n-1}\oplus0)Q^T$. Now let $b$ be any of the first $n-1$ columns of $Q$. Then $AA^+b=b$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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