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Consider the following matrix $$C=\begin{bmatrix}-A& -B^T\\ -B &0\end{bmatrix}$$ where $A>0$ and B is a matrix such that the diagonal entries of B are all zero and the rest of the entries are either zero or 1, and $(I-B) \mathbf{1}=0$ and $\mathbf{1}^T(I-B)=0$, where $\mathbf{1}$ is the vector of all ones). Do the eigenvalues of C have special properties?

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Is anything else known about $A$? –  user7530 Jan 1 '13 at 21:02
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Are you sure about $B\mathbf 1=0$? If $B$ is a $0/1$ matrix then $B\mathbf 1\ge 0$ componentwise with equality iff $B=0$. –  Hagen von Eitzen Jan 1 '13 at 21:11
    
He means that entries of B are 0, 1 or -1. 1 if there is an edge going from i to j, and -1 if there is an edge going from j to i. –  Calvin Lin Jan 1 '13 at 21:37
    
What if there is both? –  user7530 Jan 1 '13 at 21:41
    
... and $0$ if there's both an edge from $i$ to $j$ and from $j$ to $i$? –  Robert Israel Jan 1 '13 at 21:51
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1 Answer

up vote 1 down vote accepted

Here are some obvious properties (let $A$ be $n\times n$):

  1. As $C$ is real symmetric, all its eigenvalues are real.
  2. The characteristic polynomial of $C$ is $\det(\lambda^2I + \lambda A - B^TB).$
  3. As $A$ is positive definite and the Schur complement of $A$ is $BA^{-1}B^T$, the nullitiy of $C$ is equal to the nullity of $B$. The null space of $C$ consists of vectors of the form $(0,v^T)^T$, where $B^Tv=0$.
  4. $C$ is indefinite. For any $v\in\mathbb{R}^n$, $(\mathbf{1}^T,v^T)C(\mathbf{1}^T,v^T)^T =-\mathbf{1}^TA\mathbf{1}-2(\mathbf{1}^Tv)$. Therefore, one can choose $v$ to make the quantity positive or negative at will.
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can any comments be made from a perron frobenius viewpoint –  dineshdileep Jan 2 '13 at 13:08
    
@dineshdileep Oops. I thought $A>0$ meant $A$ is positive definite (actually I mistook Robert Israel's comment as the OP's). If it means $A$ is entrywise positive instead, then yes, obviously we can say talk about the Perron eigenpair for a negative matrix. –  user1551 Jan 2 '13 at 14:51
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