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This question inquired about the determinant of this matrix: $$ \begin{bmatrix} -\lambda &1 &0 &1 &0 &1 \\ 1& -\lambda &1 &0 &1 &0 \\ 0& 1& -\lambda &1 &0 &1 \\ 1& 0& 1& -\lambda &1 &0 \\ 0& 1& 0& 1& -\lambda &1 \\ 1& 0& 1& 0&1 & -\lambda \end{bmatrix} $$ and of other matrices in a sequence to which it belongs. In a comment I mentioned that if we permute the indices 1, 2, 3, 4, 5, 6 to put the odd ones first and then the even ones, thus 1, 3, 5, 2, 4, 6, then we get this: $$ \begin{bmatrix} -\lambda & 0 & 0 & 1 & 1 & 1 \\ 0 & -\lambda & 0 & 1 & 1 & 1 \\ 0 & 0 & -\lambda & 1 & 1 & 1 \\ 1 & 1 & 1 & -\lambda & 0 & 0 \\ 1 & 1 & 1 & 0 & -\lambda & 0 \\ 1 & 1 & 1 & 0 & 0 & -\lambda \end{bmatrix} $$ So this is of the form $$ \begin{bmatrix} A & B \\ B & A \end{bmatrix} $$ where $A$ and $B$ are symmetric matrices whose characteristic polynomials and eigenvalues are easily found, even if we consider not this one case of $6\times 6$ matrices, but arbitrarily large matrices following the same pattern.

Are there simple formulas for determinants, characteristic polynomials, and eigenvalues for matrices of this latter kind?

I thought of the Haynesworth inertia additivity formula because I only vaguely remembered what it said. But apparently it only counts poistive, negative, and zero eigenvalues.

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I suppose we could treat it as a symmetric block Toeplitz matrix... –  J. M. Jul 28 '11 at 2:41
    
...and if we so treat it, does that lead us toward an answer to the question? –  Michael Hardy Jul 28 '11 at 3:41
    
I suppose, but I'll have to dig through my notes to be sure (I do remember these things being well-studied, but I'm hazy on how the eigenproblem simplifies for these). –  J. M. Jul 28 '11 at 3:43
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i would suggest applying schur-decomposition for the blocks. –  Peter Sheldrick Jul 28 '11 at 11:40
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4 Answers

up vote 6 down vote accepted

We have $$ \det \left( \begin{array}{cc} A & B\\ C & D \end{array} \right) = \det(A-BD^{-1}C) \det(D), $$ where the matrix $A-BD^{-1}C$ is called a Schur complement. In your case, $A=D=-\lambda I_n$ and $B=C=J_n$ = the order $n$ matrix with all entries equal to 1. So, the RHS is equal to $\det(-\lambda I_n + \frac{n}{\lambda} J_n) \det(-\lambda I_n) = (-n)^n \det(-\frac{\lambda^2}{n}I_n + J_n)$. If I remember correctly, $\det(xI_n + J_n) \equiv x^{n-1}(x+n)$, but you should check whether this is true or not.

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Very nice. I should have thought of that identity, since I believe I used it when I took a course on the theory of the Wishart distribution. –  Michael Hardy Jul 28 '11 at 19:04
    
The only matrices whose determinants remain to be evaluated after what user1551 did above are matrices in which all diagonal entries are equal to each other and all off-diagonal entries are equal to each other. There is a standard formula, moderately easy to prove, for the determinant of such a matrix. –  Michael Hardy Jul 28 '11 at 21:28
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I am not sure whether I understand what you want to ask.. but the following are some facts on the matrix of this type

$\det\begin{bmatrix} A & B \\\\ B & A \end{bmatrix}=\det(A+B)\det(A-B)$. The eigenvalues of $\begin{bmatrix} A & B \\\\ B & A \end{bmatrix}$ are the union of eigenvalues of $A+B$ and the eigenvalues of $A-B$.

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Because the subblocks of the second matrix (let's call it $C$) commute i.e. AB=BA, you can use a lot of small lemmas given, for example here.

And also you might consider the following elimination: Let $n$ be the size of $A$ or $B$ and let,(say for $n=4$) $$ T = \left(\begin{array}{cccccccc} 1 &0 &0 &0 &0 &0 &0 &0\\ 0 &0 &0 &0 &1 &0 &0 &0\\ -1 &1 &0 &0 &0 &0 &0 &0\\ -1 &0 &1 &0 &0 &0 &0 &0\\ -1 &0 &0 &1 &0 &0 &0 &0\\ 0 &0 &0 &0 &-1 &1 &0 &0\\ 0 &0 &0 &0 &-1 &0 &1 &0\\ 0 &0 &0 &0 &-1 &0 &0 &1 \end{array} \right) $$ Then , $TCT^{-1}$ gives $$ \hat{C} = \begin{pmatrix}-\lambda &n &\mathbf{0} &\mathbf{1} \\n &-\lambda &\mathbf{1} &\mathbf{0}\\ & &-\lambda I &0\\&&0&-\lambda I \end{pmatrix} $$

from which you can identify the upper triangular block matrix. The bold face numbers indicate the all ones and all zeros rows respectively. $(1,1)$ block is the $2\times 2$ matrix and $(2,2)$ block is simply $-\lambda I$.

EDIT: So the eigenvalues are $(-\lambda-n),(-\lambda+n)$ and $-\lambda$ with multiplicity of $2(n-1)$. Thus the determinant is also easy to compute, via their product.

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Your $2n\times 2n$ matrix $M$ acts on the vector space $V=\mathbb C^n\oplus\mathbb C^n$. Now if $W_1=\{(v,v):v\in\mathbb C^n\}$ and $W_2=\{(v,-v):v\in\mathbb C^n\}$, then we also have $V=W_1\oplus W_2$. Moreover, both $W_1$ and $W_2$ are invariant under $M$, so to find the eigenvalues/eigenvectors/characteristic polynomial/etc, it is enough to do it for those restrictions: they are $A+B$ and $A-B$.

This way you obtain, for example, the facts mentioned in Sunni's answer immediately.

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