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Let $A=(a_{i,j})$ be a $n$ x $n$ matrix $(n\geq 2)$, where $a_{i,i} = |x|^2-2x_i ^2$ and $a_{i,j} = -2x_i x_j$ for $i\neq j$. Here $|x|^2 = x_1^2+x_2^2+ \cdots + x_n^2$. I'd like to compute the determinant of $A$.

Calculations suggest that $\det A = -|x|^{2n}$ and it is true for $n=2,3,4$. But I have no idea how to prove/disprove that identity for general $n$.

This matrix came up when computing the Hessian of the function $\phi(\xi) = \log |\xi| + \frac{1}{\tau}x\cdot \xi$ at its critical point $\displaystyle \xi_0 = -\frac{\tau}{|x|^2}x$. I was estimating an oscillatory integral by using the stationary phase method.

2 Answers 2

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Notice that we can write your matrix as $A(x) = |\underline{x}|^2 I - 2\underline{x}\underline{x}^t = |\underline{x}|^2(I-2\underline{v}\underline{v}^t)$

where $\underline{v} = \frac{\underline{x}}{|\underline{x}|}$. So your matrix is basically an Householder matrix, scaled by $|\underline{x}|^2$. In other words, your transformation is a reflection about the vector $\underline{v}$ plus a dilatation of a factor $|\underline{x}|^2$. Now, since an Householder matrix has has one eigenvalue equal to $-1$ and all the others equal to $1$ (easy to prove), it follows that for your matrix

$\lambda_1(A) = -|\underline{x}|^2\qquad \lambda_{2,\ldots,n}=|\underline{x}|^2$ and hence,

$det(A)=-|\underline{x}|^{2n}$

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    Interesting. I just got to know Householder matrix from this answer. Interesting. Thank you very much.2012-10-29
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Because $A$ is symmetric, by the spectral theorem, there is a real orthogonal matrix $Q$ such that $Q^{−1}AQ$ is a diagonal matrix D. The determinant of $A$ will then be the product of the diagonal entries of $D$. It should be possible to find a general formula for $Q$ and $D$ (as $n$ varies...)

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    Thanks for your interest and answer. Bartgol gave a simple proof for the problem.2012-10-29