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Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?


Approach:

The matrix of this quadratic form can be derived to be the following

$$M := \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\ \end{pmatrix}$$

It suffices to show that $\operatorname{det}M > 0$, then the claim follows.

Any hints how to show the positivity of this determinant?

  • 0
    I don't understund why $\det M > 0 $ suffices. For example the matrix : $$A=\left(\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&-1\end{array} \right)$$ has a $ > 0 $ determinant but this matrix is note poistive !2012-06-17
  • 0
    @Mohamed, I am not at all sure that the person asking knows this, but it follows by induction using http://en.wikipedia.org/wiki/Sylvester%27s_criterion2012-06-17
  • 0
    It only suffices if we prove it for all $m=1..n$2012-06-17
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    @ leonbloy : this in general but for the case thank you very much @ Will Jgay who explained the real reason: it is the nature of this matrix that allows a proof of this fact by induction.2012-06-17

7 Answers 7