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Given a matrix A, I need determine the number of positive, negative and zero eigen value of it without actually computing it?. The matrix is given below

\begin{pmatrix}0&...&0 & 1\\ 0&...&0 & 2\\ .&.&. &.\\ .&.&. &.\\ 0&...&0 &n-1\\ 1&...&n-1&n \end{pmatrix}

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up vote 3 down vote accepted

Assume $n > 1$.

Clearly vectors $v_k = -k \mathbf{e}_1 + \mathbf{e}_k$ are zero eigenvalues one for $1 < k < n$. Thus there are $n-2$ zero eigenvalues.

The trace of the matrix equals $n$, thus the sum of all eigenvalues is positive. All eigenvalues are real, because the matrix is symmetric. Thus, there is 1 positive, one negative, and $n-2$ zero eigenvalues.

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Dear Sasha: Wonderful! You may want to spell out the reason why there is a negative eigenvalue. – Pierre-Yves Gaillard Nov 2 '11 at 5:56
@Pierre-YvesGaillard, because the matrix is not positive semidefinite (there are non-zero off-diagonal elements, while the diagonal elements are zero). – Tapu Nov 2 '11 at 6:04
Dear Sasha: Thanks for your comment. I had realized that. I was just suggesting that you might include this point to your answer, for the sake of completeness... – Pierre-Yves Gaillard Nov 2 '11 at 6:18
Dear Pierre, I am shamefully admitting it was me (swapan), not Sasha, who missed your comment. – Tapu Nov 2 '11 at 6:54
Dear @Swapan: Sorry, I didn't read the signature in your first comment above. I assumed it was written by Sasha. So thank you for having answered my question. (But I still think that adding a short sentence to Sasha's already outstanding answer would make it even better.) – Pierre-Yves Gaillard Nov 2 '11 at 10:15

An alternative proof could use Sylvester's law of inertia i.e. the number of positive, negative and zero eigenvalues of a given symmetric matrix $A$ does not change under a change of coordinates $A \rightarrow T'AT$ ($\det T \neq 0$).

Since you can easily use the $1$ in the first row/ column to eliminate any non-zero entries up to the $n$ in the lower right corner you are basically left with the task to determine the number of positive and negative eigenvalues of $B=\begin{pmatrix} n && 1 \\ 1 && 0 \end{pmatrix}$. By computation or the same argument that was stated by @Sasha we get that $B$ has one positive and one negative Eigenvalue and therefor A has one positive, one negative and n-2 zero eigenvalues.

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In general, when $u,v\in\mathbb{R}^n$ are linearly independent, $uv^\top+vu^\top$ always have one positive eigenvalue, one negative eigenvalue and $n-2$ zero eigenvalues. See an explanation in my answer to another question a week ago. In your case, we have $u^\top = (1,2,\ldots,n-1,\frac{n}{2})$ and $v^\top=(0,0,\ldots,0,1)$.

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Very nice! That way I was able to make you earn 20 points instead of just 10... – Pierre-Yves Gaillard Nov 2 '11 at 11:00
Thanks for the points. ;-D – user1551 Nov 2 '11 at 11:26

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