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Given this symmetric matrix-$A$:

$\begin{pmatrix} 14 &14 & 8 &12 \\ 14 &17 &11 &14 \\ 8& 11 &11 &10 \\ 12 & 14 &10 & 12 \end{pmatrix}$

I need to find $p,q$ such that $p$ is the number of 1's and $q$ is the number of -1's in the diagonalizable matrix $D_{p,q}$ such that $D_{p,q}$= Diag {$1,1,\ldots 1,-1,-1, \ldots-1,0,0,\ldots0$}.

$D=P^{t}AP$ while $P$ is the the matrix that contains the eigenvectors of $A$ as a Columns.

I tried to use Jacobi method but I found out that $|A|=0$, so I can't use it, but I know now that $0$ is an eigenvalue of $A$, So Do I really need to compute $P$ in order to find $p$ and $q$? It's a very long and messy process.

Thank you

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up vote 0 down vote accepted

The characteristic polynomial of $A$ is $P(x)= x^4 - 54x^3 + 262x^2 - 192x $. It has $0$ as a simple root, and the other three are positive. Therefore $A$ has three positive eigenvalues and one equal to zero. Since the signature can be obtained from the signs of the eigenvalues, we are done. Therefore $p=3,q=0$.

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