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Prove that any real symmetric matrix can be expressed as the difference of two positive definite symmetric matrices.

I was trying to use the fact that real symmetric matrices are diagonalisable , but the confusion I am having is that 'if $A$ be invertible and $B$ be a positive definite diagonal matrix, then is $ABA^{-1}$ positive definite' .

Thanks for any help .

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up vote 4 down vote accepted

Let $S$ be your symmetric matrix. You can now add a large positive multiple of the identity matrix. This ensures that your matrix $S+c I$ is diagonally dominant and symmetric, and thus positive definite.


Now, you clearly have $S= (S+cI)-cI$ (and $c I$ is certainly positive definite as well).

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Thank you very much . That helped a lot . – Ester May 20 '12 at 16:18

Let $S$ a symmetric matrix; we can write it as $P^tDP$, where $P$ is orthogonal and $D$ diagonal. We can write $D=\operatorname{diag}(\lambda_j,1\leq j\leq n)$. Put $D_1:=\operatorname{diag}(\lambda_j^+,1\leq j\leq n)$, $D_2:=\operatorname{diag}(\lambda_j^-,1\leq j\leq n)$ where $\lambda_j^+$ and $\lambda_j^-$ are the positive and negative parts of $\lambda_j$. Then $P^tD_1P$ and $P^tD_2P$ are non-negative defined and $D_1-D_2=D$. Now use $P^tD_1P+\varepsilon I$ and $P^tD_2P+\varepsilon I$ for $\varepsilon$ small enough to get the result.

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By non-negative defined do you mean positive semi-definite? – Ester May 20 '12 at 15:49
$A$ is non-negative defined if $x^tAx\geq 0$ for all $x$. By positive defined I guess it's $x^tAx>0$ for $x\neq 0$. – Davide Giraudo May 20 '12 at 17:09
Yes . Thanks for your help . – Ester May 21 '12 at 2:11

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