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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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3 Answers 3

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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.

See

http://mathworld.wolfram.com/DiagonallyDominantMatrix.html

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

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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 definite 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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  • $\begingroup$ By non-negative defined do you mean positive semi-definite? $\endgroup$
    – Ester
    May 20, 2012 at 15:49
  • $\begingroup$ $A$ is non-negative defined if $x^tAx\geq 0$ for all $x$. By positive defined I guess it's $x^tAx>0$ for $x\neq 0$. $\endgroup$ May 20, 2012 at 17:09
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Let $ A^{*} $ be the adjoint of $ A $ and $S$ the positive square root of the positive self-adjoint operator $ S^{2}=A^{*}A $ (e.g. Rudin, ``Functional Analysis'', Mc Graw-Hill, New York 1973, p. 313-314, Th. 12.32 and 12.33) and write $ P=S+A $, $ N=S-A $.

Let $n$ be the finite dimension of $A$ and $\lambda_{i}, i=1\dots n$ its eigenvalues. The eigenvalues of $S$ are $|\lambda_{i}|\ge0$, those of $P$ are $0$ if $\lambda_{i}\le0$ and $2|\lambda_{i}|$ if $\lambda_{i}>0$ and those of $N$ are $0$ if $\lambda_{i}\ge0$ and $2|\lambda_{i}|$ if $\lambda_{i}<0$.

Thus $S$, $P$ and $N$ are positive definite according to the definition given by Rudin in Th. 12.32.

$ A=S-N $ and $ A=(P-N)/2 $ are two possible decomposition of $A$ into the difference of two positive definite operators.

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