# Symmetric Matrix as the Difference of Two Positive Definite Symmetric Matrices

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 .

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).

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.

• By non-negative defined do you mean positive semi-definite? May 20, 2012 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 $x\neq 0$. May 20, 2012 at 17:09

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.