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
$
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.
