Is the matrix $(A - D^{T}AD)$ positive definite if the spectral radius of $D$, $\rho(D)<1$? Here $A$ is a positive definite matrix and $D^{T}$ denotes the transpose of $D$. $A$ and $D$ are both real matrices.
My intuition says so but I am not being able to prove rigorously. One can assume $D^{T}AD = DAD^{T}$, if necessary.
On
For the first question, not in general. Indeed, take $A=I$. So, $((I- D^t D)x, x)=(x,x)-(Dx, Dx)$. This can be less than $0$, even if $D$ is nilpotent. .................................................................
THIS IS WRONG:
However, if moreover $D^t A D= DAD^t$, that means $D$ is normal for the inner product $(x,y)_A\colon =(Ax,y)$. $\ \ \rho(D)<1$ implies $(Dx,Dx)_A< (x,x)_A$ for all $x\ne 0$, which is equivalent to $A- D^t AD$ positive definite.
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CORRECTED: No, it does not mean $D$ is normal for that inner product. In fact, the conjugate of $D$ is $A^{-1}D^tA$, so the normality condition is $$DA^{-1}D^tA = A^{-1}D^tA D$$
This is not true. $$ \pmatrix{2&1\\ 1&1}-\pmatrix{1\\ &0}\pmatrix{2&1\\ 1&1}\pmatrix{1\\ &0} =\pmatrix{0&1\\ 1&1} $$ is indefinite. So, if we replace $\operatorname{diag}(1,0)$ by $\operatorname{diag}(1-\epsilon,0)$ for some sufficiently small $\epsilon>0$, the result is indefinite too.