$(I-cP)A^TA$ positive-definite?

Question

I would like to know if $(I-cP)A^TA$ is positive-definite when $A^TA$ is a positive-definite matrix, $I$ is the identity matrix, $P$ is a transition probability matrix, and $c$ a positive scalar with $c<1$. Using the Gershgorin circle theorem, I know $I-cP$ is positive-definite. By setting $B=(I-cP)A^TA$ and $\overline{B}=A(I-cP)A^T$ we have $\overline{B}A=AB$. $\overline{B}$ is semi positive-definite since $A$ is full column-rank. With the Lyapunov theorem, I can just prove $B$ has positive eigenvalues. Experimentally, I did not find counterexamples yet.

Answer 1

I would like to know if $(I-cP)A^TA$ is positive-definite when $A^TA$ is a positive-definite matrix.

No. Consider e.g. $A=I$ and $$ I-cP=I-0.9\pmatrix{1&0\\ 1&0}=\pmatrix{0.1&0\\ -0.9&1}, \\ \quad S=\frac{(I-cP)+(I-cP)^T}{2}=\pmatrix{0.1&-0.45\\ -0.45&1}. $$ The symmetric part $S$ of $I-cP$ is not positive definite, as it has a negative determinant.

Using the Gershgorin circle theorem, I know $I-cP$ is positive-definite.

No. The $I-cP$ above already offers a counterexample. Gershgorin only tells you that $I-cP$ is positive-stable (i.e. its eigenvalues has positive real parts), not that its symmetric part is positive definite. Note that the eigenvalues of the symmetric part of a matrix $M$ are in general not the real parts of the eigenvalues of $M$.