Let P be a symmetric positive definite matrix. Without losing generality, assume P is of dimension three, i.e.,
$$\mathbf{P} = \begin{bmatrix}P_1&P_2&P_3\\P_2&P_4&P_5\\P_3&P_5&P_6 \end{bmatrix} $$.
Let I be the identity matrix of the same dimension as P. Let
$$\mathbf{I}_1 = \begin{bmatrix}1 & 0 & 0\\0 & 0 & 0\\ 0 & 0 & 0\end{bmatrix},$$
$$\mathbf{I}_2 = \begin{bmatrix}0 & 0 & 0\\0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix},$$
$$\mathbf{I}_3 = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}.$$
Then, $$\lambda((\mathbf{I}+\mathbf{P}\mathbf{I}_1)^n (\mathbf{I}+\mathbf{P}\mathbf{I}_2)^n(\mathbf{I}+\mathbf{P}\mathbf{I}_3)^n) >1 \ \forall n\geq 0,$$ where $\lambda$ is an arbitrary eigenvalue of the matrix in the parentheses.
The observation above is actually valid for all symmetric positive definite matrices of all dimensions. This conclusion is arrived by running more than 20,000 random matrix tests, i.e., more than 20,000 random P matrix is generated.
I tried to prove this property by using the properties of positive definite matrix and by connecting the two facts that
- $\lambda(\mathbf{I+P})>1$
- $\mathbf{I}+\mathbf{P}\mathbf{I}_1+\mathbf{I}+\mathbf{P}\mathbf{I}_2+\mathbf{I}+\mathbf{P}\mathbf{I}_3 = 3\mathbf{I}+\mathbf{P}$.
However, I did not have any progress. Please share me with your idea once you find any approach which may probably prove that.