Determining eigenvalues with limited information

Question

The following question is from a System Theory test with only answers (no solutions). Maybe someone here knows how to tackle it.

Consider the discrete time system $$x(k+1) = Ax(k)$$ with a matrix $$P = \begin{bmatrix} 8 & 2\\ 2 & 4 \end{bmatrix}$$ such that $$A^TPA-P=- \begin{bmatrix} 4 & 1 \\ 1 & 2 \end{bmatrix}$$ Which of the following statements is correct about the location of the eigenvalues of $A$?

1. $A$ has only eigenvalues with modulus (absolute value) between $0.8$ and $1$.

2. $A$ has only eigenvalues with modulus (absolute value) between $0.6$ and $0.8$.

3. $A$ has only eigenvalues with modulus (absolute value) between $0.4$ and $0.6$.

4. $A$ has only eigenvalues with modulus (absolute value) smaller than $0.4$.

5. None of the above.

I thought about moving $P$ to the other side of the equals sign but that doesn't get me any further:

$$A^T \begin{bmatrix} 8 & 2\\ 2 & 4 \end{bmatrix}A = \begin{bmatrix} 4 & 1\\ 1 & 2 \end{bmatrix}$$

Thanks in advance for the help.

Answer 1

First note that $A^T P A-P=-\frac{1}{2}P$

So, $A^T P A=\frac{1}{2}P$

Let $Av=\lambda v$.

Then, we note that $v^T A^T=\lambda v^T$. So in particular,

$v^TA^T P A v=\frac{1}{2}v^TPv$

$\lambda^2 v^T P v=\frac{1}{2} v^T P v$

Note that $v^T P v>0$ for a non-zero vector $v$. ($v^T Pv$ defines a positive definite bilinear form or a inner product)

So $\lambda^2=\frac{1}{2}$