Let $A \in \mathbb{R}^{nxn}$. $A = \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix}$.
Given the following two initial vectors for the power method,
- $x_0 = (1, 0)$
- $y_0 = (1, -1)$
I need to find the values of $\alpha \in \mathbb{R}$ for which the power method converges.
Doing the initial product $A \cdot x_o$:
- $A(1,0) = (1, \alpha)$, and
- $A(1,-1) = (1+\alpha, 1-\alpha)$.
So I can see from there that in neither case will the first product be zero, which would make the algorithm not converge to the dominant eigenvalue.
I could extract, calculating the determinant of $A$, that it will have two eigenvalues:
- $\lambda_1 = 1 + \frac{1}{2}|\alpha|$
- $\lambda_2 = 1 - \frac{1}{2}|\alpha|$
But I don't know how to proceed to find a restriction on $\alpha$.
Are there other conditions that are needed to ensure that the power method will converge?