We have the matrix \begin{equation*}A=\begin{pmatrix}7 & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{4} & -5 & 1 \\ \frac{1}{2} & 1 & \frac{3}{2}\end{pmatrix}\end{equation*}
with the eigenvalues $\lambda_1 < \lambda_2 < \lambda_3$.
Using the Gershgorin lemma I want to estimate the smallest eigenvalue $\lambda_1$ (using the smallest center of a Gershgorin circle).
The circles are \begin{align*}&K_1=\left \{|x-7|\leq \left |\frac{1}{4}\right |+\left |\frac{1}{2}\right |\right \}=\left \{|x-7|\leq 0.75\right \} \\ &K_2=\left \{|x-(-5)|\leq \left |\frac{1}{4}\right |+|1|\right \}=\left \{|x+5|\leq 1.25\right \} \\ &K_3=\left \{\left |x-\frac{3}{2}\right |\leq \left |\frac{1}{2}\right |+|1|\right \}=\left \{\left |x-\frac{3}{2}\right |\leq 1.5\right \} \end{align*}
Therefore, we get \begin{align*} -6.25\leq &\lambda_1 \leq -3.75 \\ 0\leq &\lambda_2\leq 3 \\ 6.25\leq &\lambda_3 \leq 7.75 \end{align*} right?
Next, I want to make a shift around this estimate and calculate approximately the eigenvalue using the Inverse Iteration (3 steps) with initial vector $x^{(0)} = (−1,0,0)^T$.
We have to consider the matrix $(A-sI)^{-1}$ and apply the power method, or not?But which $s$ do we have to take? Do we maybe consider as $s$ a number in the interval $[-6.25, -3.75]$, i.e. the interval of $\lambda_1$ ?
When you choose s to to be within domain $K_1$, the inverse powermethod will converge to the eigenvektor of the eigenpair $(\lambda_1, v_1)$. The closer s to the actual value of $\lambda_1$ is, the faster is the convergence.