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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 $\lambda_1$ ?

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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.

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  • $\begingroup$ The eigenvalues are: wolframalpha.com/input/… So, can we get for example $s=-5$ ? $\endgroup$ – Mary Star Jan 19 '18 at 16:31
  • $\begingroup$ if you select s=-5, you should converge to the eigenvector (0.0967097, -6.70119, 1) $\endgroup$ – Waterhouse Jan 19 '18 at 16:35
  • $\begingroup$ So, we have to find the inverse matrix $(A+5I)^{-1}$ and then apply the power method for the matrix $(A+5I)^{-1}$ and the initial vector $x^{(0}$, right? $\endgroup$ – Mary Star Jan 19 '18 at 18:23
  • $\begingroup$ You choose a $v_0$, an initial guess for the eigenvector ( not [0,0,0] ) and then you iterate as following: $v_{i+1}= \frac{(A+5I)^{-1}v_{i}}{||(A+5I)^{-1}v_{i}||} $ and you should compute the inverse once, not for every iteration. $\endgroup$ – Waterhouse Jan 19 '18 at 22:28
  • $\begingroup$ As initial vector it is given the $x^{(0)}=(-1,0,0)^T$. What do you mean by "you should compute the inverse once, not for every iteration" ? I have calculated (if I am not mistaken) $$\|(A+5I)^{-1}x^{(3)}\|_{\infty}=\frac{9255292892}{1414521867}\approx 6.54$$ Can we conclude from that something about the eigenvalues of $A$? Does it maybe hold that the eigenvalue of $A+5I$ will then be $\frac{1}{6.5}$ and the eigenvalue of $A$ is $\lambda$ such that $|\lambda+5|=\frac{1}{6.5}$ ? $\endgroup$ – Mary Star Jan 19 '18 at 23:13

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