TLDR: Can the shifted power method to loop before finding all possible eigenvalues of a matrix.
Example of what I mean:
Given the shifted power method uses a dominant eigenvalue solver and $\lambda_j$ is the dominant eigenvalue of $A$, it means that for the first shift we find the dominant eigenvalue of $A-\lambda_j\ I$ (assigned $\mu_j$). Now we know from earlier that for all eigenvalues of $A$ (assigned $\lambda_i$) then $\lambda_i-\lambda_j$ is an eigenvalue of $A-\lambda_j\ I$. Thus, we know that $\mu_j$ as the dominant eigenvector (greatest magnitude) must maximise the magnitude output of,
$$\lambda_i-\lambda_j=\mu_j$$
Example, application of method to a matrix $A$ with eigenvalues:
$$\{3,2,-1\}$$
Applying dominant eigenvector solver to $A$ we find $\lambda_j=3$,
Apply shifted power method:
$$2-(3)=-1$$
$$-1-(3)=-4$$
Therefore $\mu_i=-4$ as it has a greater magnitude, thus we find $\lambda_j=-1$
Apply shifted power method:
$$2-(-1)=3$$
$$3-(-1)=4$$
Therefore $\mu_i=4$ as it has a greater magnitude, thus we find $\lambda_j=3$
Apply shifted power method:
$$2-(3)=-1$$
$$-1-(3)=-4$$
Therefore $\mu_i=-4$ as it has a greater magnitude, thus we find $\lambda_j=-1$
Thus, the cycle repeats and we never discover the last eigenvalue $2$.
Does this mean that the shifted power method is not guaranteed to find all the eigenvalues?
Best Answer
Turns out the question contains a reasonable counter-example to the suggestion that the power shifting method alongside a power scaling solver will guarantee that all the eigenvalues of a matrix are obtained.