Approximation of eigenvalue with inverse iteration method

[mathmari]

Hey! :giggle:

We have the matrix $\begin{pmatrix}2 & 1/2 & 1 \\ 1/2 & 3/2 & 1/2 \\ 1 & 1/2 & 2\end{pmatrix}$.
We take as initial approximation of $\lambda_2$ the $1.2$. We want to calculate this value approximately using the inverse iteration (2 steps) using as starting vector $x^{(0)}=\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}$.

At the inverse iteration method do we have to use at each step the Rayleigh-Quotient or only at the beginning and then just the power iteration ?
I think to get a better approximation that we have to use the Rayleigh-Quotient at each step. Is that correct?

:unsure:

 

[I like Serena]

Hey mathmari!

It seems to me that we cannot use the power iteration method, since it can only find the eigenvalue with the greatest magnitude.
Instead we can use the inverse iteration method, which finds an eigenvector given an approximation of the eigenvalue, which does not improve the eigenvalue.
Or we can use the Rayleigh quotient iteration method, which is an extension of the inverse iteration method. It improves the eigenvalue as well as the eigenvector.

Since the initial eigenvalue approximation is still a bit off, it seems to me that it's best to use the Rayleigh quotient iteration for every iteration.
Or else we should try some variants and see what works best. :unsure:

 

[mathmari]

It seems to me that we cannot use the power iteration method, since it can only find the eigenvalue with the greatest magnitude.
Instead we can use the inverse iteration method, which finds an eigenvector given an approximation of the eigenvalue, which does not improve the eigenvalue.
Or we can use the Rayleigh quotient iteration method, which is an extension of the inverse iteration method. It improves the eigenvalue as well as the eigenvector.

Since the initial eigenvalue approximation is still a bit off, it seems to me that it's best to use the Rayleigh quotient iteration for every iteration.
Or else we should try some variants and see what works best. :unsure:

Ok! Thank you for your answer!