##### Activity5.2.3

The key to finding the eigenvalue of \(A\) having the smallest absolute value is to note that the eigenvectors of \(A\) are the same as those of \(A^{-1}\text{.}\)

If \(\vvec\) is an eigenvector of \(A\) with associated eigenvector \(\lambda\text{,}\) explain why \(\vvec\) is an eigenvector of \(A^{-1}\) with associated eigenvalue \(\lambda^{-1}\text{.}\)

Explain why the eigenvalue of \(A\) having the smallest absolute value is the reciprocal of the dominant eigenvalue of \(A^{-1}\text{.}\)

Explain how to use the power method applied to \(A^{-1}\) to find the eigenvalue of \(A\) having the smallest absolute value.

If we apply the power method to \(A^{-1}\text{,}\) we begin with an intial vector \(\xvec_0\) and generate the sequence \(\xvec_{k+1} = A^{-1}\xvec_k\text{.}\) It is not computationally efficient to compute \(A^{-1}\text{,}\) however, so instead we solve the equation \(A\xvec_{k+1} = \xvec_k\text{.}\) Explain why an \(LU\) factorization of \(A\) is useful for implementing the power method applied to \(A^{-1}\text{.}\)

The following Sage cell defines a command called

`inverse_power`that applies the power method to \(A^{-1}\text{.}\) That is,`inverse_power(A, x0, N)`prints the vectors \(\xvec_k\text{,}\) where \(\xvec_{k+1} = A^{-1}\xvec_k\text{,}\) and multipliers \(\frac{1}{m_k}\text{,}\) which approximate the eigenvalue of \(A\text{.}\) Use it to find the eigenvalue of \(A=\left[\begin{array}{rr} -5.1 \amp 5.7 \\ -3.8 \amp 4.4 \\ \end{array}\right]\) having the smallest absolute value.The inverse power method only works if \(A\) is invertible. If \(A\) is not invertible, what is its eigenvalue having the smallest absolute value?

Use the power method and the inverse power method to find the eigenvalues and associated eigenvectors of the matrix \(A = \left[\begin{array}{rr} -0.23 \amp -2.33 \\ -1.16 \amp 1.08 \\ \end{array}\right] \text{.}\)