lizfcm/notes/Nov-3.org

• eigenvalues → power method
  • We know:
    • However, calculating $A^{-1}$ is inefficient
  • Conclusions
• Eigenvalue Shifting

eigenvalues → power method

we iterate on the x_k+1 = A x_k

y = Av_0 v_1 = \frac{1}{|| y ||} (y) λ_0 = v_0^T A v_0 = v_0^T y

Find the largest eigenvalue;

  while (error > tol && iter < max_iter) {
    v_1 = (1 / magnitude(y)) * y;
    w = m_dot_v(a, v_1);
    lambda_1 = v_dot_v(transpose(v_1), w);
    error = abs(lambda_1 - lambda_0);
    iter++;
    lambda_0 = lambda_1;
    y = v_1;
  }

  return [lambda_1, error];

Find the smallest eigenvalue:

We know:

If λ_1 is the largest eigenvalue of $A$ then \frac{1}{λ_1} is the smallest eigenvalue of $A^{-1}$.

If λ_n is the smallest eigenvalue of $A$ then \frac{1}{λ_n} is the largest eigenvalue of $A^{-1}$.

However, calculating $A^{-1}$ is inefficient

So, transform $w = A^{-1} v_1 \Rightarrow$ Solve $Aw = v_1$ with LU or GE (line 3 of above snippet).

And, transform $y = A^{-1} v_0 \Rightarrow$ Solve $Ay = v_0$ with LU or GE.

Conclusions

We have the means to compute the approximations of λ_1 and λ_n.

(λ_1 → power method)

(λ_n → inverse power method)

Eigenvalue Shifting

If (λ, v) is an eigen pair, (v ≠ 0)

Av = λv

Thus for any μ ∈ R

(Av - μ I v) = (A - μ I)v = λ v - μ I v = (λ - μ)v ⇒ λ - μ is an eigenvalue of (A - μ I)

(A - μ I)v = (λ - μ)v

Idea is to choose μ close to our eigenvalue. We can then inverse iterate to construct an approximation of λ - μ and then add μ back to get λ.

v_0 = a_1 v_1 + a_2 v_2 + ⋯ + a_n v_n A v_0 = a_1 (λ_1 v_1) + ⋯