Try demmel's implementation.
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-13
@@ -3,6 +3,7 @@ import numpy.linalg as LA
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"""
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Classic Lanczos method (without re-orthogonalization)
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Using Demmel's book version.
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Arguments
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L : Real valued NxN symmetric matrix
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@@ -23,22 +24,20 @@ beta : ndarray
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def lanczos(L, s, M):
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N = len(s)
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alp = np.zeros(M)
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beta = np.zeros(M - 1)
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V = np.zeros((N, M))
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V[:, 0] = s / LA.norm(s)
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beta = np.zeros(M)
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V = np.zeros((N, M + 1))
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V[:, 1] = s / LA.norm(s)
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for j in range(M):
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for j in range(1, M):
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w = L @ V[:, j]
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alp[j] = np.dot(V[:, j], w)
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v_tilde = w - V[:, j] * alp[j]
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if j > 0:
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v_tilde = v_tilde - V[:, j - 1] * beta[j - 1]
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w = w - V[:, j] * alp[j] - V[:, j - 1] * beta[j - 1]
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if j < M - 1:
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beta[j] = LA.norm(v_tilde)
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if beta[j] == 0:
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break
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V[:, j + 1] = v_tilde / beta[j]
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beta[j] = LA.norm(w)
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if beta[j] == 0:
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print("Breakdown")
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break
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V[:, j + 1] = w / beta[j]
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return [V, alp, beta]
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return [V[:, 1:], alp, beta[1:]]
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