Finding Schur Form¶

Copyright (C) 2026 Andreas Kloeckner

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In [1]:
import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as plt

np.set_printoptions(precision=3, linewidth=120)

Make a nice, well-behaved, diagonalizable matrix with real eigenvalues:

In [2]:
n = 6
A = np.random.randn(n, n)
U, sigma, VT = la.svd(A)
sigma = 1 + np.arange(n)
X = U + 0.1*np.random.randn(n, n)
A = la.inv(X) @ np.diag(sigma) @ X
In [11]:
def rayleigh_quotient(mat, x):
    return x @ (mat@x) / (x@x)
    
def find_an_eigenpair(mat):
    n = len(mat)
    x = np.random.randn(n)
    while True:
        sigma = rayleigh_quotient(mat, x)
        if la.norm(mat@x-sigma*x)/(la.norm(sigma*x)) < 1e-13:
            break
        try:
            x = la.solve(mat - sigma * np.eye(n), x)
        except la.LinAlgError:
            # singular? it's an eigenvalue
            return sigma, x 
        x = x / la.norm(x)

    return rayleigh_quotient(mat, x), x
In [41]:
lam, x = find_an_eigenpair(A)

print(A@x - lam*x)

del lam
del x

[ 0.000e+00  0.000e+00  0.000e+00 -7.459e-17  4.441e-16  0.000e+00]
In [20]:
i = 0
Q = np.eye(n)
R = A.copy()

Based on an eigenpair, go column-by-column to find Schur form.

NOTE: Please don't do this in real life. It's $O(n^4)$ and numerically not fantastic. But it does (constructively) support the notion that Schur form exists!

In [26]:
lam, xeig = find_an_eigenpair(R[i:, i:])

eig_onb = np.eye(n)
eig_onb[i:, i:], _ = la.qr(np.hstack((xeig.reshape(-1, 1), np.random.randn(n-i, n-i-1))))
R = eig_onb.T @ R @ eig_onb

i += 1
R.round(3)
Out[26]:
array([[ 3.   ,  0.107, -0.218, -0.15 ,  0.09 , -0.136],
       [-0.   ,  4.   ,  0.048, -0.339,  0.333,  0.644],
       [-0.   , -0.   ,  5.   , -0.024,  0.216,  1.048],
       [-0.   , -0.   , -0.   ,  6.   ,  0.982, -2.186],
       [-0.   ,  0.   , -0.   ,  0.   ,  2.   ,  0.04 ],
       [-0.   ,  0.   ,  0.   ,  0.   ,  0.   ,  1.   ]])
In [ ]: