Theory Notebook

Converted from theory.ipynb for web reading.

Systems of Equations — From Hyperplanes to Solvers

Every exact solve, every least-squares fit, every Newton step, and every constrained update in AI is a system-of-equations problem in disguise.

This notebook is the interactive companion to notes.md. It is intentionally arranged as a learning path rather than a section-by-section copy of the note.

This notebook was shaped using the local course style plus public materials from MIT 18.06, Stanford EE263, Netlib's Templates for the Solution of Linear Systems, public numerical linear algebra notebook resources, and the repository notation / chapter guides.

---

Code cell 3

import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl

try:
    import seaborn as sns
    sns.set_theme(style="whitegrid", palette="colorblind")
    HAS_SNS = True
except ImportError:
    plt.style.use("seaborn-v0_8-whitegrid")
    HAS_SNS = False

mpl.rcParams.update({
    "figure.figsize":    (10, 6),
    "figure.dpi":         120,
    "font.size":           13,
    "axes.titlesize":      15,
    "axes.labelsize":      13,
    "xtick.labelsize":     11,
    "ytick.labelsize":     11,
    "legend.fontsize":     11,
    "legend.framealpha":   0.85,
    "lines.linewidth":      2.0,
    "axes.spines.top":     False,
    "axes.spines.right":   False,
    "savefig.bbox":       "tight",
    "savefig.dpi":         150,
})
np.random.seed(42)
print("Plot setup complete.")

Code cell 4

import numpy as np
import numpy.linalg as la
import scipy.linalg as sla
from scipy import stats

COLORS = {
    "primary": "#0077BB",
    "secondary": "#EE7733",
    "tertiary": "#009988",
    "error": "#CC3311",
    "neutral": "#555555",
    "highlight": "#EE3377",
}
HAS_MPL = True
np.set_printoptions(precision=8, suppress=True)
np.random.seed(42)

def header(title):
    print("\n" + "=" * len(title))
    print(title)
    print("=" * len(title))

def check_true(name, cond):
    ok = bool(cond)
    print(f"{'PASS' if ok else 'FAIL'} - {name}")
    return ok

def check_close(name, got, expected, tol=1e-8):
    ok = np.allclose(got, expected, atol=tol, rtol=tol)
    print(f"{'PASS' if ok else 'FAIL'} - {name}: got {got}, expected {expected}")
    return ok

def check(name, got, expected, tol=1e-8):
    return check_close(name, got, expected, tol=tol)

def softmax(z, axis=-1, tau=1.0):
    z = np.asarray(z, dtype=float) / float(tau)
    z = z - np.max(z, axis=axis, keepdims=True)
    e = np.exp(z)
    return e / np.sum(e, axis=axis, keepdims=True)

def cosine_similarity(a, b):
    a = np.asarray(a, dtype=float); b = np.asarray(b, dtype=float)
    return float(a @ b / (la.norm(a) * la.norm(b) + 1e-12))

def numerical_rank(A, tol=1e-10):
    return int(np.sum(la.svd(A, compute_uv=False) > tol))

def orthonormal_basis(A, tol=1e-10):
    Q, R = la.qr(A)
    keep = np.abs(np.diag(R)) > tol
    return Q[:, keep]

def null_space(A, tol=1e-10):
    U, S, Vt = la.svd(A)
    return Vt[S.size:,:].T if S.size < Vt.shape[0] else Vt[S <= tol,:].T



# Compatibility helpers used by the Chapter 02 theory and exercise cells.
def null_space(A, tol=1e-10):
    A = np.asarray(A, dtype=float)
    U, S, Vt = la.svd(A, full_matrices=True)
    rank = int(np.sum(S > tol))
    return Vt[rank:].T

svd_null_space = null_space

def gram_schmidt(vectors, tol=1e-10):
    A = np.asarray(vectors, dtype=float)
    if A.ndim == 1:
        A = A.reshape(1, -1)
    basis = []
    for v in A:
        w = v.astype(float).copy()
        for q in basis:
            w = w - np.dot(w, q) * q
        norm = la.norm(w)
        if norm > tol:
            basis.append(w / norm)
    return np.array(basis)

def projection_matrix_from_columns(A, tol=1e-10):
    Q = orthonormal_basis(np.asarray(A, dtype=float), tol=tol)
    return Q @ Q.T


def random_unit_vectors(n, d):
    X = np.random.randn(n, d)
    return X / np.maximum(la.norm(X, axis=1, keepdims=True), 1e-12)

def pairwise_distances(X):
    X = np.asarray(X, dtype=float)
    diff = X[:, None, :] - X[None, :, :]
    return la.norm(diff, axis=-1)


def normalize(x, axis=None, tol=1e-12):
    x = np.asarray(x, dtype=float)
    norm = la.norm(x, axis=axis, keepdims=True)
    return x / np.maximum(norm, tol)

def frobenius_inner(A, B):
    return float(np.sum(np.asarray(A, dtype=float) * np.asarray(B, dtype=float)))

def outer_sum_product(A, B):
    A = np.asarray(A, dtype=float)
    B = np.asarray(B, dtype=float)
    return sum(np.outer(A[:, k], B[k, :]) for k in range(A.shape[1]))

def softmax_rows(X):
    return softmax(X, axis=1)

def col_space(A, tol=1e-10):
    return orthonormal_basis(np.asarray(A, dtype=float), tol=tol)

def row_space(A, tol=1e-10):
    return orthonormal_basis(np.asarray(A, dtype=float).T, tol=tol).T

def rref(A, tol=1e-10):
    R = np.array(A, dtype=float, copy=True)
    m, n = R.shape
    pivots = []
    row = 0
    for col in range(n):
        pivot = row + int(np.argmax(np.abs(R[row:, col]))) if row < m else row
        if row >= m or abs(R[pivot, col]) <= tol:
            continue
        if pivot != row:
            R[[row, pivot]] = R[[pivot, row]]
        R[row] = R[row] / R[row, col]
        for r in range(m):
            if r != row:
                R[r] = R[r] - R[r, col] * R[row]
        pivots.append(col)
        row += 1
        if row == m:
            break
    R[np.abs(R) < tol] = 0.0
    return R, pivots

def nullspace_basis(A, tol=1e-10):
    A = np.asarray(A, dtype=float)
    U, S, Vt = la.svd(A, full_matrices=True)
    rank = int(np.sum(S > tol))
    return Vt[rank:].T, rank

print("Chapter helper setup complete.")

---

1. Geometry, Rank, and the Three Cases

A system $Ax=b$ is really asking one question:

$$\text{Does } b \text{ lie in the column space of } A?$$

That question splits into three cases:

• unique solution: consistent and no free directions remain
• infinitely many solutions: consistent, but some directions remain unconstrained
• no solution: the right-hand side points outside the achievable column space

The quickest computational summary is:

$$\operatorname{rank}(A) \overset{?}{=} \operatorname{rank}([A\mid b]).$$

---

1. Classifying Systems by Rank

Code cell 7

# ======================================================================
# 1.1 Classify systems by rank
# ======================================================================

def classify_system(A, b):
    A = np.array(A, dtype=float)
    b = np.array(b, dtype=float).reshape(-1, 1)
    aug = np.hstack([A, b])
    rank_A = np.linalg.matrix_rank(A)
    rank_aug = np.linalg.matrix_rank(aug)
    n = A.shape[1]

    if rank_A < rank_aug:
        label = 'inconsistent'
    elif rank_A == n:
        label = 'unique'
    else:
        label = 'infinitely many'

    return {
        'rank(A)': rank_A,
        'rank([A|b])': rank_aug,
        'unknowns': n,
        'type': label,
    }


examples = {
    'unique': (
        np.array([[1, 2], [3, 4]], dtype=float),
        np.array([5, 11], dtype=float),
    ),
    'infinite': (
        np.array([[1, 2], [2, 4]], dtype=float),
        np.array([3, 6], dtype=float),
    ),
    'inconsistent': (
        np.array([[1, 1], [1, 1]], dtype=float),
        np.array([2, 3], dtype=float),
    ),
}

for name, (A, b) in examples.items():
    summary = classify_system(A, b)
    print(f"{name.upper():<12} -> {summary}")

---

2. Matrix Form and Augmented Matrices

In theory, $Ax=b$ is compact notation. In practice, the augmented matrix $[A\mid b]$ is what we manipulate during elimination.

The three elementary row operations preserve the solution set:

1. swap two rows
2. scale a row by a non-zero scalar
3. add a multiple of one row to another

This is the algebraic engine behind Gaussian elimination.

---

2. Gaussian Elimination — Row Operations

Code cell 10

# ======================================================================
# 2.1 Elementary row operations on an augmented matrix
# ======================================================================

def augment(A, b):
    A = np.array(A, dtype=float)
    b = np.array(b, dtype=float).reshape(-1, 1)
    return np.hstack([A, b])


def swap_rows(M, i, j):
    M = M.copy()
    M[[i, j]] = M[[j, i]]
    return M


def scale_row(M, i, alpha):
    M = M.copy()
    M[i] *= alpha
    return M


def add_multiple(M, target, source, alpha):
    M = M.copy()
    M[target] += alpha * M[source]
    return M


M = augment([[1, 2, 1], [2, -1, 3], [3, 1, -1]], [9, 8, 3])
print('Initial [A|b]:\n', M)
print('\nSwap R1 and R3:')
print(swap_rows(M, 0, 2))
print('\nScale R1 by 1/3:')
print(scale_row(M, 0, 1/3))
print('\nR2 <- R2 - 2R1:')
print(add_multiple(M, 1, 0, -2))

---

3. Gaussian Elimination and RREF

Row-echelon form reveals pivots and free variables. Reduced row-echelon form goes further: it exposes the solution structure directly.

We will implement a small RREF routine manually. This is not meant to outperform numpy or LAPACK. It is for understanding exactly how elimination reveals rank and solution geometry.

---

3. Reduced Row Echelon Form (RREF)

Code cell 13

# ======================================================================
# 3.1 Manual RREF with partial pivoting
# ======================================================================

def rref(M, tol=1e-12):
    M = np.array(M, dtype=float).copy()
    rows, cols = M.shape
    pivot_cols = []
    r = 0

    for c in range(cols):
        if r >= rows:
            break

        pivot = r + np.argmax(np.abs(M[r:, c]))
        if abs(M[pivot, c]) < tol:
            continue

        if pivot != r:
            M[[r, pivot]] = M[[pivot, r]]

        M[r] = M[r] / M[r, c]

        for i in range(rows):
            if i != r and abs(M[i, c]) > tol:
                M[i] -= M[i, c] * M[r]

        M[np.abs(M) < tol] = 0.0
        pivot_cols.append(c)
        r += 1

    return M, pivot_cols


A = np.array([[1, 2, 0, 1],
              [2, 4, 1, 3],
              [0, 0, 1, 1]], dtype=float)
b = np.array([1, 3, 1], dtype=float)
R, pivots = rref(augment(A, b))

print('RREF([A|b]) =')
print(R)
print('\nPivot columns:', pivots)
free_cols = [j for j in range(A.shape[1]) if j not in pivots]
print('Free-variable columns:', free_cols)

---

4. Particular Solutions and Null-Space Directions

If a system is consistent and underdetermined, the general solution is

$$x = x_p + x_h, \qquad x_h \in \operatorname{null}(A).$$

This is one of the most important structural facts in the chapter: one concrete solution plus every homogeneous direction that changes nothing.

---

4. Null Space Basis and Solution Structure

Code cell 16

# ======================================================================
# 4.1 Recover one particular solution and a null-space basis
# ======================================================================

def nullspace(A, tol=1e-12):
    U, s, Vt = np.linalg.svd(A)
    rank = np.sum(s > tol)
    return Vt[rank:].T


x_particular, *_ = np.linalg.lstsq(A, b, rcond=None)
N = nullspace(A)

print('Particular solution x_p:')
print(x_particular)
print('\nBasis vectors for null(A):')
print(N)

if N.shape[1] > 0:
    coeffs = np.array([0.7, -1.2])[:N.shape[1]]
    x_general = x_particular + N @ coeffs
    print('\nExample general-solution member x_p + N c:')
    print(x_general)
    print('Residual ||Ax - b||:', np.linalg.norm(A @ x_general - b))

print('\nRank + nullity =', np.linalg.matrix_rank(A), '+', N.shape[1], '=', np.linalg.matrix_rank(A) + N.shape[1])

---

5. Least Squares as Projection

An overdetermined system usually has no exact solution. The right question becomes:

$$x^\star = \arg\min_x \|Ax-b\|_2^2.$$

The residual at the optimum is orthogonal to the column space:

$$A^\top (Ax^\star - b) = 0.$$

That is both the normal-equations condition and the geometric projection condition.

---

5. Least Squares and Normal Equations

Code cell 19

# ======================================================================
# 5.1 Least squares, normal equations, and ridge
# ======================================================================

X = np.array([[1, 1],
              [1, 2],
              [1, 3],
              [1, 4]], dtype=float)
y = np.array([2, 3, 3.5, 5], dtype=float)

XtX = X.T @ X
Xty = X.T @ y
w_normal = np.linalg.solve(XtX, Xty)
w_lstsq, *_ = np.linalg.lstsq(X, y, rcond=None)
residual = y - X @ w_lstsq

lam = 0.5
w_ridge = np.linalg.solve(XtX + lam * np.eye(X.shape[1]), Xty)

print('Normal-equation solution:', w_normal)
print('np.linalg.lstsq solution:', w_lstsq)
print('Difference:', np.linalg.norm(w_normal - w_lstsq))
print('\nResidual vector:', residual)
print('X^T residual (should be near 0):', X.T @ residual)
print('\nRidge solution (lambda = 0.5):', w_ridge)
print('Residual norm (OLS):', np.linalg.norm(residual))
print('Residual norm (ridge):', np.linalg.norm(y - X @ w_ridge))

---

6. Conditioning and Sensitivity

A system can be exactly solvable and still be numerically dangerous. The condition number measures how much relative input perturbations can be amplified in the solution.

This is why numerical linear algebra distinguishes:

• problem difficulty: conditioning
• algorithm quality: stability

You need both.

---

6. Numerical Stability and Conditioning

Code cell 22

# ======================================================================
# 6.1 Ill-conditioning demo
# ======================================================================

A_bad = np.array([[1000, 999], [999, 998]], dtype=float)
b = np.array([1999, 1997], dtype=float)
b_perturbed = np.array([1999.001, 1997], dtype=float)

x = np.linalg.solve(A_bad, b)
x_perturbed = np.linalg.solve(A_bad, b_perturbed)

rel_rhs_change = np.linalg.norm(b_perturbed - b) / np.linalg.norm(b)
rel_x_change = np.linalg.norm(x_perturbed - x) / np.linalg.norm(x)

print('A =\n', A_bad)
print('det(A) =', np.linalg.det(A_bad))
print('cond_2(A) =', np.linalg.cond(A_bad))
print('\nExact-ish solution x      =', x)
print('Perturbed solution x\'   =', x_perturbed)
print('\nRelative RHS change  =', rel_rhs_change)
print('Relative x change    =', rel_x_change)
print('Amplification factor =', rel_x_change / rel_rhs_change)

---

7. Iterative Solvers for Large Systems

Direct solvers dominate dense moderate-size problems. Iterative solvers dominate when the system is large, sparse, or structured.

We will compare three canonical ideas:

• Jacobi: simple and parallel
• Gauss-Seidel: uses fresh information immediately
• Conjugate Gradient: the standard Krylov method for SPD systems

---

7. Iterative Solvers (Jacobi, Gauss-Seidel, CG)

Code cell 25

# ======================================================================
# 7.1 Jacobi, Gauss-Seidel, and Conjugate Gradient
# ======================================================================

A_spd = np.array([[4.0, 1.0], [1.0, 3.0]])
b_spd = np.array([1.0, 2.0])
x_star = np.linalg.solve(A_spd, b_spd)


def jacobi(A, b, x0, steps):
    D = np.diag(np.diag(A))
    R = A - D
    x = x0.copy()
    hist = []
    for _ in range(steps):
        x = np.linalg.solve(D, b - R @ x)
        hist.append((x.copy(), np.linalg.norm(b - A @ x)))
    return hist


def gauss_seidel(A, b, x0, steps):
    x = x0.copy()
    hist = []
    for _ in range(steps):
        for i in range(len(b)):
            sigma1 = A[i, :i] @ x[:i]
            sigma2 = A[i, i + 1:] @ x[i + 1:]
            x[i] = (b[i] - sigma1 - sigma2) / A[i, i]
        hist.append((x.copy(), np.linalg.norm(b - A @ x)))
    return hist


def conjugate_gradient(A, b, x0, steps):
    x = x0.copy()
    r = b - A @ x
    p = r.copy()
    hist = [(x.copy(), np.linalg.norm(r))]
    for _ in range(steps):
        Ap = A @ p
        alpha = (r @ r) / (p @ Ap)
        x = x + alpha * p
        r_new = r - alpha * Ap
        beta = (r_new @ r_new) / (r @ r)
        p = r_new + beta * p
        r = r_new
        hist.append((x.copy(), np.linalg.norm(r)))
    return hist


x0 = np.zeros(2)
jacobi_hist = jacobi(A_spd, b_spd, x0, steps=3)
gs_hist = gauss_seidel(A_spd, b_spd, x0, steps=3)
cg_hist = conjugate_gradient(A_spd, b_spd, x0, steps=2)

print('True solution:', x_star)
print('\nJacobi:')
for k, (xk, rk) in enumerate(jacobi_hist, start=1):
    print(f'  iter {k}: x = {xk}, residual = {rk:.6e}')

print('\nGauss-Seidel:')
for k, (xk, rk) in enumerate(gs_hist, start=1):
    print(f'  iter {k}: x = {xk}, residual = {rk:.6e}')

print('\nConjugate Gradient:')
for k, (xk, rk) in enumerate(cg_hist):
    print(f'  iter {k}: x = {xk}, residual = {rk:.6e}')

print('\ncond_2(A_spd) =', np.linalg.cond(A_spd))

---

8. Nonlinear Systems and Newton's Method

The nonlinear system

$$F(x,y) = \begin{pmatrix} x^2 + y^2 - 1 \\ x - y^2 \end{pmatrix} = 0$$

combines a circle and a parabola. Newton's method repeatedly replaces the nonlinear problem with a local linear system involving the Jacobian.

---

8. Nonlinear Systems — Newton's Method

Code cell 28

# ======================================================================
# 8.1 Newton iterations for a nonlinear 2D system
# ======================================================================

def F(z):
    x, y = z
    return np.array([x**2 + y**2 - 1, x - y**2], dtype=float)


def J(z):
    x, y = z
    return np.array([[2*x, 2*y], [1.0, -2*y]], dtype=float)


def newton(z0, steps):
    z = np.array(z0, dtype=float)
    hist = []
    for _ in range(steps):
        delta = np.linalg.solve(J(z), -F(z))
        z = z + delta
        hist.append((z.copy(), np.linalg.norm(F(z))))
    return hist


hist = newton((1.0, 1.0), steps=3)
for k, (zk, err) in enumerate(hist, start=1):
    print(f'iter {k}: z = {zk}, ||F(z)|| = {err:.6e}')

print('\nAt (0,0), the Jacobian is')
print(J((0.0, 0.0)))
print('det(J(0,0)) =', np.linalg.det(J((0.0, 0.0))))

---

9. Equality Constraints and KKT Systems

Constrained optimization is still system solving. For

$$\min_{x,y} \; x^2 + 2y^2 \quad \text{subject to } x+y=3,$$

the Lagrangian is

$$\mathcal{L}(x,y,\lambda) = x^2 + 2y^2 + \lambda(x+y-3),$$

and the first-order conditions become a block linear system in $(x,y,\lambda)$.

---

9. Constrained Systems and KKT Conditions

Code cell 31

# ======================================================================
# 9.1 Solve a tiny KKT saddle-point system
# ======================================================================

KKT = np.array([
    [2.0, 0.0, 1.0],
    [0.0, 4.0, 1.0],
    [1.0, 1.0, 0.0],
])
rhs = np.array([0.0, 0.0, 3.0])

sol = np.linalg.solve(KKT, rhs)
x_opt, y_opt, lam = sol

print('KKT matrix =\n', KKT)
print('RHS =', rhs)
print('\nSolution [x, y, lambda] =', sol)
print('Constraint check x + y =', x_opt + y_opt)
print('Objective value =', x_opt**2 + 2 * y_opt**2)

---

10. AI-Facing Systems

Three recurring AI patterns are worth seeing directly:

1. attention chooses simplex-constrained coefficients
2. deep equilibrium models define hidden states by fixed points
3. Gaussian processes reduce Bayesian inference to SPD solves

We will keep each example tiny and interpretable.

---

10. AI Applications — Attention and Fixed Points

Code cell 34

# ======================================================================
# 10.1 Attention as a simplex-constrained coefficient system
# ======================================================================

Q = np.array([[1, 0], [0, 1], [1, 1]], dtype=float)
K = np.array([[1, 1], [1, 0], [0, 1]], dtype=float)
V = np.array([[1, 0], [0, 1], [1, 1]], dtype=float)

scores = Q @ K.T / np.sqrt(2.0)
weights = np.exp(scores - scores.max(axis=1, keepdims=True))
weights = weights / weights.sum(axis=1, keepdims=True)
output = weights @ V

print('Scores S =\n', scores)
print('\nSoftmax weights =\n', weights)
print('Row sums =', weights.sum(axis=1))
print('\nAttention output O =\n', output)
print('\nFirst output row as convex combination coefficients:', weights[0])

---

10. AI Applications (continued) — DEQ and GP Solve

Code cell 36

# ======================================================================
# 10.2 Fixed-point iteration (DEQ intuition) and a tiny GP solve
# ======================================================================

W = np.array([[0.25, -0.10], [0.15, 0.20]], dtype=float)
u = np.array([0.4, -0.2], dtype=float)

z = np.zeros(2)
for k in range(8):
    z = np.tanh(W @ z + u)
    print(f'fixed-point iter {k + 1}: z = {z}')

print('\nTiny Gaussian-process style SPD solve:')
K_gp = np.array([[1.0, 0.7, 0.4], [0.7, 1.0, 0.6], [0.4, 0.6, 1.0]], dtype=float)
sigma2 = 0.1
y_gp = np.array([1.0, -0.5, 0.75], dtype=float)
alpha = np.linalg.solve(K_gp + sigma2 * np.eye(3), y_gp)
print('alpha =', alpha)
print('Residual norm =', np.linalg.norm((K_gp + sigma2 * np.eye(3)) @ alpha - y_gp))

---

What to Remember

• $Ax=b$ is a question about column-space membership and null-space freedom.
• Elimination is the fastest way to see consistency, pivots, and free variables.
• Least squares is projection, not just a formula.
• Conditioning governs sensitivity; stability governs how much extra numerical trouble your algorithm adds.
• Newton, KKT systems, DEQs, Gaussian processes, and second-order optimizers all reduce to repeated linear-system solves.

If you understand systems of equations at this level, you understand a large part of why modern AI computation looks the way it does.

References and Learning Sources

• MIT 18.06 Linear Algebra: https://web.mit.edu/18.06/www/
• Stanford EE263, Introduction to Linear Dynamical Systems: https://stanford.edu/class/ee263/
• Netlib, Templates for the Solution of Linear Systems: https://www.netlib.org/templates/templates.html
• UIUC CS 357 course materials on linear systems and conditioning: https://courses.grainger.illinois.edu/cs357/
• Jonathan Shewchuk, An Introduction to the Conjugate Gradient Method Without the Agonizing Pain: https://www.cs.cmu.edu/~quake-papers/painless-conjugate-gradient.pdf
• Ashish Vaswani et al., Attention Is All You Need (2017): https://arxiv.org/abs/1706.03762
• Shaojie Bai, J. Zico Kolter, Vladlen Koltun, Deep Equilibrium Models (2019): https://arxiv.org/abs/1909.01377
• James Martens and Roger Grosse, K-FAC (2015): https://proceedings.mlr.press/v37/martens15.html
• Vineet Gupta et al., Shampoo (2018): https://proceedings.mlr.press/v80/gupta18a.html
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• Jupyter documentation on notebooks as narrative documents: https://docs.jupyter.org/en/latest/use/use-cases/nb_as_docs.html