Chapter 4. Eigenvalues and Eigenforms
Recommended article: 【Linear Algebra】 Linear Algebra Contents
1. Eigenvalues and Eigenvectors
2. Diagonalization of Matrices
4. Differential Equations and Eigenvalues
1. Eigenvalues and Eigenvectors
⑴ Eigenvalues and eigenvectors
① Used in diagonalizability, spectral clustering, etc.
② Theorem 1. The necessary and sufficient condition for A - λI to have a non-zero kernel in (A - λI)x = 0 is det(A - λI) = 0.
○ Proof: rank-nullity theorem
③ Theorem 2. When the eigenvalues of A ∈ ℝ2×2 are λ1, λ2, A2 - (λ1 + λ2)A + λ1λ2I = O holds.
○ Proof
From the definition of eigenvalues, the following holds.
From this, it is easily known that λ1 + λ2 = a + d, and λ1λ2 = ad - bc.
Therefore, by the Cayley-Hamilton theorem, the above proposition holds.
Applying this to calculate An as (A - λ1I)(A - λ2I) Q(A) + kA + sE for An (for n ≥ 2).
④ Theorem 3. The determinant det(A) of a real matrix A of size n × n is equal to the product of its eigenvalues.
○ For example, for a 3 × 3 matrix, the following holds.
○ If A has three linearly independent eigenvectors, det(A) can be shown as follows:
○ If the eigenvalues are λ1, λ2 with multiplicities 1 and 2, respectively, det(A) can be shown as follows:
○ Here, w2 is a generalized eigenvector corresponding to λ2 and v2.
○ In this way, by considering each case, it can be easily shown that the product of the eigenvalues equals det(A).
⑤ Theorem 4. By applying the diagonalization of matrices, A = PDP-1, we can obtain An = PDnP-1
⑥ Theorem 5. For real matrices A and B of size n × n, the eigenvalues of matrix AB are the same as the eigenvalues of matrix BA (ref)
○ Proof: Assume λ is an eigenvalue of AB
⇔ ABv = λv, v ≠ 0
⇔ BABv = BA (Bv) = B · λv = λ (Bv)
⇔ Considering all possible cases as follows, the eigenvalues of BA are the same as those of AB
○ When Bv ≠ 0: λ is an eigenvalue of BA
○ When Bv = 0
⇔ ABv = A0 = 0 = λv
⇔ λ = 0 is an eigenvalue of AB
⇔ 0 = det(AB) = det(A) × det(B) = det(BA)
⇔ Hence, there exists v’ ≠ 0 such that BAv’ = 0
⇔ λ = 0 is an eigenvalue of BA
⑦ Theorem 6. For a real matrix A of size m × n, the nonzero eigenvalues of the matrices AAT and ATA are identical.
○ Related to Theorem 5.
⑧ Application 1. Hückel approximation and Hamiltonian
⑨ Python code (cf. WolframAlpha)
○ Using numpy
import numpy as np
matrix = np.array([
[4.56, 10.2, 7.68, 13.68],
[10.2, 33.2, 20.2, 24.8],
[7.68, 20.2, 17.84, 23.44],
[13.68, 24.8, 23.44, 45.55]
])
eigenvalues, eigenvectors = np.linalg.eigh(matrix)
sorted_indices = np.argsort(eigenvalues)[::-1]
sorted_eigenvalues = eigenvalues[sorted_indices]
sorted_eigenvectors = eigenvectors[:, sorted_indices]
print("Eigenvalues:\n", sorted_eigenvalues)
print("\nEigenvectors:\n", sorted_eigenvectors)
○ Using sympy
from sympy import Matrix
# Create SymPy matrix
matrix = Matrix([
[4.56, 10.2, 7.68, 13.68],
[10.2, 33.2, 20.2, 24.8],
[7.68, 20.2, 17.84, 23.44],
[13.68, 24.8, 23.44, 45.55]
])
# Calculate eigenvaleus and eigenvectors
eigenvalues = matrix.eigenvals() # 고유값 계산
eigenvectors = matrix.eigenvects() # 고유벡터 계산
# Output
print("Eigenvalues:")
for eigval, multiplicity in eigenvalues.items():
print(f"Value: {eigval}, Multiplicity: {multiplicity}")
print("\nEigenvectors:")
for eigval, eigen_mult, eigvecs in eigenvectors:
print(f"Eigenvalue: {eigval}")
for vec in eigvecs:
print(f"Eigenvector: {vec}")
⑵ Generalized eigenvectors
① Overview
○ Definition: When the geometric multiplicity of an asymmetric matrix is less than the algebraic multiplicity, vectors that replace the lacking eigenvectors
○ The generalized eigenvectors of matrix A of rank m are as follows
○ (A - λI)mxm = 0 & (A - λI)m-1xm ≠ 0
○ When m = 1, it is the same as the definition of eigenvector.
○ Used in Jordan canonical form.
② Example 1. The eigenvalue of the following matrix A is 1, and the corresponding eigenvector is v1 = (1, 0)T
○ The algebraic multiplicity of eigenvalue 1 is 2, and the geometric multiplicity is 1.
○ (A - λI)2 v2 = (A - λI) {(A - λI) v2} = 0 ⇔ (A - λI) v2 = v1
○ v2 = (*, 1)T (e.g., (0, 1)T) that satisfies (A - I)v2 = (1, 0)T is called a generalized eigenvector.
③ Example 2. Ref
⑶ Eigenfunctions
① Theorem of Sturm-Liouville: Different eigenfunctions are orthogonal to each other
2. Diagonalization of Matrices
⑴ Diagonal Matrix
① An n × n matrix where all elements except the diagonal are zero
② Properties
○ Property 1. The determinant is the product of the diagonal elements
○ Property 2. The inverse matrix has the reciprocal of each diagonal element of the given diagonal matrix
⑵ Block Diagonal Matrix
① Definition: When Ai ∈ ℳni,ni (F) and n1 + ··· + nk = n, the matrix A in the following form
② Properties: When Ai, Bi ∈ ℳni,ni (F),
○ Property 1. diag(A1, ···, Ak)·diag(B1, ···, Bk) = diag(A1B1, ···, AkBk)
○ Property 2. If Ai is invertible, (diag(A1, ···, Ak))-1 = diag(A1-1, ···, Ak-1)
⑶ Diagonalizability
① Definition: A matrix A is diagonalizable if there exist a diagonal matrix D and an invertible matrix P satisfying the following
② Orthogonal Diagonalization: For a matrix P composed of orthonormal basis vectors.
③ Theorem 1. The necessary and sufficient condition for an n × n matrix A to be diagonalizable is that A has n linearly independent eigenvectors.
④ Theorem 2. If an n × n matrix A has n distinct eigenvalues λ1, λ2, ···, λn, it can be diagonalized
○ If P is a matrix having linearly independent eigenvectors of A as columns, D is as follows
⑤ Theorem 3. Spectral Theorem: An n × n matrix A can be orthogonally diagonalized if and only if A is a symmetric matrix
⑥ Theorem 4. Generalized eigenvectors of an n × n matrix A always span ℝn, thus such a matrix A can be diagonalized
○ Example: n × n matrix An satisfying A2 = A has eigenvalues λ = 0, 1 and can be diagonalized.
⑦ Theorem 5. A matrix that commutes with a diagonal matrix D having three distinct diagonal elements is also a diagonal matrix.
⑷ Jordan Canonical Form
① When a matrix can’t be diagonalized, but there are still generalized eigenvectors corresponding to the eigenvalues, it can be transformed into the Jordan canonical form
② Jordan Canonical Form: A = PJP-1 (where J is not a complete diagonal matrix, but a matrix composed of Jordan blocks)
③ Theorem 1. If Ak = 0 (nilpotent), considering the Jordan form of the matrix A, A becomes a block diagonal matrix consisting of suitable Ak matrices.
⑸ Examples
① Given matrix A
② Calculate eigenvalues and eigenvectors
③ Calculate P and D
3. Quadratic Forms
⑴ Definition: 4x12 + 2x1x2 + 3x22 is a quadratic form, but 4x12 + 2x1 is not a quadratic form.
⑵ Representation using a symmetric matrix
⑶ Definition of signs of quadratic forms: For a symmetric matrix A ∈ ℳn and any vector x ≠ 0 on ℝn
① Positive definite matrix: All eigenvalues of A are greater than 0 ⇔ Q(x) > 0 ∀x ≠ 0 ⇔ xtAx = xt(λx) = λ x 2 > 0
② Positive semi-definite matrix: All eigenvalues of A are greater than or equal to 0 ⇔ Q(x) ≥ 0 ∀x ≠ 0 ⇔ xtAx = xt(λx) = λ x 2 ≥ 0
③ Negative definite matrix: All eigenvalues of A are less than 0 ⇔ Q(x) < 0 ∀x ≠ 0 ⇔ xtAx = xt(λx) = λ x 2 < 0
④ Negative semi-definite matrix: All eigenvalues of A are less than or equal to 0 ⇔ Q(x) ≤ 0 ∀x ≠ 0 ⇔ xtAx = xt(λx) = λ x 2 ≤ 0
⑤ Indefinite matrix
○ When the signs of the eigenvalues are not uniform, unlike ① to ④
⑥ Theorem 1.
○ Proof: Since a symmetric matrix A can be orthogonally diagonalizable (∵ spectral theorem)
⑷ Determination of the signs of quadratic forms: For a symmetric matrix A ∈ ℳn and a subset S =｛i1, i2, ···, ik｝ of ｛1, 2, ···, n｝
① Principal submatrix: A k × k matrix formed by selecting the rows and columns of A corresponding to S
② Leading principal submatrix: A k × k matrix obtained by omitting the last n-k row and column vectors of A
③ Theorem 1. A symmetric matrix A ∈ ℳn being positive definite is equivalent to sgn(det(Ak)) = 1, k = 1, 2, ···, n
④ Theorem 2. A symmetric matrix A ∈ ℳn being negative definite is equivalent to sgn(det(Ak)) = (-1)k, k = 1, 2, ···, n
⑤ Theorem 3. A symmetric matrix A ∈ ℳn being positive semi-definite is equivalent to all principal submatrix determinants being non-negative
○ If any leading principal submatrix determinant is 0 while attempting Theorem 1 or 2, try Theorem 3
○ Obviously, if the sign of any leading principal submatrix determinant changes, there is no need to attempt Theorem 3 or 4 as it indicates an indefinite matrix
⑥ Theorem 4. A symmetric matrix A ∈ ℳn being negative semi-definite is equivalent to all principal submatrix determinants of -A being non-negative
○ If the sign of any principal submatrix determinant is not uniform while attempting Theorem 3, try Theorem 4
⑦ sgn indicates 1 for positive values and -1 for negative values when the determinant is directly calculated
⑸ For a C2 function f: Ω → ℝ defined on a region Ω ⊂ ℝn, if at point p ∈ Ω, ∇f(p) = 0, then the following holds
① If Hf(p) is a positive definite matrix, then f has a local minimum at point x = p
② If Hf(p) is a negative definite matrix, then f has a local maximum at point x = p
③ If Hf(p) is an indefinite matrix, then point x = p is a saddle point of f
④ If Hf(p) is a positive semi-definite matrix, then f has a relative minimum at point x = p
⑤ If Hf(p) is a negative semi-definite matrix, then f has a relative maximum at point x = p
| ⑹ Determining the sign of a constrained quadratic form: For symmetric matrices A ∈ ℳn and B ∈ ℳm,n (with n ＞ m), given a n-dimensional quadratic form QA(x) = xtAx with the linear constraint ｛x ∈ ℝn | Bx = 0｝, the following statements hold when defining a new (m+n) × (m+n) matrix H as below. |
① Theorem 1. If sgn( Hm+i ) = (-1)m, i = m+1, m+2, ···, n, then QA(x) ＞ 0 holds
○ Keep in mind it is not a necessary and sufficient condition
② Theorem 2. If sgn( Hm+i ) = (-1)i, i = m+1, m+2, ···, n, then QA(x) ＜ 0 holds
○ Keep in mind it is not a necessary and sufficient condition
③ Example: The following example observes a negative definite symmetric matrix A
④ Even a quadratic form without a sign may have a sign under certain constraints
Input: 2020.05.19 23:48
Modified: 2024.11.12 09:45