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Linear Methods of AI

Characteristic Polynomial

Definition and Basic Concepts

To find the eigenvalues of a matrix, we need a very important mathematical tool in linear algebra. Imagine we want to find all values λ\lambdaλ that make the matrix A−λIA - \lambda IA−λI become singular (not invertible).

Let A∈Kn×nA \in \mathbb{K}^{n \times n}A∈Kn×n. We can form a special function:

χA(t)=det⁡(A−t⋅I)\chi_A(t) = \det(A - t \cdot I)χA​(t)=det(A−t⋅I)
χA(t)=an⋅tn+an−1⋅tn−1+⋯+a1t+a0\chi_A(t) = a_n \cdot t^n + a_{n-1} \cdot t^{n-1} + \cdots + a_1 t + a_0χA​(t)=an​⋅tn+an−1​⋅tn−1+⋯+a1​t+a0​

This function is a polynomial of degree nnn in t∈Kt \in \mathbb{K}t∈K, which we call the characteristic polynomial of AAA.

with coefficients a0,…,an−1,an∈Ka_0, \ldots, a_{n-1}, a_n \in \mathbb{K}a0​,…,an−1​,an​∈K.

In fact, χA(t)\chi_A(t)χA​(t) is indeed a polynomial of degree nnn for every matrix A∈Kn×nA \in \mathbb{K}^{n \times n}A∈Kn×n.

Matrix Trace and Polynomial Coefficients

Let A∈Kn×nA \in \mathbb{K}^{n \times n}A∈Kn×n be a square matrix. The trace of AAA is the sum of the diagonal elements:

traceA=∑i=1naii\text{trace} A = \sum_{i=1}^n a_{ii}traceA=i=1∑n​aii​

The matrix trace has a close relationship with the coefficients of the characteristic polynomial.

Relationship of Coefficients with Trace and Determinant

In the characteristic polynomial χA(t)\chi_A(t)χA​(t) of AAA, its coefficients have special meaning:

an=(−1)n,an−1=(−1)n−1⋅traceA,a0=det⁡Aa_n = (-1)^n, \quad a_{n-1} = (-1)^{n-1} \cdot \text{trace} A, \quad a_0 = \det Aan​=(−1)n,an−1​=(−1)n−1⋅traceA,a0​=detA

This means:

  • The highest coefficient is always (−1)n(-1)^n(−1)n
  • The second highest coefficient is related to the matrix trace
  • The constant term is the determinant of the matrix

Eigenvalues as Polynomial Roots

The most important concept of the characteristic polynomial is its relationship with eigenvalues.

Now, let's look at a very important relationship: λ∈K\lambda \in \mathbb{K}λ∈K is an eigenvalue of AAA if and only if det⁡(A−λ⋅I)=0\det(A - \lambda \cdot I) = 0det(A−λ⋅I)=0.

In other words, eigenvalues are the roots of the characteristic polynomial.

The equation for t∈Kt \in \mathbb{K}t∈K:

det⁡(A−t⋅I)=0\det(A - t \cdot I) = 0det(A−t⋅I)=0

is what we call the characteristic equation of AAA.

Algebraic Multiplicity

Now, what happens if an eigenvalue appears multiple times as a root of the characteristic polynomial? Let A∈Kn×nA \in \mathbb{K}^{n \times n}A∈Kn×n and λ∈K\lambda \in \mathbb{K}λ∈K. The multiplicity of the root t=λt = \lambdat=λ of the characteristic polynomial χA(t)\chi_A(t)χA​(t) is called the algebraic multiplicity μA(λ)\mu_A(\lambda)μA​(λ) of the eigenvalue λ\lambdaλ of AAA. We say λ\lambdaλ is an eigenvalue with multiplicity μA(λ)\mu_A(\lambda)μA​(λ) of AAA.

Multiplicity Bounds

For every eigenvalue λ\lambdaλ, it holds:

0≤μA(λ)≤n0 \leq \mu_A(\lambda) \leq n0≤μA​(λ)≤n

Relationship between Geometric and Algebraic Multiplicity

One important result in eigenvalue theory is the relationship between geometric and algebraic multiplicity.

For any matrix A∈Kn×nA \in \mathbb{K}^{n \times n}A∈Kn×n and λ∈K\lambda \in \mathbb{K}λ∈K, we have an interesting relationship:

0≤dim⁡EigA(λ)≤μA(λ)≤n0 \leq \dim \text{Eig}_A(\lambda) \leq \mu_A(\lambda) \leq n0≤dimEigA​(λ)≤μA​(λ)≤n

The geometric multiplicity of every eigenvalue is always less than or equal to its algebraic multiplicity.

Why does this happen? This can be explained using basis transformation and Jordan block form.

Examples of Characteristic Polynomial Calculation

After understanding the basic concepts, let's see how to apply them in concrete examples of characteristic polynomial calculation:

3x3 Matrix Example

Let A=(32−110−4301)∈K3×3A = \begin{pmatrix} 3 & 2 & -1 \\ 1 & 0 & -4 \\ 3 & 0 & 1 \end{pmatrix} \in \mathbb{K}^{3 \times 3}A=​313​200​−1−41​​∈K3×3. The characteristic polynomial of AAA is:

χA(t)=det⁡(3−t2−11−t−4301−t)\chi_A(t) = \det \begin{pmatrix} 3-t & 2 & -1 \\ 1 & -t & -4 \\ 3 & 0 & 1-t \end{pmatrix}χA​(t)=det​3−t13​2−t0​−1−41−t​​
=(3−t)⋅(−t)⋅(1−t)−2⋅(1⋅(1−t)+3⋅4)+1⋅3⋅t= (3-t) \cdot (-t) \cdot (1-t) - 2 \cdot (1 \cdot (1-t) + 3 \cdot 4) + 1 \cdot 3 \cdot t=(3−t)⋅(−t)⋅(1−t)−2⋅(1⋅(1−t)+3⋅4)+1⋅3⋅t
=−3t+3t2+t2−t3−2+2t−24−3t= -3t + 3t^2 + t^2 - t^3 - 2 + 2t - 24 - 3t=−3t+3t2+t2−t3−2+2t−24−3t
=−t3+4t2−4t−26= -t^3 + 4t^2 - 4t - 26=−t3+4t2−4t−26

For K=R\mathbb{K} = \mathbb{R}K=R, χA(t)\chi_A(t)χA​(t) only has the root λ1≈−1.8003\lambda_1 \approx -1.8003λ1​≈−1.8003 with algebraic multiplicity μA(λ1)=1\mu_A(\lambda_1) = 1μA​(λ1​)=1.

For K=C\mathbb{K} = \mathbb{C}K=C, χA(t)\chi_A(t)χA​(t) has roots λ1≈−1.8003\lambda_1 \approx -1.8003λ1​≈−1.8003, λ2≈2.9001+2.4559i\lambda_2 \approx 2.9001 + 2.4559iλ2​≈2.9001+2.4559i, and λ3≈2.9001−2.4559i\lambda_3 \approx 2.9001 - 2.4559iλ3​≈2.9001−2.4559i with algebraic multiplicities μA(λ1)=μA(λ2)=μA(λ3)=1\mu_A(\lambda_1) = \mu_A(\lambda_2) = \mu_A(\lambda_3) = 1μA​(λ1​)=μA​(λ2​)=μA​(λ3​)=1 respectively.

Simple Example

The characteristic polynomial of matrix A=(1201)A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}A=(10​21​) is:

χA(t)=det⁡(A−t⋅I)=det⁡(1−t201−t)\chi_A(t) = \det(A - t \cdot I) = \det \begin{pmatrix} 1-t & 2 \\ 0 & 1-t \end{pmatrix}χA​(t)=det(A−t⋅I)=det(1−t0​21−t​)
=(1−t)⋅(1−t)−0⋅2=(1−t)2= (1-t) \cdot (1-t) - 0 \cdot 2 = (1-t)^2=(1−t)⋅(1−t)−0⋅2=(1−t)2
=t2−2⋅t+1= t^2 - 2 \cdot t + 1=t2−2⋅t+1

This matrix has the root λ=1\lambda = 1λ=1 with algebraic multiplicity μA(1)=2\mu_A(1) = 2μA​(1)=2. λ=1\lambda = 1λ=1 is the only eigenvalue of AAA. We have calculated that dim⁡EigA(1)=1\dim \text{Eig}_A(1) = 1dimEigA​(1)=1.

Geometric Transformation Examples

Now, let's explore something fascinating: how the characteristic polynomial works on common geometric transformations we often encounter in R2→R2\mathbb{R}^2 \to \mathbb{R}^2R2→R2:

Rotation

Rotation with A=(cos⁡(α)−sin⁡(α)sin⁡(α)cos⁡(α))A = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \end{pmatrix}A=(cos(α)sin(α)​−sin(α)cos(α)​) has the characteristic polynomial:

χA(t)=(cos⁡(α)−t)2+sin⁡(α)2=t2−2⋅cos⁡(α)⋅t+1\chi_A(t) = (\cos(\alpha) - t)^2 + \sin(\alpha)^2 = t^2 - 2 \cdot \cos(\alpha) \cdot t + 1χA​(t)=(cos(α)−t)2+sin(α)2=t2−2⋅cos(α)⋅t+1

This has real roots if and only if cos⁡(α)2−1≥0\cos(\alpha)^2 - 1 \geq 0cos(α)2−1≥0, that is cos⁡(α)2=1\cos(\alpha)^2 = 1cos(α)2=1, so α=0\alpha = 0α=0 or α=π\alpha = \piα=π.

Reflection

Reflection along axes with A=(100−1)A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}A=(10​0−1​) has the characteristic polynomial:

χA(t)=(1−t)⋅(−1−t)=t2−1\chi_A(t) = (1-t) \cdot (-1-t) = t^2 - 1χA​(t)=(1−t)⋅(−1−t)=t2−1

The eigenvalues are λ1=1\lambda_1 = 1λ1​=1 and λ2=−1\lambda_2 = -1λ2​=−1 with μA(1)=μA(−1)=1\mu_A(1) = \mu_A(-1) = 1μA​(1)=μA​(−1)=1.

Scaling

Scaling with A=(s00s)A = \begin{pmatrix} s & 0 \\ 0 & s \end{pmatrix}A=(s0​0s​) has the characteristic polynomial:

χA(t)=(s−t)2=t2−2⋅s⋅t+s2\chi_A(t) = (s-t)^2 = t^2 - 2 \cdot s \cdot t + s^2χA​(t)=(s−t)2=t2−2⋅s⋅t+s2

The eigenvalue is λ=s\lambda = sλ=s with μA(s)=2\mu_A(s) = 2μA​(s)=2.

Shearing

Shearing with A=(1s01)A = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}A=(10​s1​) with s≠0s \neq 0s=0 has the characteristic polynomial:

χA(t)=(1−t)2=t2−2⋅t+1\chi_A(t) = (1-t)^2 = t^2 - 2 \cdot t + 1χA​(t)=(1−t)2=t2−2⋅t+1

The eigenvalue is λ=1\lambda = 1λ=1 with μA(1)=2\mu_A(1) = 2μA​(1)=2.

Properties of Similar Matrices

Similar matrices have a fascinating property: they have the same characteristic polynomial, and therefore have the same eigenvalues, the same trace, and the same determinant.

Let's see why this is true. Let S∈Kn×nS \in \mathbb{K}^{n \times n}S∈Kn×n be invertible and B=S−1⋅A⋅SB = S^{-1} \cdot A \cdot SB=S−1⋅A⋅S. Then:

χB(t)=det⁡(B−t⋅I)=det⁡(S−1⋅A⋅S−t⋅S−1⋅I⋅S)\chi_B(t) = \det(B - t \cdot I) = \det(S^{-1} \cdot A \cdot S - t \cdot S^{-1} \cdot I \cdot S)χB​(t)=det(B−t⋅I)=det(S−1⋅A⋅S−t⋅S−1⋅I⋅S)
=det⁡(S−1⋅(A−t⋅I)⋅S)= \det(S^{-1} \cdot (A - t \cdot I) \cdot S)=det(S−1⋅(A−t⋅I)⋅S)
=det⁡S−1⋅det⁡(A−t⋅I)⋅det⁡S=χA(t)= \det S^{-1} \cdot \det(A - t \cdot I) \cdot \det S = \chi_A(t)=detS−1⋅det(A−t⋅I)⋅detS=χA​(t)

Eigenvector Properties of Similar Matrices

Now, what about the eigenvectors of similar matrices? Let A,B∈Kn×nA, B \in \mathbb{K}^{n \times n}A,B∈Kn×n be similar matrices with B=S−1⋅A⋅SB = S^{-1} \cdot A \cdot SB=S−1⋅A⋅S and invertible matrix S∈Kn×nS \in \mathbb{K}^{n \times n}S∈Kn×n. If λ∈K\lambda \in \mathbb{K}λ∈K is an eigenvalue of both AAA and BBB, and v∈Knv \in \mathbb{K}^nv∈Kn is an eigenvector of AAA for eigenvalue λ\lambdaλ, then w=S−1⋅vw = S^{-1} \cdot vw=S−1⋅v is an eigenvector of BBB for eigenvalue λ\lambdaλ.

Let's see why this is true. Let A⋅v=λ⋅vA \cdot v = \lambda \cdot vA⋅v=λ⋅v and w=S−1⋅vw = S^{-1} \cdot vw=S−1⋅v. Then:

B⋅w=S−1⋅A⋅S⋅S−1⋅v=S−1⋅A⋅vB \cdot w = S^{-1} \cdot A \cdot S \cdot S^{-1} \cdot v = S^{-1} \cdot A \cdot vB⋅w=S−1⋅A⋅S⋅S−1⋅v=S−1⋅A⋅v
=S−1⋅λ⋅v=λ⋅S−1⋅v=λ⋅w= S^{-1} \cdot \lambda \cdot v = \lambda \cdot S^{-1} \cdot v = \lambda \cdot w=S−1⋅λ⋅v=λ⋅S−1⋅v=λ⋅w

This shows that similarity transformation not only preserves eigenvalues, but also provides a systematic way to transform eigenvectors.

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Eigenvalues of Diagonal and Triangular Matrices

  • Characteristic PolynomialExplore characteristic polynomials to find eigenvalues, understand algebraic multiplicity, and analyze matrix trace relationships in linear algebra.
On this page
  • Definition and Basic Concepts
  • Matrix Trace and Polynomial Coefficients
    • Relationship of Coefficients with Trace and Determinant
  • Eigenvalues as Polynomial Roots
  • Algebraic Multiplicity
    • Multiplicity Bounds
  • Relationship between Geometric and Algebraic Multiplicity
  • Examples of Characteristic Polynomial Calculation
    • 3x3 Matrix Example
    • Simple Example
  • Geometric Transformation Examples
    • Rotation
    • Reflection
    • Scaling
    • Shearing
  • Properties of Similar Matrices
    • Eigenvector Properties of Similar Matrices
  • Comments
  • Report
  • Source code