Characteristic Polynomial

1. Characteristic Polynomial

In the previous section, we saw that eigenvalues are scalars $\lambda$ for which $A - \lambda I$ is singular. This condition $\det(A - \lambda I) = 0$ defines a polynomial equation whose roots are precisely the eigenvalues.

Definition 6.3: Characteristic Polynomial (Matrix)

For $A \in M_n(F)$ , the characteristic polynomial is:

\chi_A(\lambda) = \det(\lambda I - A)

This is a monic polynomial of degree $n$ in $\lambda$ .

Remark 6.5: Convention

Some texts define $\chi_A(\lambda) = \det(A - \lambda I)$ instead. The two conventions differ by a factor of $(-1)^n$ , but have exactly the same roots. We use $\det(\lambda I - A)$ to ensure the polynomial is monic (leading coefficient 1).

Definition 6.4: Characteristic Polynomial (Linear Operator)

For a linear operator $T \in L(V)$ on finite-dimensional space $V$ , the characteristic polynomial is defined as the characteristic polynomial of any matrix representation:

\chi_T(\lambda) = \det(\lambda I - A)

where $A$ is the matrix of $T$ in any basis. This is well-defined since similar matrices have the same characteristic polynomial.

Theorem 6.6: Eigenvalue Characterization

$\lambda_0$ is an eigenvalue of $A$ if and only if $\chi_A(\lambda_0) = 0$ .

Proof:

$\lambda_0$ is an eigenvalue ⟺ $\exists v \neq 0: Av = \lambda_0 v$

⟺ $(A - \lambda_0 I)v = 0$ has non-trivial solution

⟺ $A - \lambda_0 I$ is singular

⟺ $\det(A - \lambda_0 I) = 0$

⟺ $\chi_A(\lambda_0) = 0$

∎

Example 6.3: 2×2 Example

For $A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$ :

\chi_A(\lambda) = \det\begin{pmatrix} \lambda - 4 & -2 \\ -1 & \lambda - 3 \end{pmatrix} = (\lambda - 4)(\lambda - 3) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda-2)(\lambda-5)

Eigenvalues: $\lambda_1 = 2$ , $\lambda_2 = 5$ .

Check: tr(A) = 7 = 2 + 5 ✓, det(A) = 10 = 2 × 5 ✓

Example 6.4: 3×3 Example

For $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$ (upper triangular):

\chi_A(\lambda) = (\lambda - 2)^3

Single eigenvalue $\lambda = 2$ with algebraic multiplicity 3.

2. Algebraic and Geometric Multiplicity

When an eigenvalue appears as a repeated root of the characteristic polynomial, we need to distinguish between how many times it appears as a root versus how many independent eigenvectors it has.

Definition 6.5: Algebraic Multiplicity

The algebraic multiplicity of eigenvalue $\lambda_0$ , denoted $a(\lambda_0)$ or $m_a(\lambda_0)$ , is the largest integer $k$ such that $(\lambda - \lambda_0)^k$ divides $\chi_A(\lambda)$ .

Equivalently, it's the exponent of $(\lambda - \lambda_0)$ in the factored form of $\chi_A$ .

Definition 6.6: Geometric Multiplicity

The geometric multiplicity of eigenvalue $\lambda_0$ , denoted $g(\lambda_0)$ or $m_g(\lambda_0)$ , is:

g(\lambda_0) = \dim(E_{\lambda_0}) = \dim\ker(A - \lambda_0 I) = n - \text{rank}(A - \lambda_0 I)

This is the number of linearly independent eigenvectors for $\lambda_0$ .

Theorem 6.7: Multiplicity Inequality

For every eigenvalue $\lambda_0$ :

1 \leq g(\lambda_0) \leq a(\lambda_0)

Proof:

Lower bound ( $g \geq 1$ ): If $\lambda_0$ is an eigenvalue, there exists at least one eigenvector, so $g \geq 1$ .

Upper bound ( $g \leq a$ ): Let $g = \dim E_{\lambda_0}$ . Choose a basis $v_1, \ldots, v_g$ of $E_{\lambda_0}$ and extend to a basis of $F^n$ .

In this basis, $A$ has the form:

A = \begin{pmatrix} \lambda_0 I_g & B \\ 0 & C \end{pmatrix}

Then $\chi_A(\lambda) = (\lambda - \lambda_0)^g \cdot \chi_C(\lambda)$ , so $a(\lambda_0) \geq g$ .

∎

Example 6.5: Equal Multiplicities

For $A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ :

$\chi_A(\lambda) = (\lambda - 3)^2$ , so $a(3) = 2$

$E_3 = \ker(A - 3I) = \ker(0) = \mathbb{R}^2$ , so $g(3) = 2$

Here $g = a$ , and the matrix is diagonalizable (already diagonal!).

Example 6.6: Unequal Multiplicities (Defective Matrix)

For $A = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ :

$\chi_A(\lambda) = (\lambda - 3)^2$ , so $a(3) = 2$

$E_3 = \ker\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \text{span}\{(1, 0)^T\}$ , so $g(3) = 1$

Here $g < a$ . This matrix is defective and cannot be diagonalized.

Remark 6.6: Significance for Diagonalization

A matrix is diagonalizable if and only if $g(\lambda) = a(\lambda)$ for every eigenvalue. When $g < a$ , there aren't enough eigenvectors to form a basis.

Corollary 6.2: Sum of Algebraic Multiplicities

The sum of all algebraic multiplicities equals $n$ (the size of the matrix):

\sum_{i} a(\lambda_i) = n

Corollary 6.3: Sum of Geometric Multiplicities

The sum of all geometric multiplicities is at most $n$ :

\sum_{i} g(\lambda_i) \leq n

Equality holds if and only if the matrix is diagonalizable.

3. Coefficients and Vieta's Formulas

The coefficients of the characteristic polynomial have beautiful interpretations in terms of the matrix entries and eigenvalues.

Theorem 6.8: Characteristic Polynomial Coefficients

For $A = (a_{ij}) \in M_n(F)$ , write:

\chi_A(\lambda) = \lambda^n + c_1\lambda^{n-1} + c_2\lambda^{n-2} + \cdots + c_{n-1}\lambda + c_n

Then:

$c_1 = -\text{tr}(A) = -(a_{11} + a_{22} + \cdots + a_{nn})$
$c_n = (-1)^n \det(A)$
$c_k = (-1)^k$ × (sum of all $k$ × $k$ principal minors)

Theorem 6.9: Vieta's Formulas for Matrices

If $\lambda_1, \ldots, \lambda_n$ are eigenvalues (with multiplicity):

\sum_{i=1}^{n} \lambda_i = \text{tr}(A), \quad \prod_{i=1}^{n} \lambda_i = \det(A)

More generally:

$\sum_{i} \lambda_i = \text{tr}(A)$ (sum of eigenvalues = trace)
$\sum_{i < j} \lambda_i \lambda_j$ = sum of 2×2 principal minors
$\prod_{i} \lambda_i = \det(A)$ (product of eigenvalues = determinant)

Proof:

Since $\chi_A(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n)$ , expanding and comparing coefficients:

Coefficient of $\lambda^{n-1}$ : $-(\lambda_1 + \cdots + \lambda_n) = -\text{tr}(A)$

Constant term: $(-1)^n \lambda_1 \cdots \lambda_n = (-1)^n \det(A)$

∎

Example 6.7: Using Vieta's Formulas

Given $\chi_A(\lambda) = \lambda^3 - 6\lambda^2 + 11\lambda - 6$ , find eigenvalues, trace, and determinant.

Solution: Factor: $(\lambda - 1)(\lambda - 2)(\lambda - 3)$

Eigenvalues: 1, 2, 3

tr(A) = 1 + 2 + 3 = 6 ✓ (check: $-c_1 = 6$ )

det(A) = 1 × 2 × 3 = 6 ✓ (check: $(-1)^3 c_3 = 6$ )

Example 6.8: Finding Missing Eigenvalue

A 3×3 matrix has eigenvalues 2, 5, and $\lambda_3$ . If tr(A) = 10, find $\lambda_3$ .

Solution: $2 + 5 + \lambda_3 = 10 \Rightarrow \lambda_3 = 3$

Remark 6.7: Practical Applications

These formulas are powerful for:

Checking eigenvalue computations
Finding missing eigenvalues when some are known
Determining if 0 is an eigenvalue (iff det = 0)
Quick estimates of eigenvalue properties

4. Characteristic Polynomials of Special Matrices

Theorem 6.10: Triangular Matrices

For triangular (upper or lower) matrix $A$ with diagonal entries $a_{11}, \ldots, a_{nn}$ :

\chi_A(\lambda) = (\lambda - a_{11})(\lambda - a_{22}) \cdots (\lambda - a_{nn})

The eigenvalues are exactly the diagonal entries.

Proof:

For upper triangular $A$ , $\lambda I - A$ is also upper triangular. The determinant of a triangular matrix is the product of diagonal entries:

\det(\lambda I - A) = (\lambda - a_{11})(\lambda - a_{22}) \cdots (\lambda - a_{nn})

∎

Theorem 6.11: Block Diagonal Matrices

For block diagonal $A = \text{diag}(A_1, A_2, \ldots, A_k)$ :

\chi_A(\lambda) = \chi_{A_1}(\lambda) \cdot \chi_{A_2}(\lambda) \cdots \chi_{A_k}(\lambda)

Example 6.9: Block Diagonal

For $A = \begin{pmatrix} 1 & 2 & 0 & 0 \\ 3 & 4 & 0 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 6 \end{pmatrix}$ :

$\chi_A(\lambda) = \chi_{\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}}(\lambda) \cdot (\lambda - 5)(\lambda - 6)$

Theorem 6.12: Similar Matrices

If $B = P^{-1}AP$ , then $\chi_B(\lambda) = \chi_A(\lambda)$ .

Proof:

\chi_B(\lambda) = \det(\lambda I - P^{-1}AP) = \det(P^{-1}(\lambda I - A)P) = \det(P^{-1})\det(\lambda I - A)\det(P) = \chi_A(\lambda)

∎

Corollary 6.4: Invariance Under Similarity

Trace, determinant, and eigenvalues are similarity invariants—they are the same for all similar matrices.

5. Existence of Eigenvalues

Theorem 6.13: Fundamental Theorem of Algebra

Every polynomial of degree $n \geq 1$ with complex coefficients has exactly $n$ roots in $\mathbb{C}$ (counting multiplicity).

Corollary 6.5: Existence Over ℂ

Every $n$ × $n$ complex matrix has exactly $n$ eigenvalues (counting algebraic multiplicity).

Theorem 6.14: Complex Eigenvalues of Real Matrices

For a real matrix, complex eigenvalues occur in conjugate pairs: if $\lambda = a + bi$ is an eigenvalue, so is $\bar{\lambda} = a - bi$ .

Proof:

If $\chi_A(\lambda) = 0$ and $A$ is real, then $\chi_A(\bar{\lambda}) = \overline{\chi_A(\lambda)} = 0$ .

∎

Example 6.10: No Real Eigenvalues

The rotation matrix $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (90° rotation):

$\chi_R(\lambda) = \lambda^2 + 1 = (\lambda - i)(\lambda + i)$

Eigenvalues: $\pm i$ (complex conjugate pair, no real eigenvalues).

Remark 6.8: Real vs Complex Analysis

When working over $\mathbb{R}$ :

The characteristic polynomial may not factor completely
Some matrices have no real eigenvalues (e.g., rotations)
Complex eigenvalues come in conjugate pairs

When working over $\mathbb{C}$ :

Every matrix has eigenvalues
The char poly factors completely into linear terms

6. Computing the Characteristic Polynomial

2×2 Formula

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ :

\chi_A(\lambda) = \lambda^2 - (a+d)\lambda + (ad - bc) = \lambda^2 - \text{tr}(A)\lambda + \det(A)

Example 6.11: Step-by-Step 3×3 Computation

Compute $\chi_A$ for $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{pmatrix}$ .

Method 1 (Direct): Since A is triangular:

\chi_A(\lambda) = (\lambda - 1)(\lambda - 4)(\lambda - 6)

Method 2 (Expansion): Expand $\det(\lambda I - A)$ and verify.

Verification: tr(A) = 11 = 1+4+6 ✓, det(A) = 24 = 1×4×6 ✓

Remark 6.9: Computational Complexity

Computing the characteristic polynomial via determinant expansion:

2×2: Simple formula, instant
3×3: Cofactor expansion, straightforward
4×4 and larger: Increasingly tedious; use software or specialized algorithms (Faddeev-LeVerrier, etc.)

7. Common Mistakes

Sign convention confusion

$\det(\lambda I - A)$ vs $\det(A - \lambda I)$ differ by $(-1)^n$ . Both give the same eigenvalues, but keep conventions consistent.

Confusing multiplicities

Algebraic (root multiplicity) ≠ Geometric (eigenspace dimension) in general. They're equal only for diagonalizable matrices.

Same char poly ⇒ same matrix

FALSE! Different matrices can have identical characteristic polynomials. The char poly doesn't uniquely determine a matrix.

Forgetting multiplicity in Vieta

When computing trace/det from eigenvalues, count repeated eigenvalues multiple times according to their algebraic multiplicity.

8. Worked Examples

Example 6.12: Complete Analysis

For $A = \begin{pmatrix} 2 & 1 & 1 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix}$ , find eigenvalues and their multiplicities.

Step 1: Characteristic polynomial

Since A is upper triangular: $\chi_A(\lambda) = (\lambda - 2)(\lambda - 3)^2$

Step 2: Eigenvalues and algebraic multiplicities

$\lambda_1 = 2$ , $a(2) = 1$
$\lambda_2 = 3$ , $a(3) = 2$

Step 3: Geometric multiplicities

For $\lambda = 2$ : $g(2) = \dim\ker(A - 2I) = \dim\ker\begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} = 1$

For $\lambda = 3$ : $g(3) = \dim\ker(A - 3I) = \dim\ker\begin{pmatrix} -1 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} = 1$

Conclusion: For $\lambda = 3$ : $g = 1 < a = 2$ , so A is NOT diagonalizable.

Example 6.13: Finding Char Poly from Properties

A 3×3 matrix has eigenvalues 1, 1, 2. Find the characteristic polynomial.

\chi_A(\lambda) = (\lambda - 1)^2(\lambda - 2) = \lambda^3 - 4\lambda^2 + 5\lambda - 2

Verify: tr(A) = 1+1+2 = 4 ✓ (coeff of $-\lambda^2$ )

det(A) = 1×1×2 = 2 ✓ (constant term with sign)

Example 6.14: Char Poly of a Projection

Let $P$ be a projection matrix ( $P^2 = P$ ) of rank $r$ on $\mathbb{R}^n$ .

Eigenvalues satisfy $\lambda^2 = \lambda$ , so $\lambda \in \{0, 1\}$ .

Since rank(P) = r, eigenvalue 1 has algebraic multiplicity r.

Since nullity(P) = n-r, eigenvalue 0 has algebraic multiplicity n-r.

\chi_P(\lambda) = \lambda^{n-r}(\lambda - 1)^r

Example 6.15: Using Matrix Equations

If $A^2 - 5A + 6I = 0$ , find possible eigenvalues of $A$ .

Solution: If $\lambda$ is an eigenvalue of $A$ , then:

\lambda^2 - 5\lambda + 6 = 0 \Rightarrow (\lambda - 2)(\lambda - 3) = 0

Possible eigenvalues: $\lambda \in \{2, 3\}$

Example 6.16: Companion Matrix

The companion matrix of polynomial $p(\lambda) = \lambda^3 - 6\lambda^2 + 11\lambda - 6$ :

C = \begin{pmatrix} 0 & 0 & 6 \\ 1 & 0 & -11 \\ 0 & 1 & 6 \end{pmatrix}

The characteristic polynomial of $C$ is exactly $p(\lambda)$ !

This is a key construction: any monic polynomial is the char poly of its companion matrix.

9. Preview: Minimal Polynomial

Definition 6.7: Minimal Polynomial

The minimal polynomial $m_A(\lambda)$ is the monic polynomial of smallest degree such that $m_A(A) = 0$ .

Theorem 6.15: Minimal vs Characteristic

The minimal polynomial:

Divides the characteristic polynomial
Has the same roots (eigenvalues) as the char poly
May have different multiplicities
Has degree ≤ n

Example 6.17: When They Differ

For $A = I_n$ :

Characteristic polynomial: $\chi_A(\lambda) = (\lambda - 1)^n$
Minimal polynomial: $m_A(\lambda) = \lambda - 1$

The minimal polynomial has degree 1 while the char poly has degree n.

Remark 6.10: Significance

The minimal polynomial is important because:

It determines when a matrix is diagonalizable (iff minimal poly has no repeated roots)
It gives the simplest polynomial relation satisfied by the matrix
It's key to the Jordan normal form theory

10. Applications

Stability Analysis

The characteristic polynomial roots determine system stability: stable if all roots have negative real part (continuous) or magnitude < 1 (discrete).

Recurrence Relations

Linear recurrences like Fibonacci have solutions determined by roots of a characteristic polynomial.

Differential Equations

Solutions to $\mathbf{x}' = A\mathbf{x}$ involve $e^{\lambda t}$ terms for each eigenvalue $\lambda$ .

Graph Theory

The characteristic polynomial of the adjacency matrix encodes graph properties like number of spanning trees.

Example 6.18: Fibonacci Recurrence

The Fibonacci sequence $F_n = F_{n-1} + F_{n-2}$ has matrix form:

\begin{pmatrix} F_{n+1} \\ F_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}

The characteristic polynomial $\lambda^2 - \lambda - 1$ has roots $\phi = \frac{1+\sqrt{5}}{2}$ and $\hat{\phi} = \frac{1-\sqrt{5}}{2}$ (golden ratio!).

This gives Binet's formula: $F_n = \frac{\phi^n - \hat{\phi}^n}{\sqrt{5}}$

Key Takeaways

Definition

$\chi_A(\lambda) = \det(\lambda I - A)$

Multiplicity Inequality

$1 \leq g(\lambda) \leq a(\lambda)$

Vieta's Formulas

$\sum\lambda_i = \text{tr}(A)$ , $\prod\lambda_i = \det(A)$

Similarity Invariance

Similar matrices have identical char poly

Quick Reference

2×2 Formula

$\chi_A(\lambda) = \lambda^2 - \text{tr}(A)\lambda + \det(A)$

$\lambda = \frac{\text{tr}(A) \pm \sqrt{\text{tr}(A)^2 - 4\det(A)}}{2}$

Key Coefficients

• Leading: 1 (monic)
• $\lambda^{n-1}$ coeff: $-\text{tr}(A)$
• Constant: $(-1)^n \det(A)$

Additional Practice

Problem 1

Find the characteristic polynomial of $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ .

Answer: $\lambda^2 - 2\lambda - 3 = (\lambda - 3)(\lambda + 1)$

Problem 2

A 4×4 matrix has eigenvalues 1, 1, 2, 2. Find its trace and determinant.

Answer: tr = 6, det = 4

Problem 3

If $\chi_A(\lambda) = \lambda^4 - 10\lambda^3 + c\lambda^2 + d\lambda + 24$ with eigenvalues 1, 2, 3, 4, find c.

Hint: c = sum of products of pairs = 1·2+1·3+1·4+2·3+2·4+3·4 = 35

Problem 4

For what values of $a$ does $A = \begin{pmatrix} a & 1 \\ 1 & a \end{pmatrix}$ have repeated eigenvalues?

Answer: Never (discriminant = 4 > 0 always)

Problem 5 (Challenge)

Prove that $A$ and $A^T$ have the same characteristic polynomial.

11. Advanced Topics

Theorem 6.16: Cayley-Hamilton (Preview)

Every matrix satisfies its own characteristic equation:

\chi_A(A) = A^n + c_1 A^{n-1} + \cdots + c_{n-1} A + c_n I = 0

This powerful result will be explored in detail in a later section.

Corollary 6.6: Matrix Polynomials

Any power $A^k$ for $k \geq n$ can be expressed as a linear combination of $I, A, A^2, \ldots, A^{n-1}$ .

Example 6.19: Using Cayley-Hamilton

For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ , express $A^{-1}$ as a linear combination of $I$ and $A$ .

$\chi_A(\lambda) = \lambda^2 - 5\lambda - 2$

By Cayley-Hamilton: $A^2 - 5A - 2I = 0$

Rearranging: $A^2 - 5A = 2I$

$A(A - 5I) = 2I \Rightarrow A^{-1} = \frac{1}{2}(A - 5I)$

Theorem 6.17: Characteristic Polynomial of Block Triangular

For block upper triangular $A = \begin{pmatrix} B & C \\ 0 & D \end{pmatrix}$ :

\chi_A(\lambda) = \chi_B(\lambda) \cdot \chi_D(\lambda)

Proof:

\det(\lambda I - A) = \det\begin{pmatrix} \lambda I - B & -C \\ 0 & \lambda I - D \end{pmatrix} = \det(\lambda I - B) \cdot \det(\lambda I - D)

∎

Remark 6.11: Computational Note

This theorem is extremely useful for computing characteristic polynomials of large matrices that have block structure.

12. Real-World Examples

Example 6.20: Population Dynamics (Leslie Matrix)

A population model with three age groups:

L = \begin{pmatrix} 0 & 2 & 1 \\ 0.5 & 0 & 0 \\ 0 & 0.3 & 0 \end{pmatrix}

The dominant eigenvalue determines long-term population growth rate.

$\chi_L(\lambda) = -\lambda^3 + 0.5 \cdot 2 \cdot \lambda + 0.5 \cdot 0.3 \cdot 1 = -\lambda^3 + \lambda + 0.15$

The largest real root (≈ 1.1) indicates ~10% population growth per generation.

Example 6.21: Markov Chain

A transition matrix for weather states (sunny, cloudy, rainy):

P = \begin{pmatrix} 0.7 & 0.2 & 0.1 \\ 0.3 & 0.4 & 0.3 \\ 0.2 & 0.3 & 0.5 \end{pmatrix}

For stochastic matrices:

$\lambda = 1$ is always an eigenvalue
All eigenvalues satisfy $|\lambda| \leq 1$
The eigenvector for $\lambda = 1$ gives the stationary distribution

Example 6.22: Coupled Oscillators

Two masses connected by springs have equations of motion determined by:

M = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}

$\chi_M(\lambda) = (\lambda - 2)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3)$

Natural frequencies: $\omega_1 = 1, \omega_2 = \sqrt{3}$

13. Computational Methods

Faddeev-LeVerrier Algorithm

Computes characteristic polynomial coefficients iteratively:

Set $B_1 = A$ , $c_1 = -\text{tr}(B_1)$
For $k = 2, \ldots, n$ : $B_k = A(B_{k-1} + c_{k-1}I)$ , $c_k = -\frac{1}{k}\text{tr}(B_k)$

Remark 6.12: Numerical Stability

In practice, directly computing the characteristic polynomial and finding its roots is numerically unstable for large matrices. Instead:

QR iteration: Preferred for dense matrices
Lanczos method: For large sparse symmetric matrices
Arnoldi iteration: For large sparse non-symmetric matrices

Example 6.23: Rational Root Test

For $\chi_A(\lambda) = \lambda^3 - 6\lambda^2 + 11\lambda - 6$ :

Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6$

Testing: $\chi_A(1) = 1 - 6 + 11 - 6 = 0$ ✓

Factor: $(\lambda - 1)(\lambda^2 - 5\lambda + 6) = (\lambda - 1)(\lambda - 2)(\lambda - 3)$

14. Deeper Theory

Theorem 6.18: Eigenspaces are Direct Sum

If $\lambda_1, \ldots, \lambda_k$ are distinct eigenvalues:

E_{\lambda_1} + E_{\lambda_2} + \cdots + E_{\lambda_k} = E_{\lambda_1} \oplus E_{\lambda_2} \oplus \cdots \oplus E_{\lambda_k}

The sum is always direct (no overlap except at 0).

Proof:

Eigenvectors corresponding to distinct eigenvalues are linearly independent (proved in previous section). If $v \in E_{\lambda_i} \cap \sum_{j \neq i} E_{\lambda_j}$ , write $v = \sum_{j \neq i} v_j$ . Then both sides are eigenvectors for $\lambda_i$ (left) and a combination for other eigenvalues (right). By linear independence, $v = 0$ .

∎

Corollary 6.7: Dimension Bound

For distinct eigenvalues $\lambda_1, \ldots, \lambda_k$ :

\sum_{i=1}^{k} g(\lambda_i) = \sum_{i=1}^{k} \dim E_{\lambda_i} \leq n

Theorem 6.19: Trace Formula

For any matrix $A$ :

\text{tr}(A^k) = \sum_{i=1}^{n} \lambda_i^k

where $\lambda_i$ are eigenvalues counted with multiplicity.

Example 6.24: Using Trace Formula

If $\chi_A(\lambda) = (\lambda - 1)^2(\lambda - 2)$ , find $\text{tr}(A^2)$ .

Solution: $\text{tr}(A^2) = 1^2 + 1^2 + 2^2 = 1 + 1 + 4 = 6$

Study Tips

Master 2×2: The formula $\chi_A = \lambda^2 - \text{tr}(A)\lambda + \det(A)$ should be instant.
Use verification: Always check trace = sum, det = product of eigenvalues.
Exploit structure: Triangular and block diagonal matrices have simple char polys.
Think about multiplicities: Understand both algebraic and geometric.
Know the bounds: $1 \leq g \leq a$ is fundamental.
Practice factoring: Rational root test and special patterns help.

Characteristic Polynomial Patterns

Matrix Type	Characteristic Polynomial	Notes
Identity $I_n$	$(\lambda - 1)^n$	Single eigenvalue 1
Scalar $cI_n$	$(\lambda - c)^n$	Single eigenvalue c
Diagonal	$\prod_i (\lambda - d_i)$	Eigenvalues = diagonal
Triangular	$\prod_i (\lambda - a_{ii})$	Same as diagonal case
Nilpotent	$\lambda^n$	Only eigenvalue is 0
Projection	$\lambda^{n-r}(\lambda - 1)^r$	r = rank(P)
2D Rotation by θ	$\lambda^2 - 2\cos\theta \cdot \lambda + 1$	Eigenvalues $e^{\pm i\theta}$
Companion matrix	Given polynomial	Char poly = input poly

Chapter Summary

The characteristic polynomial $\chi_A(\lambda) = \det(\lambda I - A)$ encodes all eigenvalue information. Its roots are eigenvalues, with algebraic multiplicity given by root multiplicity. Vieta's formulas connect coefficients to trace and determinant. The fundamental theorem of algebra guarantees existence over $\mathbb{C}$ .

19

Theorems

24

Examples

12

Quiz Questions

8

FAQs

Historical Notes

Cauchy (1826): First studied eigenvalue equations systematically for quadratic forms.

Sylvester (1852): Introduced the term "matrix" and studied determinantal equations.

Cayley (1858): Stated the Cayley-Hamilton theorem: every matrix satisfies its own characteristic equation.

Frobenius (1878): Developed the theory of characteristic and minimal polynomials systematically.

Weierstrass (1868): Introduced elementary divisors, refining the study of characteristic polynomials.

Terminology: "Characteristic" comes from the idea that this polynomial characterizes essential properties of the matrix.

15. Conceptual Questions

Q1: Can two non-similar matrices have the same characteristic polynomial?

Yes! Example: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ both have $\chi = \lambda^2$ , but are not similar.

Q2: If all eigenvalues are 0, must the matrix be 0?

No! Nilpotent matrices have all eigenvalues 0 but may be non-zero. Example: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ .

Q3: Can eigenvalues be complex for a real matrix?

Yes! But they come in conjugate pairs. Example: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ has eigenvalues $\pm i$ .

Q4: Is the characteristic polynomial unique up to scalar multiple?

No, it's completely unique. We define it as a monic polynomial (leading coefficient = 1).

Q5: Does A invertible imply all eigenvalues are non-zero?

Yes! det(A) = product of eigenvalues, so A invertible ⟺ det(A) ≠ 0 ⟺ no eigenvalue is 0.

16. Challenge Problems

Challenge 1

Prove that if $A$ is idempotent ( $A^2 = A$ ), then the only possible eigenvalues are 0 and 1.

Challenge 2

If $\chi_A(\lambda) = \lambda^n + c_1\lambda^{n-1} + \cdots + c_n$ , prove $c_2 = \sum_{i < j} \lambda_i \lambda_j$ equals the sum of all 2×2 principal minors.

Challenge 3

Let $A$ be a 3×3 matrix with $\chi_A(\lambda) = \lambda^3 - 3\lambda + 2$ . Find $\text{tr}(A^3)$ .

Hint: Eigenvalues are 1, 1, -2. Use $\text{tr}(A^3) = \sum \lambda_i^3$ .

Challenge 4

Show that for any polynomial $p(\lambda)$ of degree $n$ , there exists an $n$ × $n$ matrix whose characteristic polynomial is $p(\lambda)$ .

Hint: Use the companion matrix construction.

Challenge 5

Prove that $\chi_{AB}(\lambda) = \chi_{BA}(\lambda)$ for any $n$ × $n$ matrices $A$ and $B$ .

17. Exam Preparation

What You Should Know

• Compute $\chi_A$ for 2×2 and 3×3 matrices
• Factor simple characteristic polynomials
• Find algebraic and geometric multiplicities
• Apply Vieta's formulas
• Determine diagonalizability from multiplicities
• Handle triangular and block matrices

Common Exam Questions

• Find the characteristic polynomial
• Verify eigenvalues using trace/det
• Determine if matrix is diagonalizable
• Find missing eigenvalue given others
• True/False on multiplicity properties
• Prove similarity invariance results

Verification Checklist

After computing eigenvalues, always verify:

Sum of eigenvalues = tr(A)

Product of eigenvalues = det(A)

Sum of algebraic multiplicities = n

1 \leq g(\lambda) \leq a(\lambda)

for each

Polynomial has correct degree n

Leading coefficient is 1 (monic)

What's Next?

With the characteristic polynomial understood, you're ready for:

Diagonalization: When can we write $A = PDP^{-1}$ ? The condition $g = a$ for all eigenvalues.
Jordan Normal Form: What to do when diagonalization fails—generalized eigenvectors.
Cayley-Hamilton Theorem: The remarkable fact that $\chi_A(A) = 0$ always holds.
Minimal Polynomial: The smallest polynomial annihilating the matrix.

18. Quick Examples Gallery

Example: Symmetric 2×2

$A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$

$\chi_A = \lambda^2 - 6\lambda + 8 = (\lambda - 2)(\lambda - 4)$

Eigenvalues: 2, 4 (real, distinct)

Example: Skew-Symmetric 2×2

$A = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$

$\chi_A = \lambda^2 + 4$

Eigenvalues: $\pm 2i$ (pure imaginary)

Example: 3×3 with Zero Eigenvalue

$A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}$

rank = 1, so eigenvalue 0 has $a(0) = 2$

$\chi_A = \lambda^2(\lambda - 6)$

Example: Permutation Matrix

$P = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$ (cyclic)

$\chi_P = \lambda^3 - 1$

Eigenvalues: 1, $\omega, \omega^2$ (cube roots of 1)

Memorization Aids

2×2 Formula

"Trace minus lambda, then add det"

$\lambda^2 - \text{tr}\cdot\lambda + \det$

Multiplicity Inequality

"Geometric ≤ Algebraic, like g ≤ a alphabetically"

Vieta's Formulas

"Sum = Trace, Product = Determinant"

Triangular Matrices

"Eigenvalues on the diagonal"

Learning Path

Current

6.2 Characteristic Polynomial

Module Summary

In this module, you learned how the characteristic polynomial $\chi_A(\lambda) = \det(\lambda I - A)$ encodes all eigenvalue information. You can now:

Compute characteristic polynomials for 2×2 and 3×3 matrices
Distinguish between algebraic and geometric multiplicities
Apply Vieta's formulas to verify eigenvalue computations
Recognize special cases (triangular, block diagonal, similar matrices)
Understand when matrices are diagonalizable based on multiplicities

Characteristic Polynomial Practice

12

Questions

0

Correct

0%

Accuracy

1

The characteristic polynomial of

A

is:

Easy

Not attempted

2

The degree of the characteristic polynomial for

n \times n

matrix is:

Easy

Not attempted

3

Sum of eigenvalues (with multiplicity) equals:

Medium

Not attempted

4

Product of eigenvalues (with multiplicity) equals:

Medium

Not attempted

5

Algebraic multiplicity of

\lambda_0

is:

Medium

Not attempted

6

Geometric multiplicity satisfies:

Hard

Not attempted

7

Similar matrices

A

and

P^{-1}AP

have:

Medium

Not attempted

8

For a triangular matrix, eigenvalues are:

Easy

Not attempted

9

If

\chi_A(\lambda) = (\lambda - 2)^3(\lambda + 1)

, trace of

A

is:

Medium

Not attempted

10

If

\chi_A(\lambda) = (\lambda - 2)^3(\lambda + 1)

, det of

A

is:

Medium

Not attempted

11

An

n×n

matrix over

\mathbb{C}

has:

Medium

Not attempted

12

If geometric = algebraic multiplicity for all eigenvalues, then:

Hard

Not attempted

Frequently Asked Questions

Why is it called 'characteristic' polynomial?

It 'characterizes' the matrix's essential spectral properties - its eigenvalues. Similar matrices have the same characteristic polynomial, making it an invariant under similarity transformations.

What if the characteristic polynomial doesn't factor over ℝ?

Then there are no real eigenvalues. Over ℂ, by the Fundamental Theorem of Algebra, it always factors completely into linear terms, giving exactly n eigenvalues (counting multiplicity).

What's the difference between algebraic and geometric multiplicity?

Algebraic = exponent in char poly (how many times λ₀ is a root). Geometric = dim(ker(A - λ₀I)) = number of linearly independent eigenvectors. Always: 1 ≤ geometric ≤ algebraic.

How do I find eigenvalues of a 3×3 matrix?

Expand det(A - λI) to get a cubic polynomial. Try rational roots (±divisors of constant/leading coeff), look for special structure, or factor by grouping. Numerical methods work for harder cases.

Why is det(λI - A) sometimes used instead of det(A - λI)?

They differ by (-1)ⁿ. Using λI - A makes the polynomial monic (leading coefficient 1). Both give the same eigenvalues.

How does the characteristic polynomial relate to the minimal polynomial?

The minimal polynomial divides the characteristic polynomial. They have the same roots (eigenvalues) but possibly different multiplicities. The minimal polynomial has the smallest degree annihilating A.

What happens to the characteristic polynomial under similarity?

It's invariant! If B = P⁻¹AP, then det(λI - B) = det(P⁻¹(λI - A)P) = det(λI - A). This is why eigenvalues are similarity invariants.

Can two different matrices have the same characteristic polynomial?

Yes! Non-similar matrices can have the same char poly if their Jordan structures differ. For example, I₂ and a non-identity 2×2 with eigenvalue 1 (multiplicity 2) both have char poly (λ-1)².

1. Characteristic Polynomial

2. Algebraic and Geometric Multiplicity

3. Coefficients and Vieta's Formulas

4. Characteristic Polynomials of Special Matrices

5. Existence of Eigenvalues

6. Computing the Characteristic Polynomial

2×2 Formula

7. Common Mistakes

Sign convention confusion

Confusing multiplicities

Same char poly ⇒ same matrix

Forgetting multiplicity in Vieta

8. Worked Examples

9. Preview: Minimal Polynomial

10. Applications

Stability Analysis

Recurrence Relations

Differential Equations

Graph Theory

Key Takeaways

Definition

Multiplicity Inequality

Vieta's Formulas

Similarity Invariance

Quick Reference

2×2 Formula

Key Coefficients

Additional Practice

11. Advanced Topics

12. Real-World Examples

13. Computational Methods

Faddeev-LeVerrier Algorithm

14. Deeper Theory

Study Tips

Characteristic Polynomial Patterns

Chapter Summary

Historical Notes

15. Conceptual Questions

16. Challenge Problems

17. Exam Preparation

What You Should Know

Common Exam Questions

Verification Checklist

What's Next?

18. Quick Examples Gallery

Example: Symmetric 2×2

Example: Skew-Symmetric 2×2

Example: 3×3 with Zero Eigenvalue

Example: Permutation Matrix

Memorization Aids

Learning Path

Module Summary

Related Topics

Frequently Asked Questions

Why is it called 'characteristic' polynomial?

What if the characteristic polynomial doesn't factor over ℝ?

What's the difference between algebraic and geometric multiplicity?

How do I find eigenvalues of a 3×3 matrix?

Why is det(λI - A) sometimes used instead of det(A - λI)?

How does the characteristic polynomial relate to the minimal polynomial?

What happens to the characteristic polynomial under similarity?

Can two different matrices have the same characteristic polynomial?