Eigenvalue Definition | Eigenvalues and Eigenvectors | Linear Algebra

1. Eigenvalue and Eigenvector

Our goal in studying linear transformations is to find simple matrix representations. The simplest form is a diagonal matrix. This naturally leads us to study invariant subspaces, particularly one-dimensional ones where $T(v) = \lambda v$ for some scalar $\lambda$ .

Definition 6.1: Eigenvalue and Eigenvector (Linear Operator)

Let $T: V \to V$ be a linear operator on a vector space $V(F)$ . A scalar $\lambda \in F$ is an eigenvalue (characteristic value) of $T$ if there exists a non-zero vector $\xi \in V$ such that:

T(\xi) = \lambda \xi

The non-zero vector $\xi$ is called an eigenvector (characteristic vector) belonging to eigenvalue $\lambda$ .

Remark 6.1: Non-zero Requirement

Eigenvectors must be non-zero by definition. Otherwise, the zero vector would satisfy $T(0) = \lambda \cdot 0$ for any $\lambda$ , making the concept meaningless. However, $\lambda = 0$ is a valid eigenvalue—it simply means $T(\xi) = 0$ , so eigenvectors for $\lambda = 0$ are exactly the non-zero vectors in $\ker(T)$ .

Definition 6.2: Eigenvalue and Eigenvector (Matrix)

Let $A \in M_n(F)$ be an $n \times n$ matrix. A scalar $\lambda \in F$ is an eigenvalue of $A$ if there exists a non-zero vector $X \in F^n$ such that:

AX = \lambda X

The non-zero vector $X$ is called an eigenvector of $A$ belonging to $\lambda$ .

Remark 6.2: Connection Between Operator and Matrix

If $A$ is the matrix representation of $T$ under basis $\alpha_1, \ldots, \alpha_n$ , and $\xi = (\alpha_1, \ldots, \alpha_n)X$ , then:

T(\xi) = \lambda\xi \iff AX = \lambda X

The eigenvalue $\lambda$ is the same for both, but the eigenvector $X$ is the coordinate representation of $\xi$ in the chosen basis.

Definition 6.3: Eigenspace

For eigenvalue $\lambda$ of operator $T$ , the eigenspace is:

E_\lambda = V_\lambda = \ker(T - \lambda I) = \{v \in V : Tv = \lambda v\}

This set includes the zero vector and all eigenvectors for $\lambda$ .

Theorem 6.1: Eigenspace is an Invariant Subspace

$E_\lambda$ is a subspace of $V$ and is invariant under $T$ .

Proof:

Subspace: First, $0 \in E_\lambda$ since $T(0) = \lambda \cdot 0$ .

For $\xi_1, \xi_2 \in E_\lambda$ and scalars $k_1, k_2$ :

T(k_1\xi_1 + k_2\xi_2) = k_1 T(\xi_1) + k_2 T(\xi_2) = k_1 \lambda\xi_1 + k_2 \lambda\xi_2 = \lambda(k_1\xi_1 + k_2\xi_2)

So $k_1\xi_1 + k_2\xi_2 \in E_\lambda$ , proving closure.

Invariance: For any $\xi \in E_\lambda$ , we have $T(\xi) = \lambda\xi \in E_\lambda$ .

∎

Remark 6.3: Dimension of Eigenspace

The dimension of $E_\lambda$ is at least 1 (when $\lambda$ is an eigenvalue) but can be larger. If $\dim(E_\lambda) > 1$ , there are multiple linearly independent eigenvectors for the same eigenvalue.

Definition 6.4: Spectrum

The spectrum of $T$ (or $A$ ) is the set of all eigenvalues:

\sigma(T) = \{\lambda \in F : \lambda \text{ is an eigenvalue of } T\}

Remark 6.4: Geometric Meaning

An eigenvector $v$ lies on an invariant line: the transformation $T$ maps $v$ to a scalar multiple of itself. Geometrically:

$|\lambda| > 1$ : stretching along the eigenvector direction
$|\lambda| < 1$ : shrinking along the eigenvector direction
$\lambda < 0$ : reflection (direction reversal) plus scaling
$\lambda = 1$ : the eigenvector is fixed by $T$
$\lambda = 0$ : the eigenvector is mapped to zero (in kernel)

2. Examples

Example 6.1: Diagonal Matrix

For $A = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix}$ :

$\lambda_1 = 3$ with eigenvector $(1, 0)^T$
$\lambda_2 = -2$ with eigenvector $(0, 1)^T$

Key insight: For any diagonal matrix, the eigenvalues are the diagonal entries, and the standard basis vectors are eigenvectors.

Example 6.2: Triangular Matrix

For upper triangular $A = \begin{pmatrix} 2 & 1 & 3 \\ 0 & 5 & 7 \\ 0 & 0 & -1 \end{pmatrix}$ :

Eigenvalues are $\lambda_1 = 2, \lambda_2 = 5, \lambda_3 = -1$ (the diagonal entries).

However, the eigenvectors are NOT simply the standard basis vectors (unlike diagonal matrices).

Example 6.3: Rotation Matrix (No Real Eigenvalues)

For 90° rotation in $\mathbb{R}^2$ : $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

No real eigenvalues! Geometrically, no non-zero vector stays on its own line after 90° rotation.

Over $\mathbb{C}$ : $\lambda = \pm i$ with eigenvectors $(1, \mp i)^T$ .

Example 6.4: Projection Matrix

For projection onto the x-axis: $P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$

$\lambda_1 = 1$ : eigenvectors are all vectors on the x-axis (projected onto themselves)
$\lambda_2 = 0$ : eigenvectors are all vectors on the y-axis (projected to zero)

Example 6.5: Finding Eigenvalue from Condition

Given $A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 0 & 1 \\ 1 & a & 0 \end{pmatrix}$ with non-zero $\alpha$ such that $A\alpha = 2\alpha$ . Find $a$ .

Solution: Since 2 is an eigenvalue, $\det(2I - A) = 0$ :

\det\begin{pmatrix} 1 & 1 & 0 \\ -2 & 2 & -1 \\ -1 & -a & 2 \end{pmatrix} = 9 - a = 0

Therefore $a = 9$ .

Example 6.6: Identity and Scalar Matrices

For identity $I_n$ : eigenvalue is $\lambda = 1$ with multiplicity $n$ . Every non-zero vector is an eigenvector.

For scalar matrix $cI_n$ : eigenvalue is $\lambda = c$ with multiplicity $n$ .

Example 6.7: Reflection Matrix

Reflection across the line $y = x$ : $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

$\lambda_1 = 1$ with eigenvector $(1, 1)^T$ (vectors on $y = x$ are fixed)
$\lambda_2 = -1$ with eigenvector $(1, -1)^T$ (vectors perpendicular to $y = x$ are flipped)

3. Finding Eigenvalues

Theorem 6.2: Eigenvalue Criterion

The following are equivalent for $\lambda \in F$ :

$\lambda$ is an eigenvalue of $T$ (or $A$ )
$T - \lambda I$ is not injective
$T - \lambda I$ is not surjective
$T - \lambda I$ is not invertible
$\det(A - \lambda I) = 0$

Proof:

(1) ⇒ (2): If $\lambda$ is an eigenvalue, there exists $v \neq 0$ with $Tv = \lambda v$ . Thus $(T - \lambda I)(v) = 0$ , so $\ker(T - \lambda I) \neq \{0\}$ , meaning $T - \lambda I$ is not injective.

(2) ⇔ (3) ⇔ (4): For operators on finite-dimensional spaces, injective ⇔ surjective ⇔ invertible.

(4) ⇔ (5): A matrix is invertible iff its determinant is non-zero.

∎

Definition 6.5: Characteristic Polynomial

The characteristic polynomial of matrix $A$ is:

f(\lambda) = \det(\lambda I - A) = \lambda^n + a_1\lambda^{n-1} + \cdots + a_{n-1}\lambda + a_n

Its roots are exactly the eigenvalues of $A$ .

Remark 6.5: Convention

Some texts use $\det(A - \lambda I)$ instead of $\det(\lambda I - A)$ . They differ by a factor of $(-1)^n$ but have the same roots.

Theorem 6.3: Coefficients of Characteristic Polynomial

For $f(\lambda) = \det(\lambda I - A) = \lambda^n + a_1\lambda^{n-1} + \cdots + a_n$ :

$a_1 = -\text{tr}(A)$ (negative trace)
$a_n = (-1)^n \det(A)$
$a_k = (-1)^k$ (sum of all $k$ × $k$ principal minors)

Corollary 6.1: Vieta's Formulas for Eigenvalues

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)

Example 6.8: Computing Eigenvalues

Find eigenvalues of $A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$ :

\det(\lambda I - A) = \det\begin{pmatrix} \lambda - 4 & 2 \\ -1 & \lambda - 1 \end{pmatrix} = (\lambda - 4)(\lambda - 1) + 2 = \lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3)

Eigenvalues: $\lambda_1 = 2, \lambda_2 = 3$ .

Verification: tr(A) = 4 + 1 = 5 = 2 + 3 ✓, det(A) = 4 - (-2) = 6 = 2 × 3 ✓

Example 6.9: Finding Eigenvectors

For $\lambda = 2$ , solve $(A - 2I)X = 0$ :

\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

Row reducing: $x_1 = x_2$ , so eigenvector $v_1 = (1, 1)^T$ .

For $\lambda = 3$ : eigenvector $v_2 = (2, 1)^T$ .

4. Key Properties

Theorem 6.4: Linear Independence of Eigenvectors

Eigenvectors corresponding to distinct eigenvalues are linearly independent.

Proof:

We prove by induction on the number of distinct eigenvalues $k$ .

Base case (k=1): A single eigenvector $v_1 \neq 0$ is linearly independent.

Inductive step: Assume true for $k-1$ . Suppose $c_1 v_1 + \cdots + c_k v_k = 0$ where $Av_i = \lambda_i v_i$ with distinct $\lambda_i$ .

Apply $A$ : $c_1 \lambda_1 v_1 + \cdots + c_k \lambda_k v_k = 0$

Multiply original by $\lambda_k$ : $c_1 \lambda_k v_1 + \cdots + c_k \lambda_k v_k = 0$

Subtract: $c_1(\lambda_1 - \lambda_k)v_1 + \cdots + c_{k-1}(\lambda_{k-1} - \lambda_k)v_{k-1} = 0$

By induction hypothesis and $\lambda_i - \lambda_k \neq 0$ , we get $c_1 = \cdots = c_{k-1} = 0$ , hence $c_k = 0$ .

∎

Corollary 6.2: Maximum Distinct Eigenvalues

An $n \times n$ matrix has at most $n$ distinct eigenvalues.

Theorem 6.5: Similar Matrices Have Same Eigenvalues

If $B = P^{-1}AP$ (A and B are similar), then A and B have the same characteristic polynomial, hence the same eigenvalues (with multiplicities).

Proof:

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

∎

Remark 6.6: Eigenvectors Under Similarity

If $P^{-1}AP = B$ and $X$ is an eigenvector of $A$ for $\lambda$ , then $P^{-1}X$ is an eigenvector of $B$ for the same $\lambda$ .

Theorem 6.6: Eigenvalues of AB and BA

If $\lambda \neq 0$ is an eigenvalue of $AB$ , then $\lambda$ is also an eigenvalue of $BA$ .

Proof:

Let $ABX = \lambda X$ with $X \neq 0$ . Then $BA(BX) = B(ABX) = B(\lambda X) = \lambda(BX)$ .

We need $BX \neq 0$ : If $BX = 0$ , then $ABX = 0 = \lambda X$ . Since $\lambda \neq 0$ , this implies $X = 0$ , contradiction.

∎

5. Special Matrix Properties

Theorem 6.7: Eigenvalues of Matrix Powers and Functions

If $\lambda$ is an eigenvalue of $A$ with eigenvector $\xi$ , then:

$k\lambda$ is an eigenvalue of $kA$
$\lambda^m$ is an eigenvalue of $A^m$
For polynomial $f(x) = a_n x^n + \cdots + a_1 x + a_0$ , $f(\lambda)$ is an eigenvalue of $f(A)$

In all cases, $\xi$ remains the eigenvector.

Proof:

For (2): $A^m \xi = A^{m-1}(A\xi) = A^{m-1}(\lambda\xi) = \lambda A^{m-1}\xi = \cdots = \lambda^m \xi$

For (3): $f(A)\xi = (a_n A^n + \cdots + a_0 I)\xi = a_n \lambda^n \xi + \cdots + a_0 \xi = f(\lambda)\xi$

∎

Theorem 6.8: Eigenvalues of Inverse and Adjugate

If $A$ is invertible with eigenvalue $\lambda$ (necessarily $\lambda \neq 0$ ):

$\lambda^{-1}$ is an eigenvalue of $A^{-1}$
$\det(A) \cdot \lambda^{-1}$ is an eigenvalue of $\text{adj}(A)$

The eigenvector is unchanged.

Proof:

From $A\xi = \lambda\xi$ : $\xi = A^{-1}(\lambda\xi) = \lambda A^{-1}\xi$ , so $A^{-1}\xi = \lambda^{-1}\xi$ .

Since $\text{adj}(A) = \det(A) \cdot A^{-1}$ when $A$ is invertible: $\text{adj}(A)\xi = \det(A) \cdot \lambda^{-1} \xi$ .

∎

Theorem 6.9: Eigenvalues of Transpose

$A$ and $A^T$ have the same eigenvalues (but generally different eigenvectors).

Proof:

$\det(\lambda I - A^T) = \det((\lambda I - A)^T) = \det(\lambda I - A)$

∎

Definition 6.6: Special Matrix Types

Based on eigenvalue constraints from matrix equations:

Involution ( $A^2 = I$ ): eigenvalues can only be $\pm 1$
Idempotent ( $A^2 = A$ ): eigenvalues can only be 0 or 1
Nilpotent ( $A^k = 0$ ): all eigenvalues are 0

Example 6.10: Eigenvalue Constraints from Equations

If $A^2 = A$ (idempotent), and $A\xi = \lambda\xi$ :

A^2\xi = \lambda^2 \xi = A\xi = \lambda\xi \implies \lambda^2 = \lambda \implies \lambda(\lambda - 1) = 0

So $\lambda \in \{0, 1\}$ .

Example 6.11: Computing with Eigenvalue Properties

Given 3×3 matrix $A$ with eigenvalues 1, -2, -1. Find:

$\det(A) = 1 \times (-2) \times (-1) = 2$
$\text{adj}(A)$ has eigenvalues: $2/1 = 2, 2/(-2) = -1, 2/(-1) = -2$
$(A^{-1})^2 + 2I$ has eigenvalues: $1 + 2 = 3, 1/4 + 2 = 9/4, 1 + 2 = 3$
$A^2 - A + I$ has eigenvalues: $1 - 1 + 1 = 1, 4 + 2 + 1 = 7, 1 + 1 + 1 = 3$

6. Common Mistakes

Zero is NOT an eigenvector

Eigenvectors must be non-zero by definition. The zero vector trivially satisfies $A \cdot 0 = \lambda \cdot 0$ for any $\lambda$ .

Zero CAN be an eigenvalue

$\lambda = 0$ is valid when $\ker(A) \neq \{0\}$ . This means $A$ is singular (non-invertible).

Confusing algebraic and geometric multiplicity

Algebraic = root multiplicity in characteristic polynomial. Geometric = dim(eigenspace). Always: geometric ≤ algebraic.

Similar matrices have same eigenvectors

FALSE! Similar matrices have same eigenvalues, but eigenvectors differ by the change-of-basis matrix.

Every matrix has real eigenvalues

FALSE over $\mathbb{R}$ ! Rotation matrices often have no real eigenvalues. Over $\mathbb{C}$ , every matrix has eigenvalues.

7. Applications Preview

Differential Equations

Solutions to $\mathbf{x}' = A\mathbf{x}$ involve $e^{\lambda t}$ where $\lambda$ are eigenvalues.

Stability Analysis

System stable iff all $|\lambda| < 1$ (discrete) or $\text{Re}(\lambda) < 0$ (continuous).

Principal Component Analysis

Eigenvalues of covariance matrix reveal variance along principal directions.

Quantum Mechanics

Observable quantities are eigenvalues of Hermitian operators.

Google PageRank

Page importance is the eigenvector for eigenvalue 1 of the web link matrix.

Vibration Analysis

Natural frequencies of vibrating systems are related to eigenvalues of mass-stiffness matrices.

8. Algebraic vs Geometric Multiplicity

Definition 6.7: Algebraic Multiplicity

The algebraic multiplicity of eigenvalue $\lambda$ is its multiplicity as a root of the characteristic polynomial $\det(A - \lambda I)$ .

Definition 6.8: Geometric Multiplicity

The geometric multiplicity of eigenvalue $\lambda$ is $\dim(E_\lambda) = \dim(\ker(A - \lambda I))$ , the number of linearly independent eigenvectors.

Theorem 6.10: Multiplicity Inequality

For any eigenvalue $\lambda$ :

1 \leq \text{geometric multiplicity} \leq \text{algebraic multiplicity}

Example 6.12: Multiplicity Comparison

For $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ :

Characteristic polynomial: $(\lambda - 2)^2 = 0$

Eigenvalue $\lambda = 2$ with algebraic multiplicity = 2

Eigenspace: $\ker(A - 2I) = \ker\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \text{span}\{(1, 0)^T\}$

Geometric multiplicity = 1 < 2 = algebraic multiplicity

Remark 6.7: Defective Eigenvalues

When geometric multiplicity < algebraic multiplicity, the eigenvalue is called defective. Such matrices cannot be diagonalized.

Example 6.13: Full Geometric Multiplicity

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

Same characteristic polynomial: $(\lambda - 2)^2 = 0$

But $\ker(A - 2I) = \ker(0) = \mathbb{R}^2$ , so geometric multiplicity = 2

This matrix IS diagonalizable (it's already diagonal!).

9. Worked Examples

Example 6.14: Complete Eigenanalysis

Find all eigenvalues and eigenvectors of $A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ .

Step 1: Characteristic polynomial

\det(\lambda I - A) = \det\begin{pmatrix} \lambda - 1 & -2 \\ -3 & \lambda - 2 \end{pmatrix} = (\lambda - 1)(\lambda - 2) - 6 = \lambda^2 - 3\lambda - 4 = (\lambda - 4)(\lambda + 1)

Step 2: Eigenvalues: $\lambda_1 = 4, \lambda_2 = -1$

Step 3: Eigenvectors for $\lambda_1 = 4$

(A - 4I)v = 0 \Rightarrow \begin{pmatrix} -3 & 2 \\ 3 & -2 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \Rightarrow v_1 = \frac{2}{3}v_2

Eigenvector: $v_1 = (2, 3)^T$

Step 4: Eigenvectors for $\lambda_2 = -1$

(A + I)v = 0 \Rightarrow \begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \Rightarrow v_1 = -v_2

Eigenvector: $v_2 = (1, -1)^T$

Verification: tr(A) = 3 = 4 + (-1) ✓, det(A) = -4 = 4 × (-1) ✓

Example 6.15: Rank-1 Matrix Eigenvalues

For $A = \alpha\beta^T$ where $\alpha = (1, 0, -1)^T, \beta = (1, 1, 1)^T$ :

A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ -1 & -1 & -1 \end{pmatrix}

Key observation: rank(A) = 1, so dim(ker(A)) = 2, meaning $\lambda = 0$ has algebraic multiplicity at least 2.

Since tr(A) = 0 and we need 3 eigenvalues summing to 0 with two being 0, the third is also 0.

Actually, $\beta^T \alpha = 1 + 0 - 1 = 0$ , so $A^2 = \alpha(\beta^T\alpha)\beta^T = 0$ .

This means A is nilpotent, so all eigenvalues are 0.

Example 6.16: Block Diagonal Matrix

For $A = \begin{pmatrix} B & 0 \\ 0 & C \end{pmatrix}$ where $B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}, C = \begin{pmatrix} 5 \end{pmatrix}$ :

Eigenvalues of A = eigenvalues of B ∪ eigenvalues of C = {1, 3, 5}

$\det(\lambda I - A) = \det(\lambda I - B) \cdot \det(\lambda I - C) = (\lambda - 1)(\lambda - 3)(\lambda - 5)$

Example 6.17: Finding Matrix from Eigenvalues

Find $|A^* + 3I|$ if $A$ is 3×3 with $A^2 - A - 2I = 0$ and $|A| = 2$ .

Step 1: From $A^2 - A - 2I = 0$ , eigenvalues satisfy $\lambda^2 - \lambda - 2 = 0$

So $\lambda \in \{-1, 2\}$

Step 2: Since $|A| = 2$ and A is 3×3, eigenvalue product = 2

Possible: $(-1) \times (-1) \times 2 = 2$ ✓

So eigenvalues are -1, -1, 2

Step 3: Eigenvalues of $A^*$ : $2/(-1), 2/(-1), 2/2 = -2, -2, 1$

Step 4: Eigenvalues of $A^* + 3I$ : $1, 1, 4$

Answer: $|A^* + 3I| = 1 \times 1 \times 4 = 4$

10. Eigenvalue Estimation

Theorem 6.11: Gershgorin Circle Theorem

Every eigenvalue of $A$ lies in at least one Gershgorin disk:

D_i = \left\{ z \in \mathbb{C} : |z - a_{ii}| \leq \sum_{j \neq i} |a_{ij}| \right\}

Each disk is centered at a diagonal entry with radius equal to the sum of absolute values of off-diagonal entries in that row.

Example 6.18: Using Gershgorin

For $A = \begin{pmatrix} 4 & 1 & 0 \\ 0.5 & 5 & 0.5 \\ 0 & 1 & 6 \end{pmatrix}$ :

Disk 1: center 4, radius 1 → $|z - 4| \leq 1$ , i.e., $[3, 5]$
Disk 2: center 5, radius 1 → $|z - 5| \leq 1$ , i.e., $[4, 6]$
Disk 3: center 6, radius 1 → $|z - 6| \leq 1$ , i.e., $[5, 7]$

All eigenvalues lie in $[3, 7]$ .

Remark 6.8: Diagonally Dominant Matrices

If $|a_{ii}| > \sum_{j \neq i} |a_{ij}|$ for all $i$ (strictly diagonally dominant), then all Gershgorin disks exclude 0, so the matrix is invertible.

Additional Practice

Problem 1

If A is a 4×4 matrix with det(A) = -12 and eigenvalues 1, 2, λ, μ, find λμ.

Answer: λμ = -12/(1×2) = -6

Problem 2

Prove: If A² = 0, then all eigenvalues of A are 0.

Hint: If Av = λv, then A²v = λ²v = 0, so λ² = 0.

Problem 3

Find the eigenvalues of $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$ (cyclic permutation).

Hint: Note A³ = I. Eigenvalues satisfy λ³ = 1.

Problem 4

If A has eigenvalue 3 with eigenvector v, what are the eigenvalues and eigenvectors of 2A - 5I?

Answer: Eigenvalue 2(3) - 5 = 1, same eigenvector v.

Problem 5 (Challenge)

Prove: A and Aᵀ have the same eigenvalues but generally different eigenvectors.

Chapter Summary

This module introduced eigenvalues and eigenvectors—the characteristic values and invariant directions of linear transformations. The eigenvalue equation $Av = \lambda v$ leads to the characteristic polynomial $\det(A - \lambda I) = 0$ for finding eigenvalues.

11

Theorems

18

Examples

12

Quiz Questions

10

FAQs

Key Takeaways

Definition

$Av = \lambda v$ , $v \neq 0$

Eigenspace

$E_\lambda = \ker(A - \lambda I)$

Finding Eigenvalues

Solve $\det(A - \lambda I) = 0$

Key Relations

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

Quick Reference

Eigenvalue of...

• $kA$ : $k\lambda$
• $A^m$ : $\lambda^m$
• $A^{-1}$ : $\lambda^{-1}$
• $A^T$ : $\lambda$ (same)
• $\text{adj}(A)$ : $\det(A)/\lambda$

Special Matrices

• Diagonal: eigenvalues = diagonal entries
• Triangular: eigenvalues = diagonal entries
• Idempotent: $\lambda \in \{0, 1\}$
• Involution: $\lambda \in \{-1, 1\}$
• Nilpotent: $\lambda = 0$ only

11. Real vs Complex Eigenvalues

Theorem 6.12: Fundamental Theorem of Algebra

Over $\mathbb{C}$ , every $n$ × $n$ matrix has exactly $n$ eigenvalues (counting multiplicity).

Remark 6.9: Real Matrices

A real matrix may have:

All real eigenvalues (e.g., symmetric matrices)
Complex eigenvalues in conjugate pairs (when the characteristic polynomial has complex roots)
No real eigenvalues at all (e.g., rotation by 90°)

Theorem 6.13: Complex Conjugate Pairs

For a real matrix, if $\lambda = a + bi$ is an eigenvalue with eigenvector $v$ , then $\bar{\lambda} = a - bi$ is also an eigenvalue with eigenvector $\bar{v}$ .

Example 6.19: Rotation Matrix Eigenvalues

For rotation by angle $\theta$ : $R = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$

Characteristic polynomial: $\lambda^2 - 2\cos\theta\cdot\lambda + 1 = 0$

Solutions: $\lambda = \cos\theta \pm i\sin\theta = e^{\pm i\theta}$

$\theta = 0$ : eigenvalues 1, 1 (identity)
$\theta = \pi$ : eigenvalues -1, -1 (180° rotation)
$\theta = \pi/2$ : eigenvalues $\pm i$ (90° rotation, no real eigenvalues)

Theorem 6.14: Real Symmetric Matrices

A real symmetric matrix $A = A^T$ has:

All real eigenvalues
Orthogonal eigenvectors for distinct eigenvalues
An orthonormal basis of eigenvectors (spectral theorem)

Remark 6.10: Why This Matters

Real symmetric matrices are extremely important in applications:

Covariance matrices in statistics
Hessian matrices in optimization
Hamiltonian matrices in quantum mechanics
Stiffness matrices in structural engineering

12. Computational Aspects

Remark 6.11: Computing Eigenvalues

For matrices larger than 4×4, the characteristic polynomial approach is impractical because:

No closed-form solution for polynomials of degree ≥ 5 (Abel-Ruffini theorem)
Numerical instability in computing determinants
Root-finding for high-degree polynomials is difficult

Numerical Methods

Power Method: Find the dominant eigenvalue by repeated multiplication.
QR Algorithm: Iteratively decompose A = QR, then form A' = RQ. Converges to triangular form.
Lanczos Algorithm: For large sparse symmetric matrices.
Arnoldi Iteration: For large non-symmetric matrices.

Example 6.20: Power Method

To find the largest eigenvalue of $A$ :

Start with random $v_0$
Compute $v_{k+1} = Av_k / ||Av_k||$
Eigenvalue estimate: $\lambda \approx v_k^T A v_k / v_k^T v_k$

Converges if $|\lambda_1| > |\lambda_2|$ (dominant eigenvalue is unique).

Remark 6.12: Complexity

Characteristic polynomial via determinant: O(n!) naive, O(n³) with LU
QR iteration: O(n³) per iteration, typically O(1) iterations
For sparse matrices: can be much faster with specialized methods

13. Connections

Determinant

$\det(A) = \prod \lambda_i$

A is invertible iff no eigenvalue is 0.

Trace

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

Sum of diagonal = sum of eigenvalues.

Rank

rank(A) = n - (# of zero eigenvalues)

For diagonalizable matrices.

Matrix Norm

$||A||_2 = \sqrt{\lambda_{\max}(A^TA)}$

Spectral norm = largest singular value.

Condition Number

$\kappa(A) = |\lambda_{\max}|/|\lambda_{\min}|$

Measures numerical stability.

Matrix Exponential

$e^A = P \cdot \text{diag}(e^{\lambda_i}) \cdot P^{-1}$

For diagonalizable A = PDP⁻¹.

Study Tips

Always verify: Check that trace = sum of eigenvalues, det = product of eigenvalues.
Start simple: Practice with 2×2 matrices until the process is automatic.
Use special cases: Triangular and diagonal matrices have eigenvalues on the diagonal.
Remember constraints: For A² = A, eigenvalues ∈ {0,1}. For A² = I, eigenvalues ∈ {±1}.
Think geometrically: Eigenvectors are directions preserved by the transformation.
Connect concepts: λ = 0 ⟺ A is singular ⟺ ker(A) ≠ {0}.

Common Eigenvalue Patterns

Matrix Type	Eigenvalues	Notes
Identity I	All 1	Every vector is an eigenvector
Zero matrix	All 0	Every vector is an eigenvector
Diagonal	Diagonal entries	Standard basis are eigenvectors
Triangular	Diagonal entries	Eigenvectors may differ from standard basis
Projection (P² = P)	0 or 1 only	E₁ = image, E₀ = kernel
Reflection (P² = I)	±1 only	E₁ = mirror, E₋₁ = perpendicular
Rotation (2D)	$e^{\pm i\theta}$	Real only for θ = 0, π
Nilpotent (Aᵏ = 0)	All 0	Not diagonalizable if A ≠ 0

Historical Notes

Etymology: "Eigen" is German for "own" or "characteristic." Eigenvalues are the "characteristic values" of a transformation.

Euler (1750s): Studied principal axes of rotation of rigid bodies, which are eigenvectors of the inertia tensor.

Cauchy (1826): First systematic study of eigenvalues in the context of quadratic forms and symmetric matrices. Proved that real symmetric matrices have real eigenvalues.

Sylvester (1852): Coined the term "matrix" and studied eigenvalues in the context of invariant theory.

Cayley (1858): Introduced matrices as algebraic objects and proved the Cayley-Hamilton theorem.

Hilbert (1904): Extended eigenvalue theory to infinite-dimensional spaces, founding spectral theory.

Modern Era: Eigenvalue computation became crucial with computers. The QR algorithm (1960s) remains the standard method.

14. Matrix Equation Problems

Theorem 6.15: Eigenvalues from Matrix Equations

If $f(A) = 0$ for some polynomial $f$ , then every eigenvalue $\lambda$ of $A$ satisfies $f(\lambda) = 0$ .

Proof:

If $Av = \lambda v$ with $v \neq 0$ , then $f(A)v = f(\lambda)v$ .

Since $f(A) = 0$ , we have $f(\lambda)v = 0$ .

Since $v \neq 0$ , we must have $f(\lambda) = 0$ .

∎

Example 6.21: Using Matrix Equations

If $A^3 - 2A^2 - A + 2I = 0$ , what are the possible eigenvalues?

Solution: Eigenvalues satisfy $\lambda^3 - 2\lambda^2 - \lambda + 2 = 0$

Factor: $(\lambda - 1)(\lambda + 1)(\lambda - 2) = 0$

Possible eigenvalues: $\lambda \in \{-1, 1, 2\}$

Example 6.22: Determining Eigenvalues with Constraints

Given 3×3 matrix $A$ with $A^2 = A$ , $\text{tr}(A) = 2$ , and $|A| = 0$ .

Step 1: From $A^2 = A$ , eigenvalues ∈ {0, 1}

Step 2: From $|A| = 0$ , at least one eigenvalue is 0

Step 3: From $\text{tr}(A) = 2$ , eigenvalues sum to 2

Conclusion: Eigenvalues must be 0, 1, 1

Remark 6.13: Characteristic Polynomial as Minimal Annihilating Polynomial

The characteristic polynomial $p(\lambda) = \det(\lambda I - A)$ satisfies $p(A) = 0$ (Cayley-Hamilton theorem). But there may be lower-degree polynomials that also annihilate $A$ .

15. Eigenvalue Localization

Theorem 6.16: Eigenvalue Bounds

For any matrix $A$ and induced norm $||\cdot||$ :

|\lambda| \leq ||A||

Every eigenvalue has absolute value at most the matrix norm.

Proof:

If $Av = \lambda v$ with $v \neq 0$ :

|\lambda| \cdot ||v|| = ||\lambda v|| = ||Av|| \leq ||A|| \cdot ||v||

Dividing by $||v|| > 0$ : $|\lambda| \leq ||A||$

∎

Corollary 6.3: Row Sum Bound

Using the infinity norm: $|\lambda| \leq \max_i \sum_j |a_{ij}|$

Corollary 6.4: Column Sum Bound

Using the 1-norm: $|\lambda| \leq \max_j \sum_i |a_{ij}|$

Example 6.23: Bounding Eigenvalues

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

Row sums: 3, 5, 3. So $|\lambda| \leq 5$

Column sums: 3, 5, 3. Same bound.

Gershgorin disks: [1,3], [1,5], [1,3]. Union: [1,5].

16. Spectral Properties

Definition 6.9: Spectral Radius

The spectral radius of $A$ is:

\rho(A) = \max\{|\lambda| : \lambda \text{ is an eigenvalue of } A\}

Theorem 6.17: Spectral Radius Properties

$\rho(A) \leq ||A||$ for any matrix norm
$\rho(A^k) = \rho(A)^k$
$\rho(cA) = |c| \cdot \rho(A)$
For normal matrices: $\rho(A) = ||A||_2$

Definition 6.10: Positive Definite

A symmetric matrix $A$ is positive definite if all eigenvalues are positive, equivalently if $x^T A x > 0$ for all $x \neq 0$ .

Remark 6.14: Eigenvalue Sign Classification

For symmetric matrices:

Positive definite: all $\lambda > 0$
Positive semidefinite: all $\lambda \geq 0$
Negative definite: all $\lambda < 0$
Indefinite: some positive, some negative

Challenge Problems

Challenge 1

Prove: If A is nilpotent (Aᵏ = 0 for some k), then tr(A) = det(A) = 0.

Challenge 2

If A and B are n×n matrices with AB = BA, and v is an eigenvector of A, prove that Bv is also an eigenvector of A (for the same eigenvalue) or Bv = 0.

Challenge 3

Prove: If A is a real matrix with all real eigenvalues, it doesn't necessarily mean A is symmetric. Give a counterexample.

Challenge 4

For 3×3 matrix A with eigenvalues 1, 2, 3, find det(A³ - 6A² + 11A - 6I).

Hint: Factor the polynomial using the eigenvalues.

Learning Path

Determinants→Eigenvalues→Char. Polynomial→Diagonalization→Jordan Form

You are here! Eigenvalues are the foundation for understanding matrix structure and behavior.

What's Next?

With eigenvalue fundamentals mastered, you're ready for:

Characteristic Polynomial: Deep dive into the polynomial whose roots are eigenvalues
Diagonalization: When and how a matrix can be written as $PDP^{-1}$
Jordan Normal Form: The best we can do when diagonalization fails
Cayley-Hamilton Theorem: Every matrix satisfies its own characteristic equation

Verification Checklist

☐ Did you check that eigenvectors are non-zero?
☐ Did you verify $Av = \lambda v$ by direct multiplication?
☐ Did you confirm tr(A) = sum of eigenvalues?
☐ Did you confirm det(A) = product of eigenvalues?
☐ For each eigenspace, did you verify it's actually a subspace?
☐ Did you check geometric multiplicity ≤ algebraic multiplicity?

2×2 Matrix Quick Formulas

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

Characteristic Polynomial:

$\lambda^2 - (a+d)\lambda + (ad-bc) = 0$

Eigenvalues:

$\lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}$

Sum:

$\lambda_1 + \lambda_2 = a + d = \text{tr}(A)$

Product:

$\lambda_1 \cdot \lambda_2 = ad - bc = \det(A)$

Module Completion

Congratulations! You've completed the comprehensive introduction to eigenvalues and eigenvectors. This foundational knowledge prepares you for the rich theory of matrix analysis ahead.

17

Theorems

23

Examples

12

Quiz Questions

10

FAQs

16

Sections

Eigenvalue Definition Practice

12

Questions

0

Correct

0%

Accuracy

1

If

Av = \lambda v

with

v \neq 0

, then

\lambda

is called:

Easy

Not attempted

2

The eigenspace

E_\lambda

is:

Medium

Not attempted

3

A 2×2 matrix can have at most:

Easy

Not attempted

4

If

\lambda = 0

is an eigenvalue:

Medium

Not attempted

5

Eigenvectors for different eigenvalues are:

Medium

Not attempted

6

The identity matrix

I

has eigenvalue(s):

Easy

Not attempted

7

If

A^2 = A

(idempotent), eigenvalues can only be:

Medium

Not attempted

8

If

A

is invertible with eigenvalue

\lambda

, then

A^{-1}

has eigenvalue:

Medium

Not attempted

9

The sum of all eigenvalues (with multiplicity) equals:

Medium

Not attempted

10

If

AB

has eigenvalue

\lambda \neq 0

, then

BA

has:

Hard

Not attempted

11

A nilpotent matrix (where

A^k = 0

) has eigenvalues:

Hard

Not attempted

12

Similar matrices

A

and

B = P^{-1}AP

have:

Medium

Not attempted

Frequently Asked Questions

What's the geometric meaning of eigenvectors?

An eigenvector's direction is preserved by the linear map. It may be stretched (|λ| > 1), shrunk (|λ| < 1), or flipped (λ < 0), but its direction stays the same. Eigenvectors define 'invariant directions' of the transformation.

Can an eigenvalue be complex?

Yes! Over ℂ, every n×n matrix has exactly n eigenvalues (counting multiplicity) by the Fundamental Theorem of Algebra. Over ℝ, some matrices have no real eigenvalues, like 90° rotation matrices.

Is 0 a valid eigenvector?

No! By definition, eigenvectors must be non-zero. The zero vector satisfies Av = λv for all λ, so it would make the definition meaningless. However, 0 is a valid eigenvalue!

What if an eigenspace has dimension > 1?

Then there are multiple linearly independent eigenvectors for that eigenvalue. The dimension of the eigenspace is called the geometric multiplicity. It's always ≤ the algebraic multiplicity (root multiplicity in characteristic polynomial).

How do eigenvalues relate to det and trace?

Product of all eigenvalues (with multiplicity) = det(A). Sum of all eigenvalues (with multiplicity) = trace(A) = sum of diagonal elements. These follow from the characteristic polynomial.

Why study eigenvalues?

Eigenvalues reveal fundamental properties: invertibility (no zero eigenvalue), stability (all |λ| < 1), growth rates, and enable diagonalization for easier computation of matrix powers and exponentials.

What's the difference between algebraic and geometric multiplicity?

Algebraic multiplicity = number of times λ appears as root of det(A - λI) = 0. Geometric multiplicity = dim(ker(A - λI)) = number of linearly independent eigenvectors. Always: geometric ≤ algebraic.

Do AB and BA have the same eigenvalues?

They have the same nonzero eigenvalues! If λ ≠ 0 is an eigenvalue of AB, it's also an eigenvalue of BA. The zero eigenvalue may differ in multiplicity.

How do eigenvalues of A relate to those of A^n?

If λ is an eigenvalue of A with eigenvector v, then λⁿ is an eigenvalue of Aⁿ with the same eigenvector v. This follows from Aⁿv = Aⁿ⁻¹(Av) = Aⁿ⁻¹(λv) = λAⁿ⁻¹v = ... = λⁿv.

What are the eigenvalues of a triangular matrix?

The eigenvalues of a triangular (upper or lower) matrix are exactly its diagonal entries. This is because det(A - λI) factors as a product of (aᵢᵢ - λ) terms.

Eigenvalues and Eigenvectors

1. Eigenvalue and Eigenvector

2. Examples

3. Finding Eigenvalues

4. Key Properties

5. Special Matrix Properties

6. Common Mistakes

Zero is NOT an eigenvector

Zero CAN be an eigenvalue

Confusing algebraic and geometric multiplicity

Similar matrices have same eigenvectors

Every matrix has real eigenvalues

7. Applications Preview

Differential Equations

Stability Analysis

Principal Component Analysis

Quantum Mechanics

Google PageRank

Vibration Analysis

8. Algebraic vs Geometric Multiplicity

9. Worked Examples

10. Eigenvalue Estimation

Additional Practice

Chapter Summary

Key Takeaways

Definition

Eigenspace

Finding Eigenvalues

Key Relations

Quick Reference

Eigenvalue of...

Special Matrices

11. Real vs Complex Eigenvalues

12. Computational Aspects

Numerical Methods

13. Connections

Determinant

Trace

Rank

Matrix Norm

Condition Number

Matrix Exponential

Study Tips

Common Eigenvalue Patterns

Historical Notes

14. Matrix Equation Problems

15. Eigenvalue Localization

16. Spectral Properties

Challenge Problems

Learning Path

What's Next?

Verification Checklist

2×2 Matrix Quick Formulas

Module Completion

Related Topics

Frequently Asked Questions

What's the geometric meaning of eigenvectors?

Can an eigenvalue be complex?

Is 0 a valid eigenvector?

What if an eigenspace has dimension > 1?

How do eigenvalues relate to det and trace?

Why study eigenvalues?

What's the difference between algebraic and geometric multiplicity?

Do AB and BA have the same eigenvalues?

How do eigenvalues of A relate to those of A^n?

What are the eigenvalues of a triangular matrix?