Eigenvalues and Eigenvectors: Complete Guide with Step-by-Step Solutions

Quick Navigation

• What are Eigenvalues and Eigenvectors?
• The Fundamental Equation
• How to Find Eigenvalues
• How to Find Eigenvectors
• Multiplicity: Algebraic vs Geometric
• Diagonalization
• Real-World Applications
• Step-by-Step Examples
• Special Matrices
• Frequently Asked Questions
• Practice Problems
• Related Topics

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What are Eigenvalues and Eigenvectors?

Definition

For a square matrix $A$, an eigenvector $\mathbf{v}$ is a non-zero vector that, when multiplied by $A$, results in a scaled version of itself. The scaling factor $\lambda$ is called the eigenvalue.

$$ A \mathbf{v} = \lambda \mathbf{v} $$

Geometric Interpretation

Most vectors change direction when multiplied by a matrix. Eigenvectors are special because they do not change direction (unless $\lambda < 0$, in which case they reverse). They only stretch, shrink, or flip.

Eigenvalue ($\lambda$)	Effect on Vector $\mathbf{v}$
$\lambda > 1$	Stretches (magnitude increases)
$0 < \lambda < 1$	Shrinks (magnitude decreases)
$\lambda = 1$	Unchanged (fixed direction)
$\lambda = 0$	Collapses to zero (in null space)
$\lambda < 0$	Reverses direction

Historical Context

The word "eigen" comes from German, meaning "own," "proper," or "characteristic." Thus, eigenvalues are the "characteristic values" of a matrix.

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The Fundamental Equation

Deriving the Characteristic Equation

Starting from $A \mathbf{v} = \lambda \mathbf{v}$, we can rearrange terms:

$$ A \mathbf{v} - \lambda \mathbf{v} = \mathbf{0} \implies (A - \lambda I) \mathbf{v} = \mathbf{0} $$

For a non-zero solution $\mathbf{v}$ to exist, the matrix $(A - \lambda I)$ must be singular (non-invertible). This occurs if and only if its determinant is zero:

$$ \boxed{\det(A - \lambda I) = 0} $$

This is called the characteristic equation.

The Characteristic Polynomial

For an $n \times n$ matrix, $\det(A - \lambda I)$ is a polynomial of degree $n$.

For 2×2 matrices: $$ p(\lambda) = \lambda^2 - \text{tr}(A)\lambda + \det(A) $$

For 3×3 matrices: $$ p(\lambda) = \lambda^3 - \text{tr}(A)\lambda^2 + C_2 \lambda - \det(A) $$ *(Where $C_2$ is the sum of principal minors)*

Key Relationships

Relationship	Formula
Sum of eigenvalues	$\sum \lambda_i = \text{tr}(A)$
Product of eigenvalues	$\prod \lambda_i = \det(A)$
Symmetric Matrices	All eigenvalues are real
Skew-Symmetric	Eigenvalues are purely imaginary or zero

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How to Find Eigenvalues

Step-by-Step Method

1. Form the matrix $A - \lambda I$ by subtracting $\lambda$ from the diagonal entries.
2. Compute the determinant $\det(A - \lambda I)$ to get the characteristic polynomial.
3. Solve $\det(A - \lambda I) = 0$ for $\lambda$.
4. List all roots, including complex numbers and repetitions.

Example 1: 2×2 Matrix

Matrix: $$ A = \left[\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}\right] $$

Step 1: Form $A - \lambda I$ $$ A - \lambda I = \left[\begin{array}{cc} 4-\lambda & 1 \\ 2 & 3-\lambda \end{array}\right] $$

Step 2: Compute determinant $$ \det = (4-\lambda)(3-\lambda) - (1)(2) = \lambda^2 - 7\lambda + 10 $$

Step 3: Solve characteristic equation $$ \lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2) = 0 $$

Step 4: Eigenvalues $$ \lambda_1 = 5,\quad \lambda_2 = 2 $$

Example 2: Complex Eigenvalues

Matrix: $$ A = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] $$

Characteristic equation: $$ \det\left[\begin{array}{cc} -\lambda & -1 \\ 1 & -\lambda \end{array}\right] = \lambda^2 + 1 = 0 $$

Eigenvalues: $$ \lambda = \pm i \quad \text{(purely imaginary)} $$

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How to Find Eigenvectors

Step-by-Step Method

For each eigenvalue $\lambda$:

1. Form the matrix $A - \lambda I$.
2. Solve the homogeneous system $(A - \lambda I)\mathbf{v} = \mathbf{0}$.
3. Find the basis vectors for the null space (these are your eigenvectors).

Example 1: Distinct Real Eigenvalues

Matrix: $A = \left[\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}\right]$, with $\lambda_1 = 5$.

Step 1: Form $A - 5I$ $$ A - 5I = \left[\begin{array}{cc} -1 & 1 \\ 2 & -2 \end{array}\right] $$

Step 2: Solve $(A - 5I)\mathbf{v} = \mathbf{0}$ $$ \left[\begin{array}{cc} -1 & 1 \\ 2 & -2 \end{array}\right] \left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] $$ This reduces to $-v_1 + v_2 = 0 \implies v_1 = v_2$.

Step 3: Eigenvector Choosing $v_1 = 1$, we get: $$ \mathbf{v}_1 = \left[\begin{array}{c} 1 \\ 1 \end{array}\right] $$

Similarly for $\lambda_2 = 2$, we find $\mathbf{v}_2 = \left[\begin{array}{c} 1 \\ -2 \end{array}\right]$.

Example 2: Defective Matrix (Repeated Eigenvalue)

Matrix: $A = \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right]$, with $\lambda = 2$ (multiplicity 2).

Step 1: Form $A - 2I$ $$ A - 2I = \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] $$

Step 2: Solve $(A - 2I)\mathbf{v} = \mathbf{0}$ $$ \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right] \left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \implies v_2 = 0 $$ $v_1$ is free. We get only one independent eigenvector: $$ \mathbf{v} = \left[\begin{array}{c} 1 \\ 0 \end{array}\right] $$ Since we have only 1 eigenvector for a multiplicity of 2, this matrix is defective.

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Multiplicity: Algebraic vs Geometric

Definitions

• Algebraic Multiplicity: How many times $\lambda$ appears as a root of the characteristic polynomial.
• Geometric Multiplicity: The number of linearly independent eigenvectors for $\lambda$ (dimension of the null space of $A-\lambda I$).

The Critical Rule

$$ 1 \le \text{Geometric Multiplicity} \le \text{Algebraic Multiplicity} $$

• If Geometric = Algebraic for all $\lambda$, the matrix is Diagonalizable.
• If Geometric < Algebraic for any $\lambda$, the matrix is Defective.

Example

Matrix: $A = \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right]$

• Algebraic Multiplicity of $\lambda=2$: 2 (root of $(\lambda-2)^2$)
• Geometric Multiplicity: 1 (only one eigenvector $\left[\begin{smallmatrix} 1 \ 0 \end{smallmatrix}\right]$)
• Conclusion: Defective.

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Diagonalization

Definition

A matrix $A$ is diagonalizable if it can be written as:

$$ A = P D P^{-1} $$

Where:

• $D$ is a diagonal matrix containing the eigenvalues.
• $P$ is a matrix whose columns are the corresponding eigenvectors.

Condition for Diagonalization

An $n \times n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors.

Example: Diagonalizing a 2×2 Matrix

Matrix: $A = \left[\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}\right]$

Eigenvalues: $\lambda_1 = 5, \lambda_2 = 2$ Eigenvectors: $\mathbf{v}_1 = \left[\begin{array}{c} 1 \\ 1 \end{array}\right], \mathbf{v}_2 = \left[\begin{array}{c} 1 \\ -2 \end{array}\right]$

Construct P and D: $$ P = \left[\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right], \quad D = \left[\begin{array}{cc} 5 & 0 \\ 0 & 2 \end{array}\right] $$

Why is this useful? Computing powers of $A$ becomes easy: $A^n = P D^n P^{-1}$.

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Real-World Applications

📊 Principal Component Analysis (PCA)

PCA uses eigenvectors of the covariance matrix to find directions of maximum variance in data.

• Eigenvectors: Principal components (directions).
• Eigenvalues: Amount of variance explained by each component.

🔬 Quantum Mechanics

The Schrödinger equation $H\psi = E\psi$ is an eigenvalue problem.

• $H$: Hamiltonian operator.
• $\psi$: Wavefunction (eigenvector).
• $E$: Energy level (eigenvalue).

🌐 Google PageRank

PageRank finds the dominant eigenvector of the web's link matrix.

• $\lambda = 1$: The stationary distribution of importance.
• Eigenvector: The ranking score of each page.

🏗️ Vibration Analysis

Natural frequencies of structures are determined by eigenvalues.

• Eigenvalues: Squared natural frequencies ($\omega^2$).
• Eigenvectors: Mode shapes (how the structure deforms).

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Step-by-Step Examples

Example 1: Symmetric Matrix (Real Orthogonal Eigenvectors)

Matrix: $$ A = \left[\begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array}\right] $$

Step 1: Characteristic Equation $$ \det\left[\begin{array}{cc} 3-\lambda & 1 \\ 1 & 3-\lambda \end{array}\right] = (3-\lambda)^2 - 1 = \lambda^2 - 6\lambda + 8 = 0 $$ Roots: $\lambda_1 = 4, \lambda_2 = 2$.

Step 2: Eigenvectors

• For $\lambda_1 = 4$: $\mathbf{v}_1 = \left[\begin{array}{c} 1 \\ 1 \end{array}\right]$
• For $\lambda_2 = 2$: $\mathbf{v}_2 = \left[\begin{array}{c} 1 \\ -1 \end{array}\right]$

Note: $\mathbf{v}_1 \cdot \mathbf{v}_2 = 0$. Symmetric matrices always have orthogonal eigenvectors.

Example 2: Complex Eigenvalues (Rotation)

Matrix: $$ A = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] $$

Step 1: Eigenvalues $\lambda^2 + 1 = 0 \implies \lambda = \pm i$.

Step 2: Eigenvector for $\lambda = i$ $$ \left[\begin{array}{cc} -i & -1 \\ 1 & -i \end{array}\right] \left[\begin{array}{c} v_1 \\ v_2 \end{array}\right] = \mathbf{0} \implies v_2 = -i v_1 $$ $$ \mathbf{v} = \left[\begin{array}{c} 1 \\ -i \end{array}\right] $$

Example 3: 3×3 Triangular Matrix

Matrix: $$ A = \left[\begin{array}{ccc} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 4 \end{array}\right] $$

Step 1: Eigenvalues For triangular matrices, eigenvalues are simply the diagonal entries: $$ \lambda_1 = 2, \quad \lambda_2 = 3, \quad \lambda_3 = 4 $$

Step 2: Eigenvectors Solve $(A-\lambda I)\mathbf{v}=0$ for each. Since eigenvalues are distinct, the matrix is diagonalizable.

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Special Matrices

Matrix Type	Eigenvalue Properties	Eigenvector Properties
Symmetric ($A=A^T$)	All real	Orthogonal
Skew-Symmetric ($A=-A^T$)	Purely imaginary or 0	Orthogonal (complex)
Orthogonal ($A^T=A^{-1}$)	$|\lambda|=1$	Orthonormal
Diagonal	Diagonal entries	Standard basis vectors
Nilpotent ($A^k=0$)	All $\lambda=0$	Incomplete (defective)
Projection ($P^2=P$)	$\lambda \in {0, 1}$	Span subspaces

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Frequently Asked Questions

Q: Can a matrix have zero eigenvalues?

A: Yes. If $\lambda=0$ is an eigenvalue, the matrix is singular (non-invertible). The eigenvectors for $\lambda=0$ form the null space.

Q: Are eigenvectors unique?

A: No. If $\mathbf{v}$ is an eigenvector, then $c\mathbf{v}$ is also an eigenvector for any non-zero scalar $c$. We often normalize them to length 1.

Q: What if eigenvalues are repeated?

A: The matrix might still be diagonalizable if there are enough independent eigenvectors (Geometric = Algebraic). If not, it is defective.

Q: Can non-square matrices have eigenvalues?

A: No. Eigenvalues are only defined for square matrices. For rectangular matrices, we use Singular Value Decomposition (SVD).

Q: How do eigenvalues relate to stability?

A: In systems $\dot{\mathbf{x}} = A\mathbf{x}$:

• All $\text{Re}(\lambda) < 0$: Stable.
• Any $\text{Re}(\lambda) > 0$: Unstable.

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Practice Problems

Beginner

1. Find eigenvalues of $A = \left[\begin{array}{cc} 5 & 0 \\ 0 & 3 \end{array}\right]$.
2. Find the trace and determinant of $A = \left[\begin{array}{cc} 2 & 3 \\ 1 & 4 \end{array}\right]$.

Intermediate

3. Find eigenvalues of $A = \left[\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right]$.
4. Is $A = \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right]$ diagonalizable? Why?

Advanced

5. Find eigenvalues and eigenvectors of $A = \left[\begin{array}{ccc} 4 & 1 & 0 \\ 1 & 4 & 1 \\ 0 & 1 & 4 \end{array}\right]$.
6. Prove that eigenvalues of a symmetric matrix are real.

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Summary

Key Takeaways

• Equation: $A\mathbf{v} = \lambda\mathbf{v}$
• Characteristic Eq: $\det(A - \lambda I) = 0$
• Trace: Sum of eigenvalues.
• Determinant: Product of eigenvalues.
• Diagonalization: Possible if $n$ independent eigenvectors exist.

When to Use This Solver

• ✅ Finding eigenvalues/eigenvectors for $2\times2$ to $6\times6$ matrices.
• ✅ Checking diagonalizability.
• ✅ Understanding matrix properties (symmetry, definiteness).
• ❌ Non-square matrices (Use SVD).
• ❌ Very large matrices ($N > 10$) (Use numerical libraries).

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Related Topics

• Characteristic Polynomial Calculator
• Matrix Diagonalization Solver
• SVD Calculator
• Positive Definite Checker
• Linear ODE System Solver

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Try It Yourself!

Use the calculator above to explore eigenvalues:

Test these examples:

Symmetric (Real): $$ A = \left[\begin{array}{cc} 4 & 1 \\ 1 & 4 \end{array}\right] $$

Rotation (Complex): $$ A = \left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] $$

Defective: $$ A = \left[\begin{array}{cc} 2 & 1 \\ 0 & 2 \end{array}\right] $$

3×3 symmetric (tridiagonal): $$A = \begin{bmatrix} 2 & -1 & 0 \ -1 & 2 & -1 \ 0 & -1 & 2 \end{bmatrix}$$

3×3 upper triangular: $$A = \begin{bmatrix} 3 & 1 & 2 \ 0 & 4 & 1 \ 0 & 0 & 5 \end{bmatrix}$$

Pro tip: The sum of eigenvalues equals trace, product equals determinant—quick check for correctness!