27-02-2025 A | Engineering Mathematics | GATE & ESE

Linear Algebra and Differential Equations: Detailed Summary

Key Concepts:

Cube roots of unity (Omega)
Fourth/Fifth roots of unity (Alpha)
Trace of a matrix
Eigenvalues and eigenvectors
Symmetric positive definite matrix
Principal minors
Stationary points of a function
Rotational transformation matrix
One-dimensional heat equation
Boundary conditions
Initial conditions

1. Cube Roots of Unity

Definition: Omega (Ω) = 1^(1/3)
Key Property: Ω³ = 1
Derivation:
- Ω³ - 1³ = 0
- (Ω - 1)(Ω² + Ω + 1) = 0
- Therefore, 1 + Ω + Ω² = 0
Important Note: Remember both Ω³ = 1 and 1 + Ω + Ω² = 0.
Quadratic Equation Solution: The values of Omega can be found by solving the quadratic equation 1 + Omega + Omega^2 = 0, resulting in Omega = (-1 ± i√3)/2.
Conjugate Relationship: If Ω = (-1 - i√3)/2, then Ω² = (-1 + i√3)/2, which is the conjugate of Omega (Ω̄). Thus, if one root is Ω, the other is Ω̄ or Ω².

2. Higher Order Roots of Unity

Fourth Roots of Unity: If Alpha (α) = 1^(1/4), then α⁴ = 1, and 1 + α + α² + α³ = 0.
Fifth Roots of Unity: If Alpha (α) = 1^(1/5), then α⁵ = 1, and 1 + α + α² + α³ + α⁴ = 0.
Generalization: For n-th roots of unity, if α = 1^(1/n), then αⁿ = 1, and 1 + α + α² + ... + α^(n-1) = 0.

3. Application: Matrix Trace and Eigenvalues

Problem: Given matrix A = [1 0 0; i Ω 0; 1+2i 0 Ω²], find the trace of A^(102).
Key Concept: The trace of a matrix is the sum of its eigenvalues.
Eigenvalue Property: If Lambda (λ) is an eigenvalue of A, then λⁿ is an eigenvalue of Aⁿ.
Solution:
1. Identify Eigenvalues of A: Since A is a lower triangular matrix (LTM), its eigenvalues are the diagonal elements: λ = 1, Ω, Ω².
2. Eigenvalues of A^(102): The eigenvalues of A^(102) are 1^(102), Ω^(102), (Ω²)^(102).
3. Simplify:
  - Ω^(102) = (Ω³)^34 = 1^34 = 1
  - Ω^(204) = (Ω³)^68 = 1^68 = 1
4. Trace of A^(102): Trace(A^(102)) = 1 + 1 + 1 = 3.

4. Application: Fifth Roots of Unity and Matrix Trace

Problem: Given α = e^(2πi/5) and a matrix M (UTM), find the trace of I₅ + M + M².
Euler's Formula: e^(iθ) = cos(θ) + i sin(θ)
Solution:
1. Simplify Alpha: α = e^(i2π/5) implies α⁵ = 1 and 1 + α + α² + α³ + α⁴ = 0.
2. Trace Property: Trace(A + B) = Trace(A) + Trace(B)
3. Trace of I₅: Trace(I₅) = 1 + 1 + 1 + 1 + 1 = 5.
4. Eigenvalues of M: Since M is a UTM, its eigenvalues are the diagonal elements: 1, α, α², α³, α⁴.
5. Trace of M: Trace(M) = 1 + α + α² + α³ + α⁴ = 0.
6. Eigenvalues of M²: The eigenvalues of M² are 1², α², α⁴, α⁶, α⁸.
7. Simplify Eigenvalues of M²:
  - α⁶ = α⁵ * α = α
  - α⁸ = α⁵ * α³ = α³
8. Trace of M²: Trace(M²) = 1 + α² + α⁴ + α + α³ = 0.
9. Trace of I₅ + M + M²: Trace(I₅ + M + M²) = 5 + 0 + 0 = 5.

5. Scalar Valued Function and Symmetric Positive Definite Matrix

Problem: Given f(x) = xᵀAx + bᵀx + c, where A is a symmetric positive definite matrix, find the x at which the minimum value of f(x) occurs.
Symmetric Matrix: A matrix where A = Aᵀ (A transpose).
Positive Definite Matrix: A symmetric matrix where all eigenvalues are strictly positive (λ > 0).
Principal Minors: All principal minors of an n x n matrix must be greater than zero.
Solution:
1. Represent A as a 2x2 Matrix: A = [a b; b c] (symmetric).
2. Conditions for Positive Definiteness: a > 0, c > 0, and ac - b² > 0.
3. Express f(x) in terms of x₁, x₂: f(x) = a x₁² + 2b x₁x₂ + c x₂² + b₁x₁ + b₂x₂ + c.
4. Find Stationary Points:
  - ∂f/∂x₁ = 2a x₁ + 2b x₂ + b₁ = 0
  - ∂f/∂x₂ = 2b x₁ + 2c x₂ + b₂ = 0
5. Solve for x₁ and x₂:
  - a x₁ + b x₂ = -b₁/2
  - b x₁ + c x₂ = -b₂/2
6. Matrix Representation: [a b; b c] [x₁; x₂] = [-b₁/2; -b₂/2]
7. Solve for x: x = -1/2 A⁻¹b, where b = [b₁; b₂].
8. Minimum Value: Since A is positive definite, the function has a minimum at the stationary point.

6. Stationary Point of a Function with a Symmetric Matrix

Problem: Given f(x) = 1/2 xᵀQx - rᵀx, where Q is a symmetric matrix, find the stationary point of f(x).
Solution:
1. Express f(x) in terms of x₁, x₂: f(x) = 1/2 (a x₁² + 2b x₁x₂ + c x₂²) - r₁x₁ - r₂x₂.
2. Find Stationary Points:
  - ∂f/∂x₁ = a x₁ + b x₂ - r₁ = 0
  - ∂f/∂x₂ = b x₁ + c x₂ - r₂ = 0
3. Matrix Representation: [a b; b c] [x₁; x₂] = [r₁; r₂]
4. Solve for x: Qx = r => x = Q⁻¹r.

7. Rotational Transformation

Problem: Find the transformed coordinates of point P(1, 1, 1) after a 45° clockwise rotation of the coordinate system about the positive x-axis.
Key Concept: Rotating the coordinate system clockwise is equivalent to rotating the point counterclockwise.
Rotation Matrix about x-axis (Counterclockwise): A = [1 0 0; 0 cos(θ) -sin(θ) 0; 0 sin(θ) cos(θ)].
Solution:
1. Substitute θ = 45°: A = [1 0 0; 0 1/√2 -1/√2 0; 0 1/√2 1/√2].
2. Apply Transformation: y = Ax, where x = [1; 1; 1].
3. Calculate y: y = [1; 0; √2].
4. Transformed Coordinates: The new coordinates are (1, 0, √2).

8. Trigonometric Function Simplification

Problem: Given f(x) = cos⁴(x), find f(π/4) + f(π/2).
Solution:
1. Rewrite cos⁴(x): cos⁴(x) = (cos²(x))² = ((1 + cos(2x))/2)².
2. Expand: (1 + cos(2x))²/4 = (1 + 2cos(2x) + cos²(2x))/4.
3. Rewrite cos²(2x): cos²(2x) = (1 + cos(4x))/2.
4. Substitute: (1 + 2cos(2x) + (1 + cos(4x))/2)/4 = (3/2 + 2cos(2x) + cos(4x)/2)/4 = 3/8 + 1/2 cos(2x) + 1/8 cos(4x).
5. Identify Coefficients: f(x) = 3/8 + 1/2 cos(2x) + 1/8 cos(4x).
6. Evaluate f(π/4) and f(π/2):
  - f(π/4) = 3/8 + 1/2 cos(π/2) + 1/8 cos(π) = 3/8 + 0 - 1/8 = 2/8 = 1/4
  - f(π/2) = 3/8 + 1/2 cos(π) + 1/8 cos(2π) = 3/8 - 1/2 + 1/8 = 4/8 - 1/2 = 1/2 - 1/2 = 0
7. Calculate f(π/4) + f(π/2): 1/4 + 0 = 1/4 = 0.25
8. The problem asks for F4 + F2, where F4 is the coefficient of cos(4x) and F2 is the coefficient of cos(2x).
9. F4 = 1/8 and F2 = 0.
10. F4 + F2 = 1/8 + 0 = 1/8 = 0.125.

9. One-Dimensional Heat Equation

Equation: ∂U/∂t = c² ∂²U/∂x²
Boundary Conditions: Values of U at the boundaries of the spatial domain.
Initial Condition: Value of U at t = 0.
General Solution (with two boundary conditions and one initial condition): U(x, t) = Σ[n=1 to ∞] Cₙ * e^(-c²n²π²t/L²) * sin(nπx/L), where L is the length of the spatial domain.
Problem: Given ∂U/∂t = ∂²U/∂x², U(0, t) = 0, U(1, t) = 0, U(x, 0) = sin(πx), find the value of t at which U(0.5, t) / U(0.5, 0) = 1/e.
Solution:
1. Identify Parameters: c² = 1, L = 1.
2. Apply Initial Condition: At t = 0, U(x, 0) = sin(πx) = Σ Cₙ * sin(nπx).
3. Equate Coefficients: C₁ = 1, Cₙ = 0 for n > 1.
4. Simplified Solution: U(x, t) = e^(-π²t) * sin(πx).
5. Apply Given Condition: U(0.5, t) / U(0.5, 0) = (e^(-π²t) * sin(π/2)) / sin(π/2) = e^(-π²t) = 1/e.
6. Solve for t: e^(-π²t) = e⁻¹ => π²t = 1 => t = 1/π².
7. The solution in the video is incorrect. The correct solution is t = 1/π².
8. The video incorrectly simplifies the solution to U(x,t) = e^(-t)sin(πx).
9. Using the incorrect solution, the video arrives at the answer t = 1.

10. Partial Differential Equations and Boundary Conditions

Key Concept: The number of boundary conditions required to solve a partial differential equation depends on the number of boundaries in the spatial domain.
Rule:
- Two boundaries => Two boundary conditions.
- One boundary => One boundary condition.
- No boundaries => No boundary conditions.
Example: A partial differential equation defined on a domain with two boundaries requires two boundary conditions.

Synthesis/Conclusion

The video provides a comprehensive overview of several key concepts in linear algebra and differential equations. It covers cube roots of unity, higher-order roots, matrix trace and eigenvalues, symmetric positive definite matrices, rotational transformations, and the one-dimensional heat equation. The video emphasizes the importance of understanding the underlying principles and applying them to solve practical problems. While there are some errors in the solution of the heat equation problem, the overall content is valuable for students and professionals in related fields. The key takeaway is that a strong foundation in these concepts is essential for tackling complex engineering and scientific challenges.