Module 1: Systems of Linear Equations, Matrix Operations, and Quadratic Forms

Introduction to Matrices and Matrix Types

  • Overview & Applications:

    • Information in science, business, and mathematics is organized into rows and columns forming rectangular arrays called matrices.

    • Matrix representations of systems of equations are central to computational algorithms because computers are designed to manipulate structured arrays of numbers.

    • Matrices are formal mathematical objects governed by linear algebra, possessing structural properties and application domains across engineering, data science, and physical modeling.

  • Matrix Definition:

    • A matrix is a rectangular array of numbers enclosed in brackets.

    • A matrix containing mm rows and nn columns has order m×nm \times n (read "mm by nn").

    • General matrix structure:

A=(a11a12…a1na21a22…a2n⋮⋮⋱⋮am1am2…amn)m×nA = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{pmatrix}_{m \times n}

  • Examples of Matrices:

    • Example 1:

A=(0.31−50−0.216)2×3A = \begin{pmatrix} 0.3 & 1 & -5 \\ 0 & -0.2 & 16 \end{pmatrix}_{2 \times 3}

  • Example 2:

B=(40.5)2×1B = \begin{pmatrix} 4 \\ 0.5 \end{pmatrix}_{2 \times 1}

  • Classification of Matrix Types:

    • Row Matrix: A matrix possessing only one row (1×n1 \times n).

    • Example:

A=(2−13)A = \begin{pmatrix} 2 & -1 & 3 \end{pmatrix}

  • Column Matrix: A matrix possessing only one column (m×1m \times 1).

    • Example:

A=(−562)A = \begin{pmatrix} -5 \\ 6 \\ 2 \end{pmatrix}

  • Null Matrix (Zero Matrix): A matrix in which every entry is equal to zero.

    • Example:

A=(000000)A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}

  • Square Matrix: A matrix with an equal number of rows and columns (m=nm = n).

    • Example:

A=(2−341)A = \begin{pmatrix} 2 & -3 \\ 4 & 1 \end{pmatrix}

  • Diagonal Matrix: A square matrix in which all non-leading diagonal entries are zero (aij=0a_{ij} = 0 for all i≠ji \neq j).

    • Example:

A=(20004000−5)A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -5 \end{pmatrix}

  • Scalar Matrix: A diagonal matrix whose leading diagonal entries are all identical.

    • Example:

A=(200020002)A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}

  • Identity Matrix (Unit Matrix): A scalar matrix whose leading diagonal entries are all equal to 11.

    • Example:

A=(100010001)A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

Matrix Operations

  • Equality of Matrices:

    • Two matrices AA and BB are equal (A=BA = B) if and only if both matrices have the same order and their corresponding elements are identical (aij=bija_{ij} = b_{ij} for all i,ji, j).

  • Addition and Subtraction:

    • For matrices AA and BB of identical order m×nm \times n, the sum/difference A±BA \pm B is an m×nm \times n matrix obtained by adding or subtracting corresponding entries.

  • Scalar Multiplication:

    • For a scalar kk and matrix AA, the product kAkA is an m×nm \times n matrix produced by multiplying every individual entry of AA by kk.

  • Matrix Multiplication:

    • The product matrix ABAB of matrix AA (order m×nm \times n) and matrix BB (order p×qp \times q) is defined if and only if the number of columns in AA equals the number of rows in BB (n=pn = p).

    • The resulting product matrix ABAB has dimensions m×qm \times q.

Row Echelon Form and Reduced Row Echelon Form

  • Row Echelon Form:

    • A matrix AA is in echelon form if it meets two fundamental conditions:

    1. All zero rows (rows consisting entirely of zeros), if any, are situated at the bottom of the matrix.

    2. Each leading nonzero entry (the first nonzero element in a row) is located strictly to the right of the leading nonzero entry of the preceding row.

    • Example:

A=(232045−60001−32000000620000000)A = \begin{pmatrix} 2 & 3 & 2 & 0 & 4 & 5 & -6 \\ 0 & 0 & 0 & 1 & -3 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}

  • Reduced Row Echelon Form (Row Canonical Form):

    • A matrix AA is in row canonical form if it satisfies the echelon conditions (1) and (2) along with two additional requirements:

    1. Each pivot entry (leading nonzero entry) is equal to 11.

    2. Each pivot entry is the unique nonzero entry in its containing column.

    • Example 1:

A=(120001000)A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

  • Example 2:

B=(01300400010−3000012)B = \begin{pmatrix} 0 & 1 & 3 & 0 & 0 & 4 \\ 0 & 0 & 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 & 1 & 2 \end{pmatrix}

Elementary Row Operations and Equivalent Matrices

  • Elementary Row Operations:

    • Given a matrix AA with rows R1,R2,…,RmR_1, R_2, \dots, R_m, the permissible elementary row operations are:

    1. Row Interchange: Exchanging row RiR_i with row RjR_j, written as R_i \longleftrightarrow R_j$.\n 2. **Row Scaling:** Replacing row R_iwithanon−zeroscalarmultiplewith a non-zero scalar multiplek R_i,writtenas, written asR_i \longrightarrow k R_i$.

    2. Row Addition: Replacing row RjR_j with the linear sum of RjR_j and a scalar multiple kRik R_i of row RiR_i, written as Rj⟶Rj+kRiR_j \longrightarrow R_j + k R_i.

    • Note: The symbol "⟶\longrightarrow" denotes "changes to".

  • Equivalent Matrices:

    • Two matrices AA and BB are row equivalent if one can be transformed into the other through a finite sequence of elementary row operations.

  • Row Equivalence Examples:

    • Example 1: Show that the following matrices are row equivalent:

(12−31224−461036−6913)and(12−312002460000−2)\begin{pmatrix} 1 & 2 & -3 & 1 & 2 \\ 2 & 4 & -4 & 6 & 10 \\ 3 & 6 & -6 & 9 & 13 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 1 & 2 & -3 & 1 & 2 \\ 0 & 0 & 2 & 4 & 6 \\ 0 & 0 & 0 & 0 & -2 \end{pmatrix}

  • Example 2: Show that the following matrices are row equivalent:

(11−24522−31333−4−21)and(110−10−9001−7−700000)\begin{pmatrix} 1 & 1 & -2 & 4 & 5 \\ 2 & 2 & -3 & 1 & 3 \\ 3 & 3 & -4 & -2 & 1 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 1 & 1 & 0 & -10 & -9 \\ 0 & 0 & 1 & -7 & -7 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}

Rank of a Matrix

  • Formal Definitions:

    • Sub-matrix Definition: A matrix AA has rank rr if:

    1. It contains at least one square sub-matrix of order rr that is non-singular (determinant is non-zero).

    2. Every square sub-matrix of order greater than rr is singular (determinant equals zero).

    • Echelon Definition: The rank rr of matrix AA equals the exact number of non-zero rows in its canonical form or row echelon form.

    • Notation: rank(A)=r\text{rank}(A) = r or ρ(A)=r\rho(A) = r.

  • Worked Problems and Applications:

    • Problem 1: Find the rank of the matrix:

(23−1−11−1−2−4313−2630−7)\begin{pmatrix} 2 & 3 & -1 & -1 \\ 1 & -1 & -2 & -4 \\ 3 & 1 & 3 & -2 \\ 6 & 3 & 0 & -7 \end{pmatrix}

  • Problem 2 (Social Media Influence Analysis): A digital marketing agency analyzes interaction among four social media influencers represented by the matrix:

(01−3−11011310211−20)\begin{pmatrix} 0 & 1 & -3 & -1 \\ 1 & 0 & 1 & 1 \\ 3 & 1 & 0 & 2 \\ 1 & 1 & -2 & 0 \end{pmatrix}

* Positive entries indicate positive influence, negative entries indicate opposing influence, and zero indicates no direct influence.
* Task: Determine the matrix rank and state whether the interaction data provides four independent influence patterns or if some information is redundant.
  • Problem 3: Find the rank of the matrix:

(1230243232136875)\begin{pmatrix} 1 & 2 & 3 & 0 \\ 2 & 4 & 3 & 2 \\ 3 & 2 & 1 & 3 \\ 6 & 8 & 7 & 5 \end{pmatrix}

  • Problem 4 (Machine Learning Feature Selection): A machine learning engineer analyzes relationship metrics among four input features for customer behavior prediction using the matrix:

(12−1424341234−1−26−7)\begin{pmatrix} 1 & 2 & -1 & 4 \\ 2 & 4 & 3 & 4 \\ 1 & 2 & 3 & 4 \\ -1 & -2 & 6 & -7 \end{pmatrix}

* Task: Determine the matrix rank and conclude whether all four features contribute independent information to the predictive model or whether redundant features exist.
  • Problem 5: Find the rank of the matrix:

(567867891112131416171819)\begin{pmatrix} 5 & 6 & 7 & 8 \\ 6 & 7 & 8 & 9 \\ 11 & 12 & 13 & 14 \\ 16 & 17 & 18 & 19 \end{pmatrix}

Systems of Linear Equations

  • General Form:

    • A linear system of mm equations in nn unknowns x1,x2,…,xnx_1, x_2, \dots, x_n is expressed as:

a11x1+a12x2+⋯+a1nxn=b1a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1

a21x1+a22x2+⋯+a2nxn=b2a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2

⋮\vdots

am1x1+am2x2+⋯+amnxn=bma_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n = b_m

  • aij∈Ra_{ij} \in \mathbb{R} are the system coefficients, and bi∈Rb_i \in \mathbb{R} are the constant terms.

    • Solution and Consistency Definitions:

  • Solution: An ordered set of numbers x1,x2,…,xnx_1, x_2, \dots, x_n satisfying all mm equations simultaneously.

  • Consistent System: A system possessing at least one solution.

  • Inconsistent System: A system possessing no valid solution.

    • Matrix Form Representation:

  • System equation: AX=BAX = B, where:

A=(a11a12…a1na21a22…a2n⋮⋮⋱⋮am1am2…amn)m×n,X=(x1x2⋮xn)n×1,B=(b1b2⋮bm)m×1A = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} \end{pmatrix}_{m \times n}, \quad X = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}_{n \times 1}, \quad B = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix}_{m \times 1}

  • Augmented Matrix Representation:

    • Denoted [A:B][A : B] and defined as:

[A:B]=(a11a12…a1n:b1a21a22…a2n:b2⋮⋮⋱⋮:⋮am1am2…amn:bm)[A : B] = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} & : & b_1 \\ a_{21} & a_{22} & \dots & a_{2n} & : & b_2 \\ \vdots & \vdots & \ddots & \vdots & : & \vdots \\ a_{m1} & a_{m2} & \dots & a_{mn} & : & b_m \end{pmatrix}

Linear Homogeneous Systems

  • System Definition:

    • When b1=b2=⋯=bm=0b_1 = b_2 = \dots = b_m = 0 (B=0B = 0), the system AX=0AX = 0 is called a system of linear homogeneous equations.

    • Homogeneous systems always admit the zero or trivial solution x1=x2=⋯=xn=0x_1 = x_2 = \dots = x_n = 0 (X=0X = 0).

    • Because the trivial solution always exists, a system of linear homogeneous equations is inherently consistent.

  • Solution Procedure:

    1. Represent the system as an augmented matrix [A:0][A : 0].

    2. Transform [A:0][A : 0] into its corresponding row reduced echelon form to identify r=rank(A)r = \text{rank}(A).

    3. Evaluate system criteria against the number of unknowns nn:

    • Case 1 (r=nr = n): System possesses only the trivial solution (x1=x2=⋯=xn=0x_1 = x_2 = \dots = x_n = 0).

    • Case 2 (r<nr < n): System possesses infinitely many non-trivial solutions. Assign arbitrary parameters to n−rn - r free variables and solve for the remaining variables using echelon back-substitution.

  • Problems and Applied Scenarios:

    • Problem 1: Find the solution to the system:

x+y−z=0x + y - z = 0

2x+4y−z=02x + 4y - z = 0

3x+2y+2z=03x + 2y + 2z = 0

  • Problem 2 (Distributed Database System): A distributed database consists of three servers whose synchronization errors satisfy:

x+y−z=0x + y - z = 0

2x−3y+z=02x - 3y + z = 0

x−4y+2z=0x - 4y + 2z = 0

* Task: Determine the complete solution set and interpret the network synchronization error state.
  • Problem 3: Solve the system:

2x−y−3z=02x - y - 3z = 0

−x+2y−3z=0-x + 2y - 3z = 0

x+y+4z=0x + y + 4z = 0

  • Problem 4 (Communication Network Flow Balancing): A computer network consists of three interconnected channels with data flow rate adjustments x1,x2,x3x_1, x_2, x_3 satisfying:

3x1+5x2−4x3=03x_1 + 5x_2 - 4x_3 = 0

−3x1−2x2+4x3=0-3x_1 - 2x_2 + 4x_3 = 0

6x1+x2−8x3=06x_1 + x_2 - 8x_3 = 0

* Task: Determine the complete solution and evaluate whether data flow adjustments are independent or dependent.
  • Problem 5: Solve the system:

2x+2y+4z=02x + 2y + 4z = 0

w−y−3z=0w - y - 3z = 0

2w+3x+y+z=02w + 3x + y + z = 0

−2w+x+3y−2z=0-2w + x + 3y - 2z = 0

Linear Non-Homogeneous Systems and Gaussian Elimination

  • System Definition:

    • A linear system AX=BAX = B where at least one bi≠0b_i \neq 0 (B≠0B \neq 0) is called a non-homogeneous system.

  • Gaussian Elimination Procedure:

    1. Construct the augmented matrix [A:B][A : B].

    2. Perform elementary row operations to reduce [A:B][A : B] to row reduced echelon form.

    3. Determine rank(A)\text{rank}(A) and rank([A:B])\text{rank}([A : B]).

    4. Apply Consistency Rules:

    • Rule 1 (rank(A)=rank([A:B])=n\text{rank}(A) = \text{rank}([A : B]) = n): The system is consistent and yields a unique solution.

    • Rule 2 (rank(A)=rank([A:B])=r<n\text{rank}(A) = \text{rank}([A : B]) = r < n): The system is consistent and yields infinitely many solutions (parameterize n−rn - r free variables).

    • Rule 3 (rank(A)≠rank([A:B])\text{rank}(A) \neq \text{rank}([A : B])): The system is inconsistent and possesses no solution.

  • Problems and Applied Scenarios:

    • Problem 1: Verify consistency and solve if solutions exist:

x+y+2z=9x + y + 2z = 9

2x+4y−3z=12x + 4y - 3z = 1

3x+6y−5z=03x + 6y - 5z = 0

  • Problem 2 (Cloud Computing Resource Allocation): A cloud system allocates CPU capacity (x1x_1), memory capacity (x2x_2), and storage capacity (x3x_3) under requirements:

2x1−x2+x3=32x_1 - x_2 + x_3 = 3

4x1+x2+3x3=154x_1 + x_2 + 3x_3 = 15

x1−x2+5x3=14x_1 - x_2 + 5x_3 = 14

* Task: Solve the system and interpret the resource allocation state.
  • Problem 3: Verify consistency and solve if solutions exist:

3x1+2x2+x3=33x_1 + 2x_2 + x_3 = 3

2x1+x2+x3=02x_1 + x_2 + x_3 = 0

6x1+2x2+4x3=66x_1 + 2x_2 + 4x_3 = 6

  • Problem 4: Verify consistency and solve if solutions exist:

x+2y+z=3x + 2y + z = 3

2x+5y−z=−42x + 5y - z = -4

3x−2y−z=53x - 2y - z = 5

  • Problem 5 (Parallel Computing Load Balancing): Processor workloads x1,x2,x3x_1, x_2, x_3 across three computing units satisfy:

5x1−x2+2x3=75x_1 - x_2 + 2x_3 = 7

−2x1+6x2+9x3=0-2x_1 + 6x_2 + 9x_3 = 0

7x1−5x2−3x3=77x_1 - 5x_2 - 3x_3 = 7

* Task: Solve the system and interpret the computational load allocation.

LU Decomposition Method

  • Definition and Existence Condition:

    • The factorization of a square matrix AA into A=LUA = LU, where LL is a unit lower triangular matrix and UU is an upper triangular matrix, is called LU decomposition.

    • Decomposition requires all principal minors of AA to be non-singular (det⁡≠0\det \neq 0):

a11≠0,∣a11a12a21a22∣≠0,∣a11a12a13a21a22a23a31a32a33∣≠0a_{11} \neq 0, \quad \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} \neq 0, \quad \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \neq 0

  • Mathematical Formulation for 3×33 \times 3 Systems:

    • Given system AX=BAX = B, set A=LUA = LU:

L=(100l2110l31l321),U=(u11u12u130u22u2300u33)L = \begin{pmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{pmatrix}

  • Step 1: Write LUX=BLUX = B. Set UX=VUX = V where V=(v1v2v3)V = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}.

  • Step 2: Solve LV=BLV = B for VV using forward substitution:

v1=b1v_1 = b_1

l21v1+v2=b2l_{21}v_1 + v_2 = b_2

l31v1+l32v2+v3=b3l_{31}v_1 + l_{32}v_2 + v_3 = b_3

  • Step 3: Solve UX=VUX = V for XX using back substitution:

u11x1+u12x2+u13x3=v1u_{11}x_1 + u_{12}x_2 + u_{13}x_3 = v_1

u22x2+u23x3=v2u_{22}x_2 + u_{23}x_3 = v_2

u33x3=v3u_{33}x_3 = v_3

  • Matrix Entry Determination Formulas:

    • Equating entries in LU=ALU = A yields:

u11=a11,u12=a12,u13=a13u_{11} = a_{11}, \quad u_{12} = a_{12}, \quad u_{13} = a_{13}

l21u11=a21,l21u12+u22=a22,l21u13+u23=a23l_{21}u_{11} = a_{21}, \quad l_{21}u_{12} + u_{22} = a_{22}, \quad l_{21}u_{13} + u_{23} = a_{23}

l31u11=a31,l31u12+l32u22=a32,l31u13+l32u23+u33=a33l_{31}u_{11} = a_{31}, \quad l_{31}u_{12} + l_{32}u_{22} = a_{32}, \quad l_{31}u_{13} + l_{32}u_{23} + u_{33} = a_{33}

  • Problems and Applied Scenarios:

    • Problem 1: Solve using LU decomposition:

2x+6y+2z=22x + 6y + 2z = 2

3x+8y=−23x + 8y = -2

4x+9y+2z=34x + 9y + 2z = 3

  • Problem 2 (Traffic Signal Optimization): Intersection green light durations x,y,zx, y, z (in minutes) satisfy:

x+2y+3z=14x + 2y + 3z = 14

2x+3y+4z=202x + 3y + 4z = 20

3x+4y+z=143x + 4y + z = 14

* Task: Solve using LU decomposition and determine individual green signal durations.
  • Problem 3: Solve using LU decomposition:

2x−y+3z=92x - y + 3z = 9

x+y+z=6x + y + z = 6

x−y+z=2x - y + z = 2

  • Problem 4 (Cybersecurity Risk Assessment): Risk factor weights for vulnerability severity (x1x_1), threat frequency (x2x_2), and data exposure impact (x3x_3) satisfy:

3x1+6x2−3x3=33x_1 + 6x_2 - 3x_3 = 3

6x1+15x2−5x3=116x_1 + 15x_2 - 5x_3 = 11

−x1−2x2+6x3=9-x_1 - 2x_2 + 6x_3 = 9

* Task: Solve using LU decomposition and interpret the resulting factor weights.

Eigenvalues and Eigenvectors

  • Definitions:

    • For an n×nn \times n matrix AA, a non-zero vector X∈RnX \in \mathbb{R}^n is an eigenvector of AA if:

AX=λX  ⟹  (A−λI)X=0AX = \lambda X \quad \implies \quad (A - \lambda I)X = 0

  • The scalar λ\lambda is an eigenvalue of AA, and XX is the eigenvector corresponding to λ\lambda

    • Characteristic Equation:

  • A scalar λ\lambda is an eigenvalue of AA if and only if it satisfies:

det⁡(A−λI)=0or∣A−λI∣=0\det(A - \lambda I) = 0 \quad \text{or} \quad |A - \lambda I| = 0

  • For a 3×33 \times 3 matrix:

λ3−s1λ2+s2λ−s3=0\lambda^3 - s_1 \lambda^2 + s_2 \lambda - s_3 = 0

    where s1=trace(A)=sum of leading diagonal elementss_1 = \text{trace}(A) = \text{sum of leading diagonal elements}, s2=sum of principal minors of As_2 = \text{sum of principal minors of } A, and s3=det⁡(A)s_3 = \det(A).

  • For a 2×22 \times 2 matrix:

λ2−s1λ+s2=0\lambda^2 - s_1 \lambda + s_2 = 0

    where s1=trace(A)=sum of leading diagonal elementss_1 = \text{trace}(A) = \text{sum of leading diagonal elements} and s2=det⁡(A)s_2 = \det(A).

  • Problems and Applied Scenarios:

    • Problem 1: Find all eigenvalues and corresponding eigenvectors of:

(113151311)\begin{pmatrix} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \end{pmatrix}

  • Problem 2 (Network Adjacency Analysis): Three network nodes have adjacency matrix:

A=(011101110)A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}

* Task: Determine eigenvalues and eigenvectors, and interpret their significance regarding network connectivity.
  • Problem 3: Find all eigenvalues and corresponding eigenvectors of:

(423−1)\begin{pmatrix} 4 & 2 \\ 3 & -1 \end{pmatrix}

  • Problem 4 (Data Cluster Similarity): Similarity relationships among three clusters in data mining are given by:

A=(211121112)A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}

* Task: Determine eigenvalues and eigenvectors, interpreting their context in clustering.
  • Problem 5: Find all eigenvalues and corresponding eigenvectors of:

(2−2211113−1)\begin{pmatrix} 2 & -2 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & -1 \end{pmatrix}

  • Problem 6: Find all eigenvalues and corresponding eigenvectors of:

(8−62−67−42−43)\begin{pmatrix} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{pmatrix}

  • Problem 7: Find all eigenvalues and corresponding eigenvectors of:

(−1320)\begin{pmatrix} -1 & 3 \\ 2 & 0 \end{pmatrix}

Properties of Eigenvalues

  • Fundamental Mathematical Properties:

    • Property a: A square matrix AA and its transpose ATA^T share identical eigenvalues.

    • Property b: The sum of the eigenvalues of AA equals the trace of AA (∑λi=trace(A)\sum \lambda_i = \text{trace}(A)).

    • Property c: The product of the eigenvalues of AA equals det⁡(A)\det(A) (∏λi=det⁡(A)\prod \lambda_i = \det(A)).

    • Property d: Given eigenvalues λ1,λ2,…,λn\lambda_1, \lambda_2, \dots, \lambda_n of matrix AA:

    • (i) Matrix kAkA (k≠0k \neq 0) has eigenvalues k\lambda_1, k\lambda_2, \dots, k\lambda_n$.\n * (ii) Matrix power A^p((p > 0)haseigenvalues) has eigenvalues\lambda_1^p, \lambda_2^p, \dots, \lambda_n^p$.

    • (iii) Inverse matrix A−1A^{-1} has eigenvalues 1λ1,1λ2,…,1λn\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \dots, \frac{1}{\lambda_n} provided \lambda_r \neq 0$.\n * **Property e:** Matrix A \pm kIhaseigenvalueshas eigenvalues\lambda_1 \pm k, \lambda_2 \pm k, \dots, \lambda_n \pm k$.

    • Property f: Eigenvalues of a real symmetric matrix are strictly real numbers.

    • Property g: If zero (00) is an eigenvalue of matrix AA, then AA is singular (det⁡(A)=0\det(A) = 0).

    • Property h: For a triangular matrix or diagonal matrix, the eigenvalues are the leading diagonal entries.

    • Property i: Eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are mutually orthogonal.

  • Analytical Problems on Properties:

    • Problem 1: Find the sum and product of eigenvalues of:

(314026005)\begin{pmatrix} 3 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 5 \end{pmatrix}

  • Problem 2: Two eigenvalues of A=(466132−1−5−2)A = \begin{pmatrix} 4 & 6 & 6 \\ 1 & 3 & 2 \\ -1 & -5 & -2 \end{pmatrix} are equal and double the magnitude of the third. Find the eigenvalues of A^2$.\n * **Problem 3:** If one eigenvalue of A = \begin{pmatrix} 1 & -3 & 3 \ 3 & -5 & 3 \ 6 & -6 & 4 \end{pmatrix}isis4,andtheproductoftheremainingtwoeigenvaluesis, and the product of the remaining two eigenvalues is4,findalleigenvaluesof, find all eigenvalues ofA$.

  • Problem 4: Given a 3×33 \times 3 matrix AA with one eigenvalue equal to 22, sum of all eigenvalues equal to 44, and product of all eigenvalues equal to −26-26, determine all three eigenvalues.

  • Problem 5: Given eigenvalues 2,−1,−22, -1, -2 for matrix AA, find the eigenvalues of A^2 - 2I$.\n * **Problem 6:** Find the eigenvalues of A^2andandA^{-1} for matrix:\n\nA = \begin{pmatrix} 3 & 0 & 0 \ 5 & 4 & 0 \ 3 & 6 & 1 \end{pmatrix}\n\n# Diagonalization of Matrices\n\n* **Similarity of Matrices:**\n * Matrix Bissimilartois similar toAifthereexistsaninvertiblematrixif there exists an invertible matrixPsuchthatsuch thatB = P^{-1}AP$.

  • If BB is similar to AA, then AA is similar to BB via A=Q−1BQA = Q^{-1}BQ, where Q=P−1Q = P^{-1}.

    • Diagonalization Definition:

  • A square matrix AA is diagonalizable if there exists an invertible matrix PP such that P−1AP=DP^{-1}AP = D is a diagonal matrix. The matrix PP diagonalizes AA.

    • Diagonalization Procedure:

  • Find all eigenvalues λ1,λ2,…,λn\lambda_1, \lambda_2, \dots, \lambda_n and corresponding linear independent eigenvectors X1,X2,…,XnX_1, X_2, \dots, X_n of matrix A$.\n 2. Construct modal matrix P = \begin{pmatrix} X_1 & X_2 & \dots & X_n \end{pmatrix}_{n \times n}.\n 3. Compute inverse matrix P^{-1}.\n 4. Apply transformation P^{-1}AP = D,where, whereD is the resulting diagonal matrix containing eigenvalues along its main diagonal.\n\n* **Problems and Applied Scenarios:**\n * **Problem 1:** Transform matrix to diagonal form:\n\n\begin{pmatrix} 1 & 1 & 3 \ 1 & 5 & 1 \ 3 & 1 & 1 \end{pmatrix}\n\n * **Problem 2 (Communication Network Modes):** Diagonalize adjacency matrix A = \begin{pmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \end{pmatrix} to determine independent communication modes.\n * **Problem 3:** Express matrix in diagonal form:\n\n\begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix}\n\n * **Problem 4:** Express matrix in diagonal form:\n\n\begin{pmatrix} 1 & 4 \ 2 & 3 \end{pmatrix}\n\n * **Problem 5 (Traffic Flow Modes):** Diagonalize road junction interaction matrix:\n\n\begin{pmatrix} 8 & -6 & 2 \ -6 & 7 & -4 \ 2 & -4 & 3 \end{pmatrix}\n\n * Task: Use diagonal form to identify independent traffic-flow modes.\n\n * **Problem 6:** Express matrix in diagonal form:\n\n\begin{pmatrix} 2 & 1 \ 1 & 2 \end{pmatrix}\n\n * **Problem 7:** Express matrix in diagonal form:\n\n\begin{pmatrix} 7 & 2 \ -4 & 1 \end{pmatrix}\n\n * **Problem 8:** Express matrix in diagonal form:\n\n\begin{pmatrix} 2 & 3 \ 4 & 1 \end{pmatrix}\n\n# Quadratic Forms\n\n* **Mathematical Definition:**\n * A homogeneous expression of the second degree in any number of variables is called a quadratic form.\n * Examples: 2x^2 + 5xy - 7y^2,,3x_1^2 - 4x_1x_2 + 7x_2^2$.

    • General Algebraic Form:

  • For variable vector X=(x1x2…xn)TX = \begin{pmatrix} x_1 & x_2 & \dots & x_n \end{pmatrix}^T:

Q=∑i=1n∑j=1naijxixj(aij=aji)Q = \sum_{i=1}^n \sum_{j=1}^n a_{ij} x_i x_j \quad (a_{ij} = a_{ji})

Q=a11x12+a22x22+⋯+annxn2+2a12x1x2+2a13x1x3+⋯+2a(n−1)nxn−1xnQ = a_{11}x_1^2 + a_{22}x_2^2 + \dots + a_{nn}x_n^2 + 2a_{12}x_1x_2 + 2a_{13}x_1x_3 + \dots + 2a_{(n-1)n}x_{n-1}x_n

Q=∑i=1naiixi2+2∑i=1n−1∑j=i+1naijxixjQ = \sum_{i=1}^n a_{ii} x_i^2 + 2 \sum_{i=1}^{n-1} \sum_{j=i+1}^n a_{ij} x_i x_j

  • Matrix Representation:

    • System expression: Q=XTAXQ = X^T A X, where AA is a real symmetric matrix (A=ATA = A^T) of order n×nn \times n, called the matrix of the quadratic form.

    • Example: Q=2x2+6xy−5y2=(xy)(233−5)(xy)Q = 2x^2 + 6xy - 5y^2 = \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 3 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.

  • Problems:

    • Problem 1: Find matrix of quadratic form 2x_1^2 - 2x_2^2 + 4x_3^2 + 2x_1x_2 - 6x_1x_3 + 6x_2x_3$.\n * **Problem 2:** Find matrix of quadratic form 10x_1^2 + 2x_2^2 + 5x_3^2 - 4x_1x_2 + 6x_2x_3 - 10x_1x_3$.

    • Problem 3: Find quadratic form corresponding to matrix:

(0−12−114243)\begin{pmatrix} 0 & -1 & 2 \\ -1 & 1 & 4 \\ 2 & 4 & 3 \end{pmatrix}

  • Problem 4: Find quadratic form corresponding to matrix:

(125246563)\begin{pmatrix} 1 & 2 & 5 \\ 2 & 4 & 6 \\ 5 & 6 & 3 \end{pmatrix}

Orthogonality and Vector Normalization

  • Orthogonal Matrix Definition:

    • A square matrix AA is orthogonal if A−1=ATA^{-1} = A^T, which implies A A^T = A^T A = I$.\n\n* **Pairwise Orthogonality:**\n * Vectors X_1, X_2, X_3 are pairwise orthogonal if:\n\nX_1 \cdot X_2^T = X_2 \cdot X_3^T = X_3 \cdot X_1^T = 0\n\n * Note: A matrix whose columns are pairwise orthogonal and normalized is an orthogonal matrix.\n\n* **Vector Normalization:**\n * The normalized vector of X = \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} is given by:\n\nX^N = \begin{pmatrix} \frac{x_1}{d} \ \frac{x_2}{d} \ \frac{x_3}{d} \end{pmatrix} \quad \text{where } d = \sqrt{x_1^2 + x_2^2 + x_3^2}\n\n# Canonical Form of Quadratic Forms and Nature Classification\n\n* **Canonical Form Transformation:**\n * For quadratic form Q = X^T A X,transformation, transformationX = PY(where(wherePisthenormalizedmodalmatrixformedbynormalizedorthogonaleigenvectorsofis the normalized modal matrix formed by normalized orthogonal eigenvectors ofA)reduces) reducesQ to:\n\nQ = X^T A X = (PY)^T A (PY) = Y^T (P^T A P) Y = Y^T D Y\n\nQ = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \dots + \lambda_n y_n^2\n\n * where D = \begin{pmatrix} \lambda_1 & 0 & \dots & 0 \ 0 & \lambda_2 & \dots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \dots & \lambda_n \end{pmatrix}.\n\n![Canonical Form of Quadratic Form Derivation](https://assets.knowt.com/pdf-flow-prod/bc4b1a05-2d41-4866-8dd4-507a68fec4dc-figures/0.png)\n\n* **Nature Classification of Quadratic Form:**\n * A real quadratic form Q = X^T A Xininn variables is classified as:\n * **Positive Definite:** All eigenvalues of Aarestrictlypositive(are strictly positive (\lambda_i > 0).\n * **Negative Definite:** All eigenvalues of Aarestrictlynegative(are strictly negative (\lambda_i < 0).\n * **Positive Semidefinite:** All eigenvalues of Aarenon−negative(are non-negative (\lambda_i \ge 0) with at least one eigenvalue equal to zero.\n * **Negative Semidefinite:** All eigenvalues of Aarenon−positive(are non-positive (\lambda_i \le 0) with at least one eigenvalue equal to zero.\n * **Indefinite:** Eigenvalues of A consist of both positive and negative values.\n\n* **Reduction Procedure Step-by-Step:**\n 1. Obtain symmetric matrix Afromquadraticformfrom quadratic formQ = X^T A X.\n 2. Compute all eigenvalues \lambda_1, \lambda_2, \dots, \lambda_nandcorrespondingeigenvectorsand corresponding eigenvectorsX_1, X_2, \dots, X_n.\n 3. Verify pairwise orthogonality of eigenvectors (orthogonalize if degenerate) and compute normalized eigenvectors X_1^N, X_2^N, \dots, X_n^N.\n 4. Form normalized modal matrix P = \begin{pmatrix} X_1^N & X_2^N & \dots & X_n^N \end{pmatrix}.\n 5. Apply substitution X = PYtoexpresscanonicalformto express canonical formQ = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \dots + \lambda_n y_n^2$.

    1. Determine nature classification based on eigenvalue signs.

  • Problems and Applied Scenarios:

    • Problem 1 (Traffic Congestion Model): Congestion across three junctions is modeled by x2+5y2+z2+2xy+2yz+6xzx^2 + 5y^2 + z^2 + 2xy + 2yz + 6xz.

    • Task: Find canonical form and determine nature of traffic network.

    • Problem 2: Find canonical form and determine nature of quadratic form 2x_1x_2 + 2x_1x_3 + 2x_2x_3$.\n * **Problem 3:** Find canonical form and determine nature of quadratic form 2x^2 + 2y^2 + 2z^2 + 2xy + 2yz + 2xz$.

    • Problem 4 (Network Router Performance): Performance across three routers is measured by 8x_1^2 + 7x_2^2 + 3x_3^2 - 12x_1x_2 + 4x_1x_3 - 8x_2x_3$.\n * Task: Determine canonical form and classify nature of network performance model.\n * **Problem 5:** Find canonical form and determine nature of quadratic form -2x^2 + 4xy - 2y^2$.

    • Problem 6: Find canonical form and determine nature of quadratic form $$x^2 - 4xy + y^2$.