Math 2109: Vector, Matrix and Transform

Math Class Notes (Classes 4-8)

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3rd Cycle | 27.07.26 | Class-04

Normal Form of a Matrix

Definition: Normal Form (Canonical Form)

A square or rectangular matrix $A$ of order $m\times n$ can be reduced by a finite sequence of elementary row and column operations (EROs and ECOs) to one of the following four standard block structures, known as the Normal Form of the matrix:

1. $I_r$
2. $\left[\begin{array}{cc}{I}_r& O\end{array}\right]$
3. $\left[\begin{array}{c}{I}_r\\ O\end{array}\right]$
4. $\left[\begin{array}{cc}{I}_r& O\\ O& O\end{array}\right]$

where:

• $I_r$ is the Identity Matrix of order $r$.
• $O$ represents zero submatrices of conformable dimensions.
• The integer $r$ represents the Rank of the matrix $A$.

Rank and Normal Form

The size $r$ of the identity block $I_r$ corresponds exactly to the Rank of the matrix: $\text{rank}(A)=r$.

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28.07.26 | Class-05

Rank of a Matrix & Normal Form Conversions

Concept of Rank

The Rank of a matrix $A$ is the maximum number of linearly independent rows (or columns) in the matrix.

Key Properties of Row-Echelon Rank:

• A zero row (a row consisting entirely of zeros) is always linearly dependent.
• In a matrix reduced to Row-Echelon Form, every non-zero row is linearly independent.
• Consequently, the rank of a matrix is equal to the number of non-zero rows in its Row-Echelon Form.
• In any matrix, the number of independent rows always equals the number of independent columns.
• All rows of an Identity Matrix $I_n$ are linearly independent, so $\text{rank}(I_n)=n$.

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Solved Problems: Rank and Normal Form Conversions

Solved Problem 1: Converting a Matrix into Normal Form

Convert the following matrix $A$ into its Normal Form and find its rank: $A=\left[\begin{array}{cccc}1& 2& -1& 3\\ 4& 1& 2& 1\\ 3& -1& 1& 2\\ 1& 2& 0& 1\end{array}\right]$

Solution: Apply elementary row operations to create zeros in the first column below $a_{11}$: $A\underset{R_4^{\prime }\to R_4-R_1}{\overset{R_2^{\prime }\to R_2-4R_1,R_3^{\prime }\to R_3-3R_1}{\to }}\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& -7& 6& -11\\ 0& -7& 4& -7\\ 0& 0& 1& -2\end{array}\right]$

Perform ERO to eliminate the $-7$ in the third row, column 2: $\stackrel{R_3^{\prime \prime }\to R_3^{\prime }-R_2^{\prime }}{\to }\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& -7& 6& -11\\ 0& 0& -2& 4\\ 0& 0& 0& 0\end{array}\right]$

Apply elementary column operations to clear the entries in the first row to the right of $a_{11}$: $\underset{C_3^{\prime }\to C_3+C_1,C_4^{\prime }\to C_4-3C_1}{\overset{C_2^{\prime }\to C_2-2C_1}{\to }}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -7& 6& -11\\ 0& 0& -2& 4\\ 0& 0& 0& 0\end{array}\right]$

Apply column operations using the pivot in column 2 to clear entries in row 2: $\underset{C_4^{\prime }\to C_4-\frac{11}{7}{C}_2}{\overset{C_3^{\prime }\to C_3+\frac{6}{7}{C}_2}{\to }}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -7& 0& 0\\ 0& 0& -2& 4\\ 0& 0& 0& 0\end{array}\right]$

Clear the entry $4$ in row 3 using the column pivot in column 3: $\stackrel{C_4^{\prime \prime }\to C_4^{\prime }+2C_3^{\prime }}{\to }\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& -7& 0& 0\\ 0& 0& -2& 0\\ 0& 0& 0& 0\end{array}\right]$

Scale column 2 and column 3 to normalize the diagonal pivots: $\underset{C_3^{\prime \prime }\to -\frac{1}{2}{C}_3^{\prime }}{\overset{C_2^{\prime \prime }\to -\frac{1}{7}{C}_2}{\to }}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right]=\left[\begin{array}{cc}{I}_3& O\\ O& O\end{array}\right]$

Final Answer: The Normal Form of the matrix is $\left[\begin{array}{cc}{I}_3& O\\ O& O\end{array}\right]$, and its rank is $r=3$.

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Solved Problem 2: Finding Rank via Row-Echelon Form

Find the rank of the following matrix: $B=\left[\begin{array}{cccc}2& 1& 3& 2\\ 3& 0& 1& 4\\ 5& 1& 2& 3\\ -11& -4& -5& 0\end{array}\right]$

Solution: (Note: The first row of the matrix is written as $2132$ in the notes, but ERO calculations in the student’s draft treat the first entry as $1$, i.e. $R_1=\left[\begin{array}{cccc}1& 1& 3& 2\end{array}\right]$.)

Initial EROs using $R_1$ as a pivot: $\left[\begin{array}{cccc}1& 1& 3& 2\\ 3& 0& 1& 4\\ 5& 1& 2& 3\\ -11& -4& -5& 0\end{array}\right]\underset{R_3^{\prime }\to R_3-5R_1,R_4^{\prime }\to R_4+11R_1}{\overset{R_2^{\prime }\to R_2-3R_1}{\to }}\left[\begin{array}{cccc}1& 1& 3& 2\\ 0& -3& -8& -2\\ 0& -4& -13& -7\\ 0& 7& 28& 22\end{array}\right]$

Draft Method (Fractional EROs): Reducing using fractional operations: $\underset{R_4^{\prime \prime }\to R_4^{\prime }+\frac{7}{3}{R}_2^{\prime }}{\overset{R_3^{\prime \prime }\to R_3^{\prime }-\frac{4}{3}{R}_2^{\prime }}{\to }}\left[\begin{array}{cccc}1& 1& 3& 2\\ 0& -3& -8& -2\\ 0& 0& -7/3& -13/3\\ 0& 0& 28/3& 52/3\end{array}\right]$

Alternative Method (Integer-based EROs): To avoid complex fractions, we can subtract Row 3 from Row 2 to obtain a leading $1$ at $(2,2)$: $\left[\begin{array}{cccc}1& 1& 3& 2\\ 0& -3& -8& -2\\ 0& -4& -13& -7\\ 0& 7& 28& 22\end{array}\right]\stackrel{R_2^{\prime \prime }\to R_2^{\prime }-R_3^{\prime }}{\to }\left[\begin{array}{cccc}1& 1& 3& 2\\ 0& 1& 5& 5\\ 0& -4& -13& -7\\ 0& 7& 28& 22\end{array}\right]$

Eliminate column 2 entries below the new pivot: $\underset{R_4^{\prime \prime }\to R_4^{\prime }-7R_2^{\prime \prime }}{\overset{R_3^{\prime \prime }\to R_3^{\prime }+4R_2^{\prime \prime }}{\to }}\left[\begin{array}{cccc}1& 1& 3& 2\\ 0& 1& 5& 5\\ 0& 0& 7& 13\\ 0& 0& -7& -13\end{array}\right]$

Eliminate the fourth row completely: $\stackrel{R_4^{\prime \prime \prime }\to R_4^{\prime \prime }+R_3^{\prime \prime }}{\to }\left[\begin{array}{cccc}1& 1& 3& 2\\ 0& 1& 5& 5\\ 0& 0& 7& 13\\ 0& 0& 0& 0\end{array}\right]$

Since there are exactly $3$ non-zero rows in the Row-Echelon Form: Final Answer: The rank of the matrix is $\text{rank}(B)=3$.

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Solved Problem 3: Finding Rank via Row Interchanges

Find the rank of the following matrix: $C=\left[\begin{array}{cccc}3& 1& 4& 3\\ 2& 0& 1& 2\\ 1& 2& 3& 5\\ 0& 1& 2& 4\end{array}\right]$

Solution: Interchange Row 1 and Row 3 to bring the leading $1$ to the top: $C\stackrel{R_1\leftrightarrow R_3}{\to }\left[\begin{array}{cccc}1& 2& 3& 5\\ 2& 0& 1& 2\\ 3& 1& 4& 3\\ 0& 1& 2& 4\end{array}\right]$

Create zeros in column 1: $\underset{R_3^{\prime }\to R_3-3R_1}{\overset{R_2^{\prime }\to R_2-2R_1}{\to }}\left[\begin{array}{cccc}1& 2& 3& 5\\ 0& -4& -5& -8\\ 0& -5& -5& -12\\ 0& 1& 2& 4\end{array}\right]$

Interchange Row 2 and Row 4 to place the leading $1$ as the second row pivot: $\stackrel{R_2^{\prime }\leftrightarrow R_4}{\to }\left[\begin{array}{cccc}1& 2& 3& 5\\ 0& 1& 2& 4\\ 0& -5& -5& -12\\ 0& -4& -5& -8\end{array}\right]$

Create zeros in column 2 below the new pivot: $\underset{R_4^{\prime \prime }\to R_4^{\prime }+4R_2^{\prime }}{\overset{R_3^{\prime \prime }\to R_3^{\prime }+5R_2^{\prime }}{\to }}\left[\begin{array}{cccc}1& 2& 3& 5\\ 0& 1& 2& 4\\ 0& 0& 5& 8\\ 0& 0& 3& 8\end{array}\right]$

Subtract Row 4 from Row 3 to create a smaller integer pivot ($5-3=2$): $\stackrel{R_3^{\prime \prime \prime }\to R_3^{\prime \prime }-R_4^{\prime \prime }}{\to }\left[\begin{array}{cccc}1& 2& 3& 5\\ 0& 1& 2& 4\\ 0& 0& 2& 0\\ 0& 0& 3& 8\end{array}\right]$

Eliminate the entry $3$ in row 4 using Row 3: $\stackrel{R_4^{\prime \prime \prime }\to R_4^{\prime \prime }-\frac{3}{2}{R}_3^{\prime \prime \prime }}{\to }\left[\begin{array}{cccc}1& 2& 3& 5\\ 0& 1& 2& 4\\ 0& 0& 2& 0\\ 0& 0& 0& 8\end{array}\right]$

This matrix is in Row-Echelon Form. Since there are $4$ non-zero rows: Final Answer: The rank of the matrix is $\text{rank}(C)=4$.

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03.08.26 | Class-06

Systems of Linear Equations & Elimination Methods

Notation and Representation

A system of $m$ linear equations in $n$ variables can be compactly written as: $Ax=b$ where:

• $A$ is the coefficient matrix of order $m\times n$.
• $x$ is the solution vector of size $n\times 1$.
• $b$ is the RHS constant vector of size $m\times 1$.

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} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} $$

Classification of Systems of Equations

          System of L.E.
         /              \
  Consistent          Inconsistent
 (Soln. exists)      (No solution exists)
    /       \
Unique   Infinite
Solution  Solutions

Theorem: Rank Criterion for Consistency

Let $[A|b]$ denote the augmented matrix of a system of equations.

1. The system is consistent if and only if: $\text{rank}(A)=\text{rank}([A|b])$
2. The system is inconsistent if and only if: $\text{rank}(A)\ne \text{rank}([A|b])$

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Solved Problems: Solving SLEs

Solved Problem 1: Gaussian Elimination & Back-Substitution

Solve the system of linear equations using Gaussian elimination: $\{\begin{array}{l}x+y-z=7\\ 2x-y+2z=5\\ 3x+y-z=3\end{array}$

Solution: Construct the augmented matrix $[A|b]$: $[A|b]=\left[\begin{array}{cccc}1& 1& -1& 7\\ 2& -1& 2& 5\\ 3& 1& -1& 3\end{array}\right]$

Perform EROs to create zeros in column 1: $\underset{R_3^{\prime }\to R_3-3R_1}{\overset{R_2^{\prime }\to R_2-2R_1}{\to }}\left[\begin{array}{cccc}1& 1& -1& 7\\ 0& -3& 4& -9\\ 0& -2& 2& -18\end{array}\right]$

Create a zero in row 3, column 2: $\stackrel{R_3^{\prime \prime }\to R_3^{\prime }-\frac{2}{3}{R}_2^{\prime }}{\to }\left[\begin{array}{cccc}1& 1& -1& 7\\ 0& -3& 4& -9\\ 0& 0& -2/3& -12\end{array}\right]$

Consistency Analysis: The coefficient matrix $A$ and augmented matrix $[A|b]$ are in row-echelon form.

• $\text{rank}(A)=3$ and $\text{rank}([A|b])=3$.
• Since $\text{rank}(A)=\text{rank}([A|b])=\text{number of variables }(3)$, a unique solution exists.

Back-Substitution:

1. From Row 3: $-\frac{2}{3}z=-12⟹\mathbf{z}=18$
2. From Row 2: $-3y+4z=-9⟹-3y+4(18)=-9⟹-3y=-9-72=-81⟹\mathbf{y}=27$
3. From Row 1: $x+y-z=7⟹x+27-18=7⟹x+9=7⟹\mathbf{x}=-2$

Final Answer: The unique solution is $(x,y,z)=(-2,27,18)$.

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Solved Problem 2: Matrix Inversion by Elementary Transformations

Find the inverse of the following matrix $A$ using elementary row transformations: $A=\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 3\\ 1& 0& 8\end{array}\right]$

Solution: Setup the partition matrix $[A|I]$: $[A|I]=\left[\begin{array}{cccccc}1& 2& 3& 1& 0& 0\\ 2& 5& 3& 0& 1& 0\\ 1& 0& 8& 0& 0& 1\end{array}\right]$

Apply EROs to clear column 1: $\underset{R_3^{\prime }\to R_3-R_1}{\overset{R_2^{\prime }\to R_2-2R_1}{\to }}\left[\begin{array}{cccccc}1& 2& 3& 1& 0& 0\\ 0& 1& -3& -2& 1& 0\\ 0& -2& 5& -1& 0& 1\end{array}\right]$

Clear row 3, column 2: $\stackrel{R_3^{\prime \prime }\to R_3^{\prime }+2R_2^{\prime }}{\to }\left[\begin{array}{cccccc}1& 2& 3& 1& 0& 0\\ 0& 1& -3& -2& 1& 0\\ 0& 0& -1& -5& 2& 1\end{array}\right]$ (Note: The entry at row 3, column 5 in the RHS is corrected to $2$ here, resolving an arithmetic typo of $0$ in the student’s original handwritten draft).

Normalize row 3: $\stackrel{R_3^{\prime \prime \prime }\to -R_3^{\prime \prime }}{\to }\left[\begin{array}{cccccc}1& 2& 3& 1& 0& 0\\ 0& 1& -3& -2& 1& 0\\ 0& 0& 1& 5& -2& -1\end{array}\right]$

Clear column 3 entries above the pivot: $\underset{R_2^{\prime \prime \prime }\to R_2^{\prime \prime }+3R_3^{\prime \prime \prime }}{\overset{R_1^{\prime }\to R_1-3R_3^{\prime \prime \prime }}{\to }}\left[\begin{array}{cccccc}1& 2& 0& -14& 6& 3\\ 0& 1& 0& 13& -5& -3\\ 0& 0& 1& 5& -2& -1\end{array}\right]$

Clear the remaining row 1, column 2 entry to obtain the identity matrix on the LHS: $\stackrel{R_1^{\prime \prime }\to R_1^{\prime }-2R_2^{\prime \prime \prime }}{\to }\left[\begin{array}{cccccc}1& 0& 0& -40& 16& 9\\ 0& 1& 0& 13& -5& -3\\ 0& 0& 1& 5& -2& -1\end{array}\right]$

Final Answer: $A^{-1}=\left[\begin{array}{ccc}-40& 16& 9\\ 13& -5& -3\\ 5& -2& -1\end{array}\right]$

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Solved Problem 3: Gauss-Jordan Elimination (Inconsistent System)

Solve the system of equations using Gauss-Jordan elimination: $\{\begin{array}{l}{x}_2-4x_3=8\\ 2x_1-3x_2+2x_3=-1\\ 5x_1-8x_2+7x_3=1\end{array}$

Solution: Construct the augmented matrix: $[A|b]=\left[\begin{array}{cccc}0& 1& -4& 8\\ 2& -3& 2& -1\\ 5& -8& 7& 1\end{array}\right]$

Interchange Row 1 and Row 2 to place a non-zero pivot on the main diagonal: $\stackrel{R_1\leftrightarrow R_2}{\to }\left[\begin{array}{cccc}2& -3& 2& -1\\ 0& 1& -4& 8\\ 5& -8& 7& 1\end{array}\right]$

Eliminate entry in Row 3, Column 1: $\stackrel{R_3^{\prime }\to R_3-\frac{5}{2}{R}_1}{\to }\left[\begin{array}{cccc}2& -3& 2& -1\\ 0& 1& -4& 8\\ 0& -0.5& 2& 3.5\end{array}\right]$

Eliminate entry in Row 3, Column 2: $\stackrel{R_3^{\prime \prime }\to R_3^{\prime }+0.5R_2}{\to }\left[\begin{array}{cccc}2& -3& 2& -1\\ 0& 1& -4& 8\\ 0& 0& 0& 7.5\end{array}\right]$

Consistency Analysis: The echelon form has:

• $\text{rank}(A)=2$ (the third row of $A$ is all zeros).
• $\text{rank}([A|b])=3$ (the augmented matrix has a non-zero pivot $7.5$ in row 3).
• Since $\text{rank}(A)\ne \text{rank}([A|b])$, the system of equations is inconsistent.

Final Answer: The system has no solution.

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Solved Problem 4: Gauss-Jordan Elimination (Infinite Solutions Case)

Solve the system of equations: $\{\begin{array}{l}{x}_1+2x_2-x_3=1\\ 3x_1-2x_2+2x_3=2\\ 7x_1-2x_2+3x_3=5\end{array}$

Solution: Construct the augmented matrix: $[A|b]=\left[\begin{array}{cccc}1& 2& -1& 1\\ 3& -2& 2& 2\\ 7& -2& 3& 5\end{array}\right]$

Eliminate entries in Column 1 below the pivot: $\underset{R_3^{\prime }\to R_3-7R_1}{\overset{R_2^{\prime }\to R_2-3R_1}{\to }}\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -8& 5& -1\\ 0& -16& 10& -2\end{array}\right]$

Eliminate the entry in Row 3, Column 2: $\stackrel{R_3^{\prime \prime }\to R_3^{\prime }-2R_2^{\prime }}{\to }\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -8& 5& -1\\ 0& 0& 0& 0\end{array}\right]$

Consistency Analysis:

• $\text{rank}(A)=2$ and $\text{rank}([A|b])=2$.
• Since $\text{rank}(A)=\text{rank}([A|b])=2<\text{number of variables }(3)$, the system is consistent and has infinitely many solutions.

Dimension Analysis for Free Variables: $\text{Number of Free Variables}=\text{Variables}-\text{Rank}=3-2=1$

General Solution Derivation: Re-write the equations from the echelon matrix: $\{\begin{array}{l}{x}_1+2x_2-x_3=1\\ -8x_2+5x_3=-1\end{array}$

Define the free variable: Let $x_3=k$ (where $k$ is any real parameter).

Substitute $x_3=k$ into the second equation: $-8x_2+5k=-1⟹-8x_2=-1-5k⟹{\mathbf{x}}_2=\frac{1+5k}{8}$

Substitute $x_2$ and $x_3$ into the first equation: $x_1+2\left(\frac{1+5k}{8}\right)-k=1$ $\Rightarrow x_1+\frac{1+5k}{4}-k=1$ $\Rightarrow x_1=1-\frac{1}{4}-\frac{5k}{4}+k=\frac{3}{4}-\frac{k}{4}=\frac{3-k}{4}$

Final Answer: The general solution is $\left(\frac{3-k}{4},\frac{1+5k}{8},k\right)$ for $k\in \mathbb{R}$.

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04.08.26 | Class-07

Parameter Analysis in Systems of Equations

Underdetermined Systems & Free Variables

When a consistent system has fewer independent equations (after reduction) than variables, the remaining variables are designated as free variables.

Example Matrix Structure: $\left(\begin{array}{cccccc}1& 2& 3& 4& 1& 6\\ 0& 1& 2& 0& 2& 3\\ 0& 0& 0& 1& 3& 5\end{array}\right)$

• Number of variables: $5$ ($x_1,x_2,x_3,x_4,x_5$)
• Rank of matrix: $3$ (pivots in column 1, 2, and 4)
• Free variables: $\text{Variables}-\text{Rank}=5-3=2$ (the columns without pivots: column 3 and column 5, representing $x_3$ and $x_5$ as free variables).

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Solved Problem: The $\lambda$ and $\mu$ Consistency Analysis

Solved Problem: Consistency of Parametric Systems

Determine the values of $\lambda$ and $\mu$ for which the following system of linear equations: $\{\begin{array}{l}x+y+z=6\\ x+2y+3z=10\\ x+2y+\lambda z=\mu \end{array}$ has: (i) a unique solution, (ii) no solution, and (iii) infinitely many solutions.

Solution: Setup the augmented matrix $[A|b]$ and apply EROs to reduce it to echelon form: $[A|b]=\left[\begin{array}{cccc}1& 1& 1& 6\\ 1& 2& 3& 10\\ 1& 2& \lambda & \mu \end{array}\right]\underset{R_3^{\prime }\to R_3-R_2}{\overset{R_2^{\prime }\to R_2-R_1}{\to }}\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& 2& 4\\ 0& 0& \lambda -3& \mu -10\end{array}\right]$

Case Analysis:

i. Unique Solution: A unique solution requires $\text{rank}(A)=\text{rank}([A|b])=3$. This occurs if the pivot in the third row is non-zero: $\lambda -3\ne 0⟹\lambda \ne 3(\mu \text{ can be any real value})$ (Note: If $\lambda \ne 3$, the rank of both the coefficient matrix $A$ and the augmented matrix $[A|b]$ is $3$, which matches the number of variables, ensuring a unique solution).

ii. No Solution (Inconsistent System): No solution occurs if the augmented row states $0=\text{non-zero}$, which means $\text{rank}(A)=2\ne \text{rank}([A|b])=3$: $\lambda -3=0\text{ and }\mu -10\ne 0⟹\lambda =3\text{ and }\mu \ne 10$ (Note: If $\lambda =3$, the bottom row becomes $[000|\mu -10]$. Since $\mu \ne 10$, this gives a contradiction $0=\text{non-zero}$, so rank$(A)=2\ne \text{rank}([A|b])=3$).

iii. Infinitely Many Solutions: Infinitely many solutions occur if the third row vanishes completely, yielding a free variable ($\text{rank}(A)=\text{rank}([A|b])=2<3$): $\lambda -3=0\text{ and }\mu -10=0⟹\lambda =3\text{ and }\mu =10$ (Note: The bottom row becomes all zeros, so rank$(A)=\text{rank}([A|b])=2$, which is less than the number of variables $3$, yielding $3-2=1$ free variable).

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06.08.26 | Class-08

Eigenvalues and Eigenvectors

Fundamental Definitions

Let $A$ be an $n\times n$ square matrix. A scalar $\lambda$ is called an Eigenvalue of $A$ if there exists a non-zero vector $x\ne 0$ (the corresponding Eigenvector) such that: $Ax=\lambda x$

Physical Interpretation: Multiplying vector $x$ by matrix $A$ scales the vector by factor $\lambda$ without changing its spatial direction.

Homogeneous Formulation: $\lambda x-Ax=0⟹(\lambda I-A)x=0$ Since $x$ must be non-zero ($x\ne 0$) for a valid eigenvector, the homogeneous system must have non-trivial solutions.

Homogeneous Systems and Non-Trivial Solutions

A homogeneous system of linear equations $Ax=0$ is always consistent because the trivial solution ($x=0$) always exists.

• Trivial Solution: $x_1=x_2=\cdots =x_n=0$.
• Non-Trivial Solution: At least one variable is non-zero. A non-trivial solution exists if and only if $\text{rank}(A)<n$, which for square matrices is equivalent to: $\det (A)=0$ Therefore, the homogeneous eigenvalue system $(\lambda I-A)x=0$ has a non-trivial solution if and only if: $\det (\lambda I-A)=0$

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Solved Problems: Homogeneous Systems and Eigenvalues

Solved Problem 1: Solving a Large Homogeneous System

Find the general solution to the following homogeneous system of 5 equations in 4 variables: $\{\begin{array}{l}{x}_1-x_2-x_3-x_4=0\\ x_1+3x_2-x_3+x_4=0\\ 3x_1-7x_2-x_3-6x_4=0\\ 2x_1+2x_2-2x_3=0\\ 6x_1-2x_2-4x_3-5x_4=0\end{array}$

Solution: Construct the augmented coefficient matrix: $[A|0]=\left[\begin{array}{ccccc}1& -1& -1& -1& 0\\ 1& 3& -1& 1& 0\\ 3& -7& -1& -6& 0\\ 2& 2& -2& 0& 0\\ 6& -2& -4& -5& 0\end{array}\right]$

Perform EROs to create zeros in Column 1 below the pivot: $\underset{R_5^{\prime }\to R_5-6R_1}{\overset{R_2^{\prime }\to R_2-R_1,R_3^{\prime }\to R_3-3R_1,R_4^{\prime }\to R_4-2R_1}{\to }}\left[\begin{array}{ccccc}1& -1& -1& -1& 0\\ 0& 4& 0& 2& 0\\ 0& -4& 2& -3& 0\\ 0& 4& 0& 2& 0\\ 0& 4& 2& 1& 0\end{array}\right]$

Eliminate entries in Column 2 below the row 2 pivot: $\underset{R_5^{\prime \prime }\to R_5^{\prime }-R_2^{\prime }}{\overset{R_3^{\prime \prime }\to R_3^{\prime }+R_2^{\prime },R_4^{\prime \prime }\to R_4^{\prime }-R_2^{\prime }}{\to }}\left[\begin{array}{ccccc}1& -1& -1& -1& 0\\ 0& 4& 0& 2& 0\\ 0& 0& 2& -1& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 2& -1& 0\end{array}\right]$

Eliminate the fifth row using row 3: $\stackrel{R_5^{\prime \prime \prime }\to R_5^{\prime \prime }-R_3^{\prime \prime }}{\to }\left[\begin{array}{ccccc}1& -1& -1& -1& 0\\ 0& 4& 0& 2& 0\\ 0& 0& 2& -1& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

Back-Substitution: Reconstruct the system equations: $\{\begin{array}{l}{x}_1-x_2-x_3-x_4=0\\ 4x_2+2x_4=0\\ 2x_3-x_4=0\end{array}$

Let the free variable $x_4=a$ (where $a\in \mathbb{R}$):

1. From the third equation: $2x_3-a=0⟹{\mathbf{x}}_3=\frac{\mathbf{a}}{2}$
2. From the second equation: $4x_2+2a=0⟹4x_2=-2a⟹{\mathbf{x}}_2=-\frac{\mathbf{a}}{2}$
3. From the first equation: $x_1-\left(-\frac{a}{2}\right)-\left(\frac{a}{2}\right)-a=0⟹x_1+\frac{a}{2}-\frac{a}{2}-a=0⟹{\mathbf{x}}_1=\mathbf{a}$

Final Answer: The general solution is $\left(a,-\frac{a}{2},\frac{a}{2},a\right)$ for $a\in \mathbb{R}$.

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Solved Problem 2: Finding Eigenvalues of a $2\times 2$ Matrix

Solve the characteristic equation to find the eigenvalues of the matrix: $A=\left(\begin{array}{cc}1& 2\\ 3& 2\end{array}\right)$

Solution: Formulate the characteristic equation $|\lambda I-A|=0$: $\left|\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)-\left(\begin{array}{cc}1& 2\\ 3& 2\end{array}\right)\right|=0$ $\Rightarrow \left|\begin{array}{cc}\lambda -1& -2\\ -3& \lambda -2\end{array}\right|=0$

Expand the determinant: $(\lambda -1)(\lambda -2)-(-2)(-3)=0$ $\Rightarrow {\lambda }^2-3\lambda +2-6=0$ $\Rightarrow {\lambda }^2-3\lambda -4=0$ $\Rightarrow (\lambda -4)(\lambda +1)=0$ $\therefore \lambda =4,-1$

Final Answer: The characteristic roots (eigenvalues) of $A$ are ${\lambda }_1=4$ and ${\lambda }_2=-1$. (Note: The calculation was left blank or scribbled in the original notes; it has been mathematically solved and completed here).

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Properties of Eigenvalues

Properties of Eigenvalues

For any $n\times n$ matrix $A$ with eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$:

1. Sum of Eigenvalues (Trace): ${\sum }_{i=1}^n{\lambda }_i=\text{tr}(A)=a_{11}+a_{22}+\cdots +a_{nn}$
2. Product of Eigenvalues (Determinant): ${\prod }_{i=1}^n{\lambda }_i=\det (A)$
3. Eigenvalues of the Inverse Matrix ($A^{-1}$): If $A$ is invertible ($\det (A)\ne 0$), then the eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$: $\frac{1}{{\lambda }_1},\frac{1}{{\lambda }_2},\dots ,\frac{1}{{\lambda }_n}$
4. Eigenvalues of a Scaled Matrix ($kA$): The eigenvalues of $kA$ (where $k$ is a scalar) are scaled proportionally: $k{\lambda }_1,k{\lambda }_2,\dots ,k{\lambda }_n$
5. Eigenvalues of Matrix Powers ($A^m$): For any integer exponent $m\ge 1$, the eigenvalues of $A^m$ are: ${\lambda }_1^m,{\lambda }_2^m,\dots ,{\lambda }_n^m$

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Advanced Polynomial & Diagonalization Concepts (Exclusively from PYQ, not taught in class)

Minimal Polynomial (PYQ: 2025)

Minimal Polynomial Definition

The minimal polynomial $m(x)$ of an $n\times n$ matrix $A$ is the unique monic polynomial of lowest degree such that: $m(A)=O$

Key Properties:

• The minimal polynomial $m(x)$ divides any polynomial $p(x)$ for which $p(A)=O$. In particular, it divides the characteristic polynomial of $A$.
• The roots of the minimal polynomial $m(x)$ are exactly the distinct eigenvalues of $A$.

Diagonalization Terms (PYQ: 2023, 2022)

Diagonalization Concepts

A square matrix $A$ of order $n$ is diagonalizable if it is similar to a diagonal matrix $D$, i.e., there exists an invertible matrix $P$ such that $P^{-1}AP=D$.

• Modal Matrix ($P$): The matrix whose columns are the linearly independent eigenvectors of $A$.
• Spectral Matrix ($D$): The diagonal matrix whose diagonal elements are the eigenvalues of $A$ corresponding to the eigenvectors in $P$.
• Spectral Radius ($\rho (A)$): The maximum of the absolute values of the eigenvalues of $A$: $\rho (A)=\max \left\{|{\lambda }_1|,|{\lambda }_2|,\dots ,|{\lambda }_k|\right\}$

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Geometric Interpretation of Eigenvalues

The linear transformation $Ax=\lambda x$ scales the eigenvector $x$ by a factor of $\lambda$. There are four geometric cases depending on the value of $\lambda$:

Case 1: Contraction ($0<\lambda <1$)

The vector scales down in length but maintains its direction.

       λx     x
o----->----->

$\lambda x<x$

Case 2: Dilation ($\lambda >1$)

The vector stretches in length while maintaining its direction.

       x      λx
o----->----->

$\lambda x>x$

Case 3: Negative Contraction ($-1<\lambda <0$)

The vector scales down in length and reverses its direction.

   λx     o     x
<-----o----->

$|\lambda x|<x$

Case 4: Negative Dilation ($\lambda <-1$)

The vector stretches in length and reverses its direction.

     λx          o     x
<----------o----->

$|\lambda x|>x$