BSc CSIT (TU) Science Mathematics II (BSc CSIT, MTH163) Question Paper 2077 Nepal

Gauss-Jordan Method and Consistency

Method

The Gauss-Jordan method reduces the augmented matrix $[A \mid b]$ to reduced row echelon form (RREF) using elementary row operations, so the solution can be read off directly.

Worked Example

Solve

Augmented matrix:

Step 1. $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - R_1$:

Step 2. $R_2 \to -\tfrac{1}{3}R_2$, then $R_1 \to R_1 - R_2$, $R_3 \to R_3 - R_2$:

Step 3. $R_3 \to -\tfrac{3}{7}R_3 \Rightarrow z = 3$. Back-eliminate column 3: $R_1 \to R_1 - \tfrac{2}{3}R_3$, $R_2 \to R_2 - \tfrac{1}{3}R_3$:

Solution: $x = 1,\ y = 2,\ z = 3$.

Consistency via Rank

Let $r(A)$ be the rank of the coefficient matrix, $r([A\mid b])$ the rank of the augmented matrix, and $n$ the number of unknowns.

• If $r(A) \neq r([A\mid b])$ — the system is inconsistent (no solution).
• If $r(A) = r([A\mid b]) = n$ — consistent with a unique solution.
• If $r(A) = r([A\mid b]) < n$ — consistent with infinitely many solutions (with $n - r$ free parameters).

Here $r(A) = r([A\mid b]) = 3 = n$, so the system is consistent with a unique solution.

Orthogonal and Orthonormal Sets; Gram-Schmidt

Definitions

Let $V$ be an inner product space with inner product $\langle\,,\rangle$.

• A set $\{v_1,\dots,v_k\}$ of non-zero vectors is orthogonal if $\langle v_i, v_j\rangle = 0$ for all $i \neq j$.
• It is orthonormal if it is orthogonal and each vector is a unit vector: $\langle v_i, v_j\rangle = \delta_{ij}$ (i.e. $\|v_i\| = 1$).

Gram-Schmidt Process

Given linearly independent $\{u_1,\dots,u_k\}$, construct an orthogonal set $\{w_1,\dots,w_k\}$:

Normalize: $e_i = \dfrac{w_i}{\|w_i\|}$ to get an orthonormal set.

Worked Example

Let $u_1 = (1,1,0),\ u_2 = (1,0,1),\ u_3 = (0,1,1)$ in $\mathbb{R}^3$.

$w_1 = (1,1,0)$, $\langle w_1,w_1\rangle = 2$.

$w_2$: $\langle u_2, w_1\rangle = 1$, so

$\langle w_2, w_2\rangle = \tfrac14 + \tfrac14 + 1 = \tfrac32.$

$w_3$: $\langle u_3, w_1\rangle = 1$, $\langle u_3, w_2\rangle = -\tfrac12 + 1 = \tfrac12$, so

Orthonormal set (normalizing):

Diagonalization of a Matrix

Conditions for Diagonalizability

An $n\times n$ matrix $A$ is diagonalizable if there exists an invertible $P$ with $P^{-1}AP = D$ diagonal. This holds iff:

• $A$ has $n$ linearly independent eigenvectors, equivalently
• For every eigenvalue, its geometric multiplicity equals its algebraic multiplicity.

(Sufficient condition: if $A$ has $n$ distinct eigenvalues, it is diagonalizable.) The columns of $P$ are the eigenvectors and $D$ holds the corresponding eigenvalues.

Worked Example

Let $A = \begin{bmatrix} 4 & 1\\ 2 & 3 \end{bmatrix}$.

Eigenvalues: $\det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda-5)(\lambda-2)$. So $\lambda_1 = 5,\ \lambda_2 = 2$ (distinct $\Rightarrow$ diagonalizable).

Eigenvector for $\lambda = 5$: $(A-5I)v=0 \Rightarrow \begin{bmatrix}-1 & 1\\ 2 & -2\end{bmatrix}v = 0 \Rightarrow v_1 = (1,1)^T.$

Eigenvector for $\lambda = 2$: $(A-2I)v=0 \Rightarrow \begin{bmatrix}2 & 1\\ 2 & 1\end{bmatrix}v = 0 \Rightarrow v_2 = (1,-2)^T.$

Result:

Verification: $\det P = -3 \neq 0$, so $P$ is invertible and the diagonalization is valid.

A set of vectors $\{v_1, v_2, \dots, v_n\}$ is linearly dependent if there exist scalars $c_1, c_2, \dots, c_n$, not all zero, such that

This means at least one vector can be written as a linear combination of the others.

Example: In $\mathbb{R}^2$, the vectors $v_1 = (1,2)$ and $v_2 = (2,4)$ are linearly dependent because $2v_1 - v_2 = 0$ (here $c_1 = 2,\ c_2 = -1$), and $v_2 = 2v_1$.

An augmented matrix of a linear system $Ax = b$ is the matrix $[A \mid b]$ formed by appending the column of constants $b$ to the coefficient matrix $A$.

For the system

the augmented matrix is

It is used to solve the system by row reduction (Gaussian / Gauss-Jordan elimination).

The dimension of a vector space $V$ is the number of vectors in any basis of $V$ — that is, the maximum number of linearly independent vectors in $V$. It is denoted $\dim(V)$.

All bases of a given vector space contain the same number of vectors, so the dimension is well-defined.

Examples: $\dim(\mathbb{R}^n) = n$; the space of $2\times 2$ matrices has dimension $4$; $\dim(\{0\}) = 0$.

An $n\times n$ matrix $A$ is diagonalizable if and only if any of the following equivalent conditions holds:

1. $A$ has $n$ linearly independent eigenvectors.
2. For every eigenvalue, its geometric multiplicity equals its algebraic multiplicity.
3. The sum of the dimensions of the eigenspaces equals $n$.

Sufficient (but not necessary) condition: if $A$ has $n$ distinct eigenvalues, then it is diagonalizable. (Real symmetric matrices are always diagonalizable.)

An inner product space is a vector space $V$ over $\mathbb{R}$ (or $\mathbb{C}$) equipped with an inner product $\langle\,,\rangle : V \times V \to \mathbb{R}$ (or $\mathbb{C}$) satisfying, for all $u,v,w \in V$ and scalars $\alpha$:

1. Linearity: $\langle \alpha u + v, w\rangle = \alpha\langle u,w\rangle + \langle v,w\rangle$.
2. Symmetry (conjugate symmetry): $\langle u,v\rangle = \overline{\langle v,u\rangle}$ (just $\langle u,v\rangle = \langle v,u\rangle$ in the real case).
3. Positive-definiteness: $\langle v,v\rangle \ge 0$, with equality iff $v = 0$.

Example: $\mathbb{R}^n$ with the dot product $\langle x,y\rangle = \sum_i x_i y_i$.

For $A = \begin{bmatrix} 4 & 1\\ 2 & 3 \end{bmatrix}$, solve the characteristic equation $\det(A - \lambda I) = 0$:

Factoring: $(\lambda - 5)(\lambda - 2) = 0$.

Eigenvalues: $\lambda_1 = 5,\quad \lambda_2 = 2.$

(Check: trace $= 4+3 = 7 = 5+2$; determinant $= 12-2 = 10 = 5\times 2$.)

The kernel (or null space) of a linear transformation $T: V \to W$ is the set of all vectors in $V$ that map to the zero vector in $W$:

The kernel is a subspace of the domain $V$. Its dimension is called the nullity of $T$. The transformation $T$ is injective (one-to-one) if and only if $\ker(T) = \{0\}$.

By the rank-nullity theorem, $\dim(\ker T) + \dim(\operatorname{Im} T) = \dim V$.

A square complex matrix $A$ is Hermitian if it is equal to its own conjugate transpose:

The diagonal entries of a Hermitian matrix are real. Its eigenvalues are always real, and eigenvectors for distinct eigenvalues are orthogonal. (A real Hermitian matrix is just a symmetric matrix.)

Example: $\begin{bmatrix} 2 & 3-i\\ 3+i & 5 \end{bmatrix}$ is Hermitian.

Cramer's Rule gives the solution of a system $Ax = b$ of $n$ linear equations in $n$ unknowns, provided $\det(A) \neq 0$.

The solution is

where $A_i$ is the matrix obtained from $A$ by replacing its $i$-th column with the constant vector $b$.

If $\det(A) = 0$, the rule does not apply (the system has either no solution or infinitely many).

Example (2×2): for $\begin{bmatrix} a & b\\ c & d \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} e\\ f \end{bmatrix}$,