BSc CSIT (TU) Science Mathematics II (BSc CSIT, MTH163) Question Paper 2078 Nepal

Inner Product Space

Let $V$ be a vector space over the field $\mathbb{R}$ (or $\mathbb{C}$). An inner product on $V$ is a function $\langle \cdot, \cdot \rangle : V \times V \to \mathbb{R}$ that assigns to every pair of vectors $u, v$ a scalar $\langle u, v\rangle$ satisfying, for all $u, v, w \in V$ and scalars $\alpha$:

1. Symmetry: $\langle u, v\rangle = \langle v, u\rangle$ (conjugate-symmetry $\langle u,v\rangle=\overline{\langle v,u\rangle}$ over $\mathbb{C}$).
2. Linearity: $\langle \alpha u + w, v\rangle = \alpha\langle u,v\rangle + \langle w,v\rangle$.
3. Positive definiteness: $\langle u, u\rangle \ge 0$, with $\langle u,u\rangle = 0 \iff u = 0$.

A vector space equipped with such an inner product is called an inner product space. The induced norm is $\|u\| = \sqrt{\langle u, u\rangle}$.

Cauchy–Schwarz Inequality (Statement)

For all $u, v$ in an inner product space $V$,

with equality if and only if $u$ and $v$ are linearly dependent.

Proof

If $v = 0$, both sides are $0$ and the inequality holds. So assume $v \neq 0$.

For any scalar $\lambda \in \mathbb{R}$, positive definiteness gives

Choose $\lambda = \dfrac{\langle u, v\rangle}{\langle v, v\rangle}$ (valid since $\langle v,v\rangle > 0$). Substituting,

Rearranging,

Taking square roots gives $|\langle u,v\rangle| \le \|u\|\,\|v\|$.

Equality condition: Equality holds iff $\langle u - \lambda v, u - \lambda v\rangle = 0$, i.e. $u = \lambda v$, meaning $u$ and $v$ are linearly dependent. $\blacksquare$

Eigenvalues, Eigenvectors and Cayley–Hamilton Verification

Since the paper does not display a specific matrix, take the standard representative

(The same method applies to any given matrix.)

Step 1 — Characteristic equation

Step 2 — Eigenvalues

Step 3 — Eigenvectors

For $\lambda_1 = 1$: $(A - I)x = 0 \Rightarrow \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}x = 0 \Rightarrow x_1 + x_2 = 0$. Eigenvector $v_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

For $\lambda_2 = 3$: $(A - 3I)x = 0 \Rightarrow \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}x = 0 \Rightarrow x_1 = x_2$. Eigenvector $v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

Step 4 — Verify Cayley–Hamilton Theorem

The theorem states every matrix satisfies its own characteristic equation: $A^2 - 4A + 3I = 0$.

Hence $A$ satisfies its characteristic equation, verifying the Cayley–Hamilton theorem. As a bonus, $A^{-1} = \tfrac{1}{3}(4I - A) = \tfrac{1}{3}\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$.

Kernel and Range of a Linear Transformation

Let $T : V \to W$ be a linear transformation between vector spaces.

• Kernel (null space): $\ker(T) = \{ v \in V : T(v) = 0_W \}$ — the set of all vectors mapped to the zero vector. It is a subspace of $V$; its dimension is the nullity, $\text{nullity}(T)$.
• Range (image): $\text{Im}(T) = \{ T(v) : v \in V \} \subseteq W$ — the set of all images. It is a subspace of $W$; its dimension is the rank, $\text{rank}(T)$.

Rank–Nullity Theorem (Statement)

If $V$ is finite-dimensional and $T : V \to W$ is linear, then

Verification with an Example

Let $T : \mathbb{R}^3 \to \mathbb{R}^3$ be defined by

Its standard matrix is

Row reduce: $R_3 \to R_3 - R_1$ gives $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & -1 & -1 \end{pmatrix}$, then $R_3 \to R_3 + R_2$ gives $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}$.

There are 2 pivots, so $\text{rank}(T) = 2$.

Kernel: Solve $Ax = 0$. From the reduced form: $y + z = 0 \Rightarrow y = -z$, and $x + y = 0 \Rightarrow x = -y = z$. So $(x,y,z) = z(1, -1, 1)$, a 1-dimensional space; $\text{nullity}(T) = 1$.

Check: $\text{rank} + \text{nullity} = 2 + 1 = 3 = \dim \mathbb{R}^3.$ ✓

The rank–nullity theorem is verified.

Let $A$ be an $m \times n$ matrix.

• Row space: the subspace of $\mathbb{R}^n$ spanned by the row vectors of $A$, i.e. all linear combinations of the rows. Written $\text{Row}(A)$.
• Column space: the subspace of $\mathbb{R}^m$ spanned by the column vectors of $A$, i.e. all linear combinations of the columns. Written $\text{Col}(A)$; it equals the set of all $b$ for which $Ax = b$ is consistent.

A key fact: $\dim(\text{Row}(A)) = \dim(\text{Col}(A)) = \text{rank}(A)$.

Write the system $x + y = 3,\ 2x - y = 0$ in matrix form $AX = B$:

$\det(A) = (1)(-1) - (1)(2) = -3 \neq 0$, so $A^{-1}$ exists.

Solution: $x = 1,\ y = 2$. (Check: $1+2=3$ ✓, $2(1)-2=0$ ✓.)

The span of a set of vectors $\{v_1, v_2, \dots, v_k\}$ in a vector space $V$ is the set of all possible linear combinations of those vectors:

The span is always a subspace of $V$ — in fact the smallest subspace containing all the $v_i$. If $\text{span}\{v_1,\dots,v_k\} = V$, the set is said to span (generate) $V$. For example, $\text{span}\{(1,0),(0,1)\} = \mathbb{R}^2$.

For any two vectors $u, v$ in an inner product space, the Cauchy–Schwarz inequality states

with equality if and only if $u$ and $v$ are linearly dependent.

In $\mathbb{R}^n$ with the standard dot product this reads

The geometric multiplicity of an eigenvalue $\lambda$ of a matrix $A$ is the dimension of its eigenspace — that is, the number of linearly independent eigenvectors associated with $\lambda$:

It always satisfies $1 \le \text{geometric multiplicity}(\lambda) \le \text{algebraic multiplicity}(\lambda)$ (the multiplicity of $\lambda$ as a root of the characteristic polynomial). A matrix is diagonalizable iff, for every eigenvalue, the geometric multiplicity equals the algebraic multiplicity.

A real symmetric $n \times n$ matrix $A$ is positive definite if

Equivalent characterizations:

• All eigenvalues of $A$ are strictly positive ($\lambda_i > 0$).
• All leading principal minors of $A$ are positive (Sylvester's criterion).

Example: $A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ is positive definite since $x^T A x = 2x_1^2 + 3x_2^2 > 0$ for $x \neq 0$.

The standard basis of $\mathbb{R}^3$ is the set of the three unit coordinate vectors

These vectors are linearly independent and span $\mathbb{R}^3$, since any vector $(x, y, z) = x e_1 + y e_2 + z e_3$. Hence $\{e_1, e_2, e_3\}$ is a basis and $\dim \mathbb{R}^3 = 3$.

A square matrix $A$ is called idempotent if it equals its own square:

Properties: the only possible eigenvalues are $0$ and $1$; idempotent matrices represent projections.

Example: $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ satisfies $A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = A$, so it is idempotent. (The identity $I$ and the zero matrix are also idempotent.)

The norm (Euclidean length) of a vector $v = (v_1, v_2)$ in $\mathbb{R}^2$ is $\|v\| = \sqrt{v_1^2 + v_2^2}$.

For $v = (3, 4)$:

Norm $= 5$.