BSc CSIT (TU) Science Mathematics II (BSc CSIT, MTH163) Question Paper 2081 Nepal

Consistency of a Linear System by the Rank Method

Consider $AX = B$, where $A$ is the $m \times n$ coefficient matrix, $X$ the column of $n$ unknowns, $B$ the column of constants, and $[A\,|\,B]$ the augmented matrix. Let $r = \operatorname{rank}(A)$, $r' = \operatorname{rank}([A\,|\,B])$, and $n$ the number of unknowns.

(i) Unique solution

Condition: $\operatorname{rank}(A) = \operatorname{rank}([A\,|\,B]) = n$ (the system is consistent and the rank equals the number of unknowns).

Example: $x + y = 3,\ x - y = 1$.

Here $r = r' = 2 = n$, so there is a unique solution $x = 2,\ y = 1$.

(ii) Infinitely many solutions

Condition: $\operatorname{rank}(A) = \operatorname{rank}([A\,|\,B]) = r < n$ (consistent, but rank less than the number of unknowns; $n - r$ free parameters).

Example: $x + y + z = 2,\ 2x + 2y + 2z = 4$.

Here $r = r' = 1 < n = 3$, so there are infinitely many solutions (two free parameters).

(iii) No solution

Condition: $\operatorname{rank}(A) \neq \operatorname{rank}([A\,|\,B])$, i.e. $\operatorname{rank}(A) < \operatorname{rank}([A\,|\,B])$ (the system is inconsistent).

Example: $x + y = 1,\ x + y = 3$.

Here $\operatorname{rank}(A) = 1$ but $\operatorname{rank}([A\,|\,B]) = 2$, so the system has no solution.

Summary: Consistency requires $r = r'$; then $r = n$ gives a unique solution and $r < n$ gives infinitely many, while $r \neq r'$ gives no solution.

Symmetric and Skew-Symmetric Matrices

Symmetric matrix: A square matrix $A$ is symmetric if $A^T = A$, i.e. $a_{ij} = a_{ji}$ for all $i, j$.

Skew-symmetric matrix: A square matrix $A$ is skew-symmetric if $A^T = -A$, i.e. $a_{ij} = -a_{ji}$ for all $i, j$. In particular every diagonal entry satisfies $a_{ii} = -a_{ii}$, so $a_{ii} = 0$.

Every Square Matrix = Symmetric + Skew-Symmetric (uniquely)

Let $A$ be any $n \times n$ matrix. Define

Existence. Clearly

$P$ is symmetric because

$Q$ is skew-symmetric because

Thus $A = P + Q$ is a sum of a symmetric and a skew-symmetric matrix.

Uniqueness. Suppose $A = S + K$ with $S^T = S$ (symmetric) and $K^T = -K$ (skew-symmetric). Then

Adding and subtracting,

Hence the decomposition is unique, $S = P$ and $K = Q$. $\blacksquare$

Example: For $A = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$, $P = \begin{bmatrix} 1 & 3 \\ 3 & 3 \end{bmatrix}$ (symmetric) and $Q = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ (skew-symmetric), and $P + Q = A$.

Eigenvalues, Eigenvectors, Diagonalization and $A^4$

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

The method below applies to any diagonalizable matrix.

Step 1: Characteristic equation

So $(\lambda - 1)(\lambda - 3) = 0$, giving eigenvalues $\lambda_1 = 1,\ \lambda_2 = 3$.

Step 2: Eigenvectors

For $\lambda_1 = 1$: $(A - I)X = 0 \Rightarrow \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}X = 0 \Rightarrow x_1 + x_2 = 0$, eigenvector $v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$.

For $\lambda_2 = 3$: $(A - 3I)X = 0 \Rightarrow \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}X = 0 \Rightarrow x_1 = x_2$, eigenvector $v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

Step 3: Diagonalization

Let

Then $A = PDP^{-1}$ (equivalently $P^{-1}AP = D$).

Step 4: Compute $A^4$

Because $A^4 = P D^4 P^{-1}$ and $D^4 = \operatorname{diag}(1^4, 3^4) = \operatorname{diag}(1, 81)$,

Result: eigenvalues $1, 3$; eigenvectors $(1,-1)^T,\ (1,1)^T$; $A = PDP^{-1}$; and $A^4 = \begin{bmatrix} 41 & 40 \\ 40 & 41 \end{bmatrix}$.

A square matrix $A$ is skew-symmetric if $A^T = -A$, equivalently $a_{ij} = -a_{ji}$ for all $i, j$. This forces every diagonal entry to be zero ($a_{ii} = 0$).

Example:

Expanding along the first row,

Determinant $= 8$.

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

Image (range): the set of all output vectors,

It is a subspace of $W$; its dimension is the rank of $T$.

Kernel (null space): the set of all inputs mapped to the zero vector,

It is a subspace of $V$; its dimension is the nullity of $T$.

They are linked by the rank–nullity theorem: $\dim V = \dim \ker(T) + \dim \operatorname{Im}(T)$.

The algebraic multiplicity of an eigenvalue $\lambda$ of a matrix $A$ is the number of times $\lambda$ occurs as a root of the characteristic polynomial $\det(A - \lambda I) = 0$, i.e. the multiplicity of the factor $(\lambda - \lambda_0)$ in that polynomial.

For example, if $\det(A - \lambda I) = (\lambda - 2)^3(\lambda - 5)$, then the eigenvalue $\lambda = 2$ has algebraic multiplicity $3$. It is always greater than or equal to the geometric multiplicity (the dimension of the corresponding eigenspace).

An inner product on $\mathbb{R}^n$ is a function $\langle \cdot, \cdot \rangle : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ that assigns a real number to each pair of vectors and satisfies, for all $u, v, w \in \mathbb{R}^n$ and scalar $c$:

1. Symmetry: $\langle u, v \rangle = \langle v, u \rangle$.
2. Linearity: $\langle cu + w, v \rangle = c\langle u, v \rangle + \langle w, v \rangle$.
3. Positive-definiteness: $\langle u, u \rangle \geq 0$, with equality iff $u = 0$.

The standard (dot) inner product is

The vectors are linearly dependent.

Note that $(2,4) = 2\,(1,2)$, so one is a scalar multiple of the other. Equivalently, the equation $c_1(1,2) + c_2(2,4) = (0,0)$ has the non-trivial solution $c_1 = 2,\ c_2 = -1$. The determinant test also confirms this:

Since the determinant is $0$, the vectors are not linearly independent.

A vector space $V$ is said to be finite-dimensional if it has a finite spanning set, i.e. there exist finitely many vectors $v_1, v_2, \dots, v_n \in V$ such that every vector in $V$ can be written as a linear combination $a_1 v_1 + a_2 v_2 + \cdots + a_n v_n$.

Equivalently, $V$ is finite-dimensional if it possesses a basis with a finite number of elements; that number is the dimension $\dim V$. For example, $\mathbb{R}^n$ is finite-dimensional with $\dim \mathbb{R}^n = n$. A space with no finite basis (e.g. the space of all polynomials) is infinite-dimensional.

The canonical form of a quadratic form $Q(x) = x^T A x$ is the expression obtained, after a suitable change of variables, in which only squared terms appear and all cross-product terms are eliminated:

This is achieved by an orthogonal transformation $x = Py$ (with $P$ the matrix of orthonormal eigenvectors of the symmetric matrix $A$), where the coefficients $\lambda_i$ are the eigenvalues of $A$. The number of positive, negative and zero coefficients gives the signature and rank of the form (which are invariant by Sylvester's law of inertia).

For $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$:

Sum of eigenvalues $=$ trace$(A) = 1 + 4 = 5$.

Product of eigenvalues $=$ $\det(A) = (1)(4) - (2)(3) = 4 - 6 = -2$.

Thus $\lambda_1 + \lambda_2 = 5$ and $\lambda_1 \lambda_2 = -2$ (the eigenvalues themselves are $\tfrac{5 \pm \sqrt{33}}{2}$).