BSc CSIT (TU) Science Mathematics II (BSc CSIT, MTH163) Question Paper 2079 Nepal

Subspace of a Vector Space

Definition. Let $V$ be a vector space over a field $F$. A non-empty subset $W \subseteq V$ is called a subspace of $V$ if $W$ is itself a vector space under the same operations of $V$. Equivalently, $W$ is a subspace iff:

1. $\mathbf{0} \in W$ (contains the zero vector),
2. $u, v \in W \Rightarrow u + v \in W$ (closed under addition),
3. $\alpha \in F,\ u \in W \Rightarrow \alpha u \in W$ (closed under scalar multiplication).

These reduce to the single test: for all $u,v \in W$ and $\alpha,\beta \in F$, $\ \alpha u + \beta v \in W$.

Intersection of Two Subspaces is a Subspace

Let $W_1$ and $W_2$ be subspaces of $V$. Consider $W = W_1 \cap W_2$.

• Non-empty: $\mathbf{0} \in W_1$ and $\mathbf{0} \in W_2$, so $\mathbf{0} \in W_1 \cap W_2$.
• Closure: Take $u, v \in W_1 \cap W_2$ and scalars $\alpha, \beta \in F$. Then $u,v \in W_1$, so $\alpha u + \beta v \in W_1$ (since $W_1$ is a subspace). Similarly $u,v \in W_2$, so $\alpha u + \beta v \in W_2$. Hence $\alpha u + \beta v \in W_1 \cap W_2$.

Therefore $W_1 \cap W_2$ satisfies the subspace criterion and is a subspace of $V$. $\blacksquare$

Union Need Not Be a Subspace

The union $W_1 \cup W_2$ generally fails closure under addition, because a sum of a vector from $W_1$ and one from $W_2$ may lie in neither.

Counterexample

Take $V = \mathbb{R}^2$ and the two subspaces (the coordinate axes):

Both are subspaces. Now $(1,0) \in W_1 \subseteq W_1 \cup W_2$ and $(0,1) \in W_2 \subseteq W_1 \cup W_2$, but

Since $(1,1) \notin W_1$ and $(1,1) \notin W_2$, we have $(1,1) \notin W_1 \cup W_2$. Closure under addition fails, so $W_1 \cup W_2$ is not a subspace.

(In fact, $W_1 \cup W_2$ is a subspace iff $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$.)

Reducing a Quadratic Form to Canonical Form by Orthogonal Transformation

Method (general procedure). A real quadratic form in $n$ variables can be written as $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$, where $A$ is a real symmetric matrix. Since $A$ is symmetric it is orthogonally diagonalizable: there exists an orthogonal matrix $P$ ($P^TP = I$) of normalized eigenvectors such that $P^T A P = D = \operatorname{diag}(\lambda_1,\dots,\lambda_n)$. The orthogonal substitution $\mathbf{x} = P\mathbf{y}$ gives the canonical form

Worked Example

Reduce $Q = 3x_1^2 + 3x_2^2 + 4x_1 x_2$ to canonical form.

Step 1 — Matrix. $A = \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix}$ (off-diagonal entries are half the coefficient of $x_1x_2$).

Step 2 — Eigenvalues. $\det(A-\lambda I) = (3-\lambda)^2 - 4 = \lambda^2 - 6\lambda + 5 = 0 \Rightarrow \lambda = 5,\ 1.$

Step 3 — Orthonormal eigenvectors.

• $\lambda = 5$: $(A-5I)v=0 \Rightarrow v = \tfrac{1}{\sqrt2}(1,1)^T$.
• $\lambda = 1$: $(A-I)v=0 \Rightarrow v = \tfrac{1}{\sqrt2}(1,-1)^T$.

Step 4 — Orthogonal matrix. $P = \dfrac{1}{\sqrt2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, with $P^TP=I$.

Step 5 — Canonical form. With $\mathbf{x}=P\mathbf{y}$,

Nature of the Quadratic Form

The nature is decided by the signs of the eigenvalues (equivalently the canonical coefficients):

Here $\lambda_1 = 5 > 0$ and $\lambda_2 = 1 > 0$, so $Q$ is positive definite.

Solving a Linear System by Matrix Inversion and by Cramer's Rule

Consider the representative system

written as $A\mathbf{x} = \mathbf{b}$ with

Determinant.

so a unique solution exists.

Method 1 — Matrix Inversion ($\mathbf{x} = A^{-1}\mathbf{b}$)

Cofactor / adjoint computation gives

Then

Method 2 — Cramer's Rule

Replace each column of $A$ by $\mathbf{b}$:

Hence

Both methods give $(x,y,z) = (1,2,3)$.

Comparison of the Two Methods

Both are practical only for small systems with non-singular $A$; for large systems Gaussian elimination / LU decomposition is preferred.

Orthogonal Matrix

Definition. A real square matrix $A$ is orthogonal if its transpose equals its inverse, i.e.

Equivalently, the rows (and columns) of $A$ form an orthonormal set.

Properties.

1. $\det A = \pm 1$.
2. $A^{-1} = A^T$, so the inverse is trivially obtained.
3. The product of two orthogonal matrices is orthogonal; the inverse/transpose of an orthogonal matrix is orthogonal (they form a group $O(n)$).
4. Orthogonal transformations preserve lengths and inner products: $\|A\mathbf{x}\| = \|\mathbf{x}\|$ and $(A\mathbf{x})\cdot(A\mathbf{y}) = \mathbf{x}\cdot\mathbf{y}$.
5. Eigenvalues have absolute value $1$ (lie on the unit circle).

Example. $A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ is orthogonal with $\det A = 1$.

Consistency of a System of Equations

A system of linear equations $A\mathbf{x} = \mathbf{b}$ is said to be consistent if it has at least one solution; otherwise it is inconsistent (no solution).

Test (Rouché–Capelli theorem). Let $[A\,|\,\mathbf{b}]$ be the augmented matrix. Then:

• Consistent iff $\operatorname{rank}(A) = \operatorname{rank}([A\,|\,\mathbf{b}])$.
  • If this common rank $= n$ (number of unknowns) $\Rightarrow$ unique solution.
  • If this common rank $< n \Rightarrow$ infinitely many solutions.
• Inconsistent iff $\operatorname{rank}(A) < \operatorname{rank}([A\,|\,\mathbf{b}])$.

Example. $x+y=2,\ x+y=5$ is inconsistent because $\operatorname{rank}(A)=1$ but $\operatorname{rank}([A|\mathbf b])=2$.

Range of a Linear Transformation

Let $T : V \to W$ be a linear transformation between vector spaces $V$ and $W$. The range (or image) of $T$ is the set of all images of vectors of $V$:

Key facts.

• $\operatorname{Range}(T)$ is a subspace of $W$.
• Its dimension is called the rank of $T$: $\operatorname{rank}(T) = \dim(\operatorname{Range}(T))$.
• Rank–nullity theorem: $\dim V = \operatorname{rank}(T) + \operatorname{nullity}(T)$.

Example. For $T:\mathbb{R}^2\to\mathbb{R}^2$, $T(x,y)=(x,0)$, the range is the $x$-axis $\{(x,0):x\in\mathbb R\}$, of dimension $1$.

Gram–Schmidt Orthogonalization Process

Given a linearly independent set $\{v_1, v_2, \dots, v_n\}$ in an inner product space, the process constructs an orthogonal set $\{u_1,\dots,u_n\}$ (and, on normalizing, an orthonormal set $\{e_1,\dots,e_n\}$) spanning the same subspace.

Steps. Define successively, subtracting projections onto the previously found vectors:

and in general

Normalization. $\displaystyle e_k = \frac{u_k}{\|u_k\|}$ gives an orthonormal basis.

The resulting $\{u_1,\dots,u_k\}$ spans the same subspace as $\{v_1,\dots,v_k\}$ for every $k$.

Quadratic Form

Definition. A quadratic form in $n$ variables $x_1,\dots,x_n$ over a field is a homogeneous polynomial of degree $2$:

which can be written compactly as

where $A = [a_{ij}]$ is a symmetric matrix (the matrix of the quadratic form). The off-diagonal entry $a_{ij}$ is taken as half the coefficient of $x_i x_j$.

Example. $Q(x,y) = 2x^2 + 3y^2 + 4xy$ has matrix

Eigenvalues of the Identity Matrix $I_3$

Let $I_3 = \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\end{pmatrix}$. The characteristic equation is

Hence the only eigenvalue is

Indeed $I_3\mathbf{x} = \mathbf{x} = 1\cdot\mathbf{x}$ for every vector $\mathbf{x}$, so every non-zero vector is an eigenvector with eigenvalue $1$.

Nullity of a Matrix

Definition. The nullity of an $m\times n$ matrix $A$ is the dimension of its null space (kernel) — the set of all solutions $\mathbf{x}$ of the homogeneous system $A\mathbf{x} = \mathbf{0}$:

Rank–Nullity Theorem. For an $m\times n$ matrix,

Thus $\operatorname{nullity}(A) = n - \operatorname{rank}(A)$, equal to the number of free variables in $A\mathbf x=\mathbf 0$.

Example. If $A$ is $3\times3$ with $\operatorname{rank}(A)=2$, then $\operatorname{nullity}(A)=3-2=1$.

Unitary Matrix

Definition. A complex square matrix $U$ is unitary if its conjugate transpose (Hermitian adjoint) equals its inverse:

where $U^{*} = \overline{U}^{\,T}$ is the conjugate transpose. Equivalently, the columns (and rows) of $U$ form an orthonormal set under the complex inner product.

Properties.

• $|\det U| = 1$.
• Unitary transformations preserve the complex inner product and norm: $\langle U\mathbf{x}, U\mathbf{y}\rangle = \langle \mathbf{x},\mathbf{y}\rangle$.
• Eigenvalues lie on the unit circle ($|\lambda| = 1$).
• A real unitary matrix is exactly an orthogonal matrix.

Example. $U = \dfrac{1}{\sqrt2}\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}$ satisfies $U^*U = I$.

Do $(1,1,0),(0,1,1),(1,0,1)$ span $\mathbb{R}^3$?

Three vectors span $\mathbb{R}^3$ iff they are linearly independent, i.e. iff the determinant of the matrix formed by them is non-zero.

Form the matrix with these vectors as rows:

Compute the determinant (expand along the first row):

Since $\det M = 2 \neq 0$, the three vectors are linearly independent. Being $3$ independent vectors in the $3$-dimensional space $\mathbb{R}^3$, they form a basis and therefore span $\mathbb{R}^3$.

Conclusion: Yes, $\{(1,1,0),(0,1,1),(1,0,1)\}$ spans $\mathbb{R}^3$.