BSc CSIT (TU) Science Mathematics II (BSc CSIT, MTH163) Question Paper 2080 Nepal

Basis and Dimension

Basis. A set $B = \{v_1, v_2, \dots, v_n\}$ of vectors in a vector space $V$ is called a basis of $V$ if:

1. $B$ is linearly independent, and
2. $B$ spans $V$ (i.e. every vector of $V$ is a linear combination of vectors in $B$).

Dimension. The dimension of a finite-dimensional vector space $V$, written $\dim V$, is the number of vectors in any basis of $V$. (This is well-defined precisely because of the theorem proved below.)

Example: $\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis of $\mathbb{R}^3$, so $\dim \mathbb{R}^3 = 3$.

Theorem: Any two bases have the same number of elements

Statement. If $V$ is a finite-dimensional vector space and $B = \{u_1,\dots,u_m\}$ and $C = \{w_1,\dots,w_n\}$ are both bases of $V$, then $m = n$.

Key Lemma (Replacement / Steinitz Exchange Lemma). In a vector space, if $\{u_1,\dots,u_m\}$ spans $V$ and $\{w_1,\dots,w_n\}$ is linearly independent, then $n \le m$.

Proof of Lemma. Since $\{u_1,\dots,u_m\}$ spans $V$, we can write

As $w_1 \ne 0$ (it belongs to an independent set), some $a_i \ne 0$; relabel so $a_1 \ne 0$. Then $u_1$ can be solved in terms of $w_1, u_2,\dots,u_m$, so $\{w_1, u_2,\dots,u_m\}$ still spans $V$.

Now express $w_2$ in this new spanning set:

Not all of $b_2,\dots,b_m$ are zero (otherwise $w_2$ would be a multiple of $w_1$, contradicting independence of the $w$'s). Relabel so $b_2 \ne 0$ and swap out $u_2$, giving the spanning set $\{w_1, w_2, u_3,\dots,u_m\}$.

Repeating this exchange, each $w_k$ replaces one $u_i$. If $n > m$ we would run out of $u$'s before placing all the $w$'s, and the remaining $w$'s would be expressible in terms of $w_1,\dots,w_m$ — contradicting linear independence of the $w$'s. Hence $n \le m$. $\blacksquare$

Proof of Theorem. Apply the lemma twice:

• $B$ spans $V$ and $C$ is independent $\Rightarrow n \le m$.
• $C$ spans $V$ and $B$ is independent $\Rightarrow m \le n$.

Therefore $m = n$. So any two bases of $V$ contain the same number of elements, and the dimension is well-defined. $\blacksquare$

Matrix of $T$ with respect to standard bases

$T:\mathbb{R}^3 \to \mathbb{R}^2$, $T(x,y,z) = (x+y,\; y-z)$.

Apply $T$ to the standard basis vectors of $\mathbb{R}^3$:

These images form the columns of the matrix:

Check: $[T]\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix} x+y \\ y-z \end{pmatrix}$. ✓

Kernel of $T$

The kernel is the solution set of $T(x,y,z) = (0,0)$:

So $y = z$ and $x = -y$. Letting $z = t$:

Range of $T$

The range is spanned by the columns of $[T]$: $(1,0)$ and $(1,1)$, which are linearly independent. They span all of $\mathbb{R}^2$, so

Rank–Nullity check

Thus $T$ is surjective but not injective.

Gram–Schmidt Orthonormalisation

Given a basis $\{v_1, v_2, v_3\}$ of $\mathbb{R}^3$ with the standard inner product $\langle u,v\rangle = u\cdot v$, the Gram–Schmidt process produces an orthonormal basis $\{e_1,e_2,e_3\}$.

General formulas. First build orthogonal vectors $u_k$, then normalise:

and $e_k = u_k / \lVert u_k\rVert$.

Worked Example

Let $v_1 = (1,1,1)$, $v_2 = (0,1,1)$, $v_3 = (0,0,1)$.

Step 1. $u_1 = (1,1,1)$, $\lVert u_1\rVert = \sqrt{3}$, so

Step 2. $\langle v_2,u_1\rangle = 0+1+1 = 2$, $\langle u_1,u_1\rangle = 3$.

$\lVert u_2\rVert = \sqrt{\tfrac49+\tfrac19+\tfrac19} = \sqrt{\tfrac{6}{9}} = \tfrac{\sqrt6}{3}$, so

Step 3. $\langle v_3,u_1\rangle = 1$, $\langle v_3,u_2\rangle = \tfrac13$, $\langle u_2,u_2\rangle = \tfrac69 = \tfrac23$.

Computing: $-\tfrac13(1,1,1) = (-\tfrac13,-\tfrac13,-\tfrac13)$ and $-\tfrac12(-\tfrac23,\tfrac13,\tfrac13) = (\tfrac13,-\tfrac16,-\tfrac16)$. Adding to $(0,0,1)$:

$\lVert u_3\rVert = \sqrt{0+\tfrac14+\tfrac14} = \tfrac{1}{\sqrt2}$, so

Orthonormal Basis

One can verify $e_i\cdot e_j = \delta_{ij}$ (each pair is orthogonal and each vector is a unit vector).

Basis. A subset $B$ of a vector space $V$ is a basis if it is linearly independent and spans $V$. Every vector of $V$ is then a unique linear combination of the basis vectors.

Example: $\{(1,0),(0,1)\}$ is the standard basis of $\mathbb{R}^2$.

Dimension. The dimension of $V$, $\dim V$, is the number of vectors in any basis of $V$ (this number is the same for every basis).

Example: $\dim \mathbb{R}^2 = 2$, $\dim \mathbb{R}^3 = 3$, and the space $P_2$ of polynomials of degree $\le 2$ has basis $\{1,x,x^2\}$, so $\dim P_2 = 3$.

The $3\times 3$ identity matrix is

It is already in row-echelon form with three non-zero rows (three pivots), and its three rows/columns are linearly independent. Equivalently $\det(I_3) = 1 \ne 0$, so it is non-singular.

In general, $\operatorname{rank}(I_n) = n$.

A linear operator is a linear transformation $T : V \to V$ from a vector space $V$ into itself (the domain and codomain are the same space). That is, $T$ satisfies, for all $u,v \in V$ and all scalars $c$:

1. Additivity: $T(u+v) = T(u) + T(v)$,
2. Homogeneity: $T(cv) = c\,T(v)$.

Equivalently, $T(au+bv) = aT(u)+bT(v)$ for all scalars $a,b$.

Example: The rotation of $\mathbb{R}^2$ by an angle $\theta$, $T(x,y) = (x\cos\theta - y\sin\theta,\; x\sin\theta + y\cos\theta)$, is a linear operator on $\mathbb{R}^2$.

Spectral Theorem (for real symmetric matrices)

Let $A$ be a real $n \times n$ symmetric matrix ($A^{T} = A$). Then:

1. All eigenvalues of $A$ are real.
2. $A$ is orthogonally diagonalizable: there exists an orthogonal matrix $Q$ (with $Q^{T}Q = I$, i.e. $Q^{-1} = Q^{T}$) and a diagonal matrix $D = \operatorname{diag}(\lambda_1,\dots,\lambda_n)$ of eigenvalues such that

3. Eigenvectors corresponding to distinct eigenvalues are orthogonal, and $\mathbb{R}^n$ has an orthonormal basis consisting of eigenvectors of $A$.

Equivalently, $A$ admits the spectral decomposition $A = \sum_{i=1}^{n} \lambda_i\, q_i q_i^{T}$, where $\{q_i\}$ are orthonormal eigenvectors.

An orthogonal projection of a vector $v$ onto a subspace $W$ of an inner product space is the unique vector $\operatorname{proj}_W v \in W$ such that the difference $v - \operatorname{proj}_W v$ is orthogonal to every vector in $W$. It is the point of $W$ closest to $v$ (the best approximation).

Onto a single vector / line. The projection of $v$ onto a nonzero vector $u$ is

Onto a subspace with orthonormal basis $\{e_1,\dots,e_k\}$ of $W$:

The associated projection matrix $P$ satisfies $P^2 = P$ and $P^{T} = P$ (idempotent and symmetric).

The matrix is $A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I$.

Eigenvalues. $\det(A - \lambda I) = (3-\lambda)^2 = 0 \Rightarrow \lambda = 3$ (with algebraic multiplicity $2$).

Eigenvectors. Solve $(A - 3I)x = 0$:

so every nonzero vector $x \in \mathbb{R}^2$ satisfies the equation. Hence every nonzero vector is an eigenvector with eigenvalue $3$.

A convenient basis of the eigenspace is

so the eigenspace for $\lambda = 3$ is all of $\mathbb{R}^2$. (This is expected, since $3I$ acts as scalar multiplication by $3$.)

Let $B = \{b_1, b_2, \dots, b_n\}$ be an ordered basis of a vector space $V$. Since $B$ is a basis, every $v \in V$ can be written uniquely as

The scalars $c_1,\dots,c_n$ are called the coordinates of $v$ relative to $B$, and the column vector

is the coordinate vector of $v$ relative to the basis $B$.

Example: In $\mathbb{R}^2$ with $B = \{(1,1),(1,-1)\}$, the vector $v=(3,1)$ equals $2(1,1)+1(1,-1)$, so $[v]_B = \begin{pmatrix}2\\1\end{pmatrix}$.

A square matrix $A$ of order $n$ is diagonalizable if it is similar to a diagonal matrix — that is, there exists an invertible matrix $P$ and a diagonal matrix $D$ such that

Condition. $A$ is diagonalizable iff it has $n$ linearly independent eigenvectors; these form the columns of $P$, and the corresponding eigenvalues form the diagonal entries of $D$. (In particular, an $n\times n$ matrix with $n$ distinct eigenvalues is always diagonalizable.)

Use: powers become easy, $A^k = P D^k P^{-1}$, since $D^k$ is just the diagonal entries raised to the $k$-th power.

Let $A$ be an $n \times n$ matrix with eigenvalues $\lambda_1,\dots,\lambda_n$ (counted with multiplicity). Two key properties are:

1. Sum of eigenvalues = trace.

2. Product of eigenvalues = determinant.

(Other acceptable properties:) the eigenvalues of $A^{T}$ equal those of $A$; the eigenvalues of $A^{k}$ are $\lambda_i^{k}$; a real symmetric matrix has only real eigenvalues; and $A$ is singular iff $0$ is an eigenvalue.