MA353 - Linear Algebra II

# Vectorspaces and Subspaces

vectorspace

A set of vectors V is a vectorspace if:

• closed under addition and scalar multiplication - $$\overrightarrow{v} + \overrightarrow{w} \in V$$, $$c \overrightarrow{v} \in V$$
• an identity is defined - $$1 \cdot \overrightarrow{v} = \overrightarrow{v}$$

## Properties of Vectorspaces

• commutative - $$\overrightarrow{v} + \overrightarrow{w} = \overrightarrow{w} + \overrightarrow{v}$$
• distributive - $$(a + b)\overrightarrow{v} = a\overrightarrow{v} + b \overrightarrow{v}$$ and $$a(\overrightarrow{v} + \overrightarrow{w}) = a \overrightarrow{v} + b \overrightarrow{w}$$
• $$- \overrightarrow{v} + \overrightarrow{v} = 0$$
• $$0 + \overrightarrow{v} = \overrightarrow{v}$$

0 in a vectorspace is unique.

Proof

## Subspace

subspace

Let $$W,V$$ be vectorspaces. If $$W \subseteq V$$, then $$W$$ is a subspace.

Equivalently, $$W$$ is a subspace $$W \subseteq V$$ and:

1. $$0 \in W$$
2. $$\overrightarrow{v} + \overrightarrow{w} \in W$$ for $$\overrightarrow{v},\overrightarrow{w} \in W$$
3. $$a \overrightarrow{v} \in W$$ for $$\overrightarrow{v} \in W, a \in \mathbb{R}$$

The intersection of subspaces is a subspace.

Proof

Sum of vectorspaces is also a vectorspace.

$$W_1 + W_2 \equiv \{w_1 + w_2 | w_1 \in W_1, w_2 \in W_2\}$$

Proof

# Span and Linear Dependence

span

The span of a set $$S$$ is all possible linear combinations of the set.

Let $$S = {v_1, ..., v_n}$$

$$\text{span} (S) = \{c_1 v_1 + ... + c_n v_n | c_1,...,c_n \in \mathbb{R}\}$$

Let $$S$$ be a set, $$W$$ be a vectorspace.

If $$S \subseteq W$$, then $$\text{span} (S) \subseteq W$$

Proof

The span of subspaces $$W_1 \cup W_2 = W_1 + W_2$$.

Proof

linear independence

A set $$S = \{s_1, ..., s_n\}$$ is linearly independent if $$c_1, ..., c_n = 0$$ is the only solution to $$c_1 s_1 + ... + c_n s_n = 0$$

Equivalently, a set is linearly independent if every element cannot be expressed as a combination of the other elements.

1. Is the set $$S = \{1 + x + x^2 + x^3, x + x^2 + x^3, x^2 + x^3, x^3\}$$ linearly independent?

$\begin{eqnarray*} c_1(1 + x + x^2 + x^3) + c_2 ( x + x^2 + x^3) + c_3 ( x^2 + x^3) + c_4 x^3 = 0 \\ c_1 + (c_1 + c_2)x + (c_1 + c_2 + c_3)x^2 + (c_1 + c_2 + c_3 + c_4)x^3 = 0 \\ c_1 = 0 \\ (c_1 + c_2)x = 0 \\ (c_1 + c_2 + c_3)x^2 = 0 \\ (c_1 + c_2 + c_3 + c_4)x^3 = 0 \end{eqnarray*}$

## Subset Dependence

If $$S_1 \subseteq S_2$$ and $$S_1$$ is dependent, then $$S_2$$ is dependent.

Proof

If $$S_1 \subseteq S_2$$ and $$S_2$$ is independent, then $$S_1$$ is also independent.

Proof

# Basis and Dimension

A set $$B$$ is called a basis of vectorspace $$V$$ if

• it is linearly independent and a generating set ($$V \subseteq \text{span} (B)$$)

or

• $$B$$ is the smallest set of $$V$$ such that $$\text{span} (B) = V$$

or

• every vector in $$V$$ can be expressed uniquely from linear combinations of $$B$$

Proof

Let $$L$$ be an independent set of $$V$$ with $$m$$ elements. An indepenent set can be extended to create a generating set

Every basis of a finite vectorspace has the same number of elements.

Proof

For every vectorspace of dimension $$n$$

1. $$\dim(\text{generating set}) \geq n$$

2. $$\dim(\text{independent set}) \leq n$$

Proof

For a vectorspace $$V$$ of dimension $$n$$

1. A generating set with $$n$$ elements is a basis of $$V$$

2. An independent set with $$n$$ elements is a basis of $$V$$

Proof

For a vectorspace $$V$$ of dimension $$n$$

1. A generating set with dimension greater than $$n$$ can be reduced to be a basis.

2. An independent set with dimension less than $$n$$ can be extended to be a basis.

Proof

# Dimension of Subspace

Let $$W$$ be a subspace of $$V$$. If the dimensions of $$W$$ equals $$V$$, then $$V = W$$. If the dimension of $$W$$ is zero, then $$W = \{0\}$$

Corollary

A basis $$B$$ of $$W$$ can be extended to be a basis of $$V$$.

## Properties

### Sum

$$\dim(W_1 + W_2) = \dim (W_1) + \dim (W_2) - \dim (W_1 \cap W_2)$$

1. Line and coincident plane

$$\dim(\text{line + plane}) = 2 = \dim(\text{line}) + \dim(\text{plane}) - \text{dim(line} \cap \text{plane)}$$

where $$\text{dim(plane)} - \text{dim(line} \cap \text{plane)} = 1$$

2. Line and noncoincident plane

$$\dim(\text{line + plane}) = 3 = \dim(\text{line}) + \dim(\text{plane}) - \dim(\text{line} \cap \text{plane)}$$

where $$\dim(\text{plane}) - \dim(\text{line} \cap \text{plane)} = 0$$

### Direct Sum

$$W_1 + W_2 = W_1 \oplus W_2$$ iff $$\text{dim} (W_1 + W_2) = \text{dim} (W_1) + \text{dim} (W_2)$$

Proof

# Linear Transformations

A linear transformation is a mapping $$T:V \rightarrow W$$ from one vectorspace to another such that

• Let $$a,b \in V$$. $$T(a) + T(b) = T(a + b)$$
• scalar multiplication is preserved

• Let $$a \in \mathbb{R}, x \in V$$, $$T(ax) = aT(x)$$

## Properties

• $$T(\overrightarrow{0}) = \overrightarrow{0}$$
• proof: $$T(\overrightarrow{0}) = T(0*\over{v}) = 0T(\overrightarrow{v}) = \overrightarrow{0}$$
• $$T(\overrightarrow{u} - \overrightarrow{v}) = T(\overrightarrow{u}) - T(\overrightarrow{v})$$
• proof: $$T(\overrightarrow{u} - \overrightarrow{v}) = T(\overrightarrow{u}) + T(-1 \overrightarrow{v}) = T(\overrightarrow{u}) - T(\overrightarrow{v})$$
1. For $$T:P_n(\mathbb{R}) \rightarrow P_{n-1}(\mathbb{R})$$ where $$T(f) = f'$$ for $$f \in P_n(\mathbb{R})$$. Show $$T$$ is linear.

Let $$a,b \in \mathbb{R}$$, $$f,g \in P_n(\mathbb{R})$$.

$$T(af + bg) = (af + bg)' = af' + bg' = aT(f) + bT(g)$$

Identity Transformation

A mapping $$T:v \rightarrow W$$ such that $$T(\overrightarrow{v} = \overrightarrow{v}$$ for $$\overrightarrow{v} \in V$$ #enddefinition

#+begindefinition Zero Transformation

A mapping $$T:V \rightarrow W$$ such that $$T(\overrightarrow{v}) = \overrightarrow{0}$$ for $$\overrightarrow{v} in V$$

# Kernel and Image

Kernel

The kernel ($$\ker(T)$$ or $$N(T)$$) of a transformation $$T:V \rightarrow W$$ is the set of vectors in $$V$$ that map to $$\overrightarrow{0}$$ in $$W$$

Image

The image ($$Im(T)$$ or $$R(T)$$) of a transformation $$T:V \rightarrow W$$ is the set of vectors in $$W$$ mapped to by vecotrs in $$V$$.

1. Prove the kernel of $$T$$ is a subspace of $$V$$. (show $$0 \in \ker(T)$$, $$\ker(T)$$ closed for addition, $$\ker(T)$$ closed for scalar multiplication

1. By definition $$T(0) = 0$$, so $$0 \in \ker(T)$$

2. For $$\overrightarrow{u}, \overrightarrow{v} \in \ker(T)$$, we have $$T(\overrightarrow{u}) = T(\overrightarrow{v}) = 0$$

$$T(\overrightarrow{u} + \overrightarrow{v}) = T(\overrightarrow{u}) + T(\overrightarrow{v}) = 0$$

so $$\overrightarrow{u} + \overrightarrow{v} \in \ker(T)$$

3. For $$a \in R$$, $$v \in \ker(T)$$ we have $$T(\overrightarrow{v}) = 0$$

$$T(a\overrightarrow{v}) = aT(\overrightarrow{v}) = 0$$, so $$a \overrightarrow{v} \in \ker(T)$$
2. Let $$T:\mathbb{R}_3 \to \mathbb{R}_2$$, where $T \left( \begin{matrix} a\\b\\c \end{matrix} \right) = \left( \begin{matrix} a-b \\ 0 \\ 2c \end{matrix} \right)$. Find the kernel of $$T$$

$\ker(T) = \left\{ \left( \begin{matrix} a\\b\\c \end{matrix} \right) \middle| T \left( \begin{matrix} a\\b\\c \end{matrix} \right) = 0 \right\}$

$\left( \begin{matrix} a-b \\ 0 \\ 2c \end{matrix} \right) = 0$, so $$a = b$$ and $$c = 0$$

$\ker(T) = \left\{ \left( \begin{matrix} a\\a\\0 \end{matrix} \right) \middle| a \in \mathbb{R} \right\}$

3. Find the image of $$T$$.

$Im(T) = \left\{ T(\overrightarrow{v}) \middle| \overrightarrow{v} \in \mathbb{R}_3 \right\} = \left\{ \left( \begin{matrix} a-b \\ 0 \\ 2c \end{matrix} \right) \middle| a,b,c \in \mathbb{R} \right\} = \left\{ \left( \begin{matrix} a\\ 0 \\ b \end{matrix} \right) \middle| a,b \in \mathbb{R} \right\}$