Vector Spaces

Notations

• $\mathbb{F}$ can be used instead of $\mathbb{C}$ or $\mathbb{R}$

• $V$ is the vector space of $\mathbb{F}$.

Remark

The letter $\mathbb{F}$ is used because $\mathbb{R}$ and $\mathbb{C}$ are examples of what are called fields.

Lists and length

Definition

Suppose $n$ is a nonnegative integer. A list of length $n$ is an ordered collection of $n$-elements, of some mathematical objects, is separated by commas and surrounded by parentheses. A list of length $n$-looks like this:

$$ (x_1, \cdots , x_n) $$

Definition

Two lists are equal if and only if they have the same length and the same elements in the same order.

Field $\mathbb{F}^n$

Definition

$\mathbb{F}^n$ is the set of all lists of length $n$ of elements of $\mathbb{F}:

\[ \mathbb{F}^n = \{(x_1, \ldots, x_n) : x_j \in \mathbb{F} \text{ for } j = 1, \ldots, n\} \]

For \((x_1, \ldots, x_n) \in \mathbb{F}^n\) and \(j \in \{1, \ldots, n\}\), we say that \(x_j\) is the \(j^{\text{th}}\) coordinate of \((x_1, \ldots, x_n)\)

Addition in $\mathbb{F}$

Definition

Addition in \( \mathbb{F}^n \) is defined by adding corresponding coordinates:

\[ (x_1, \ldots, x_n) + (y_1, \ldots, y_n) = (x_1 + y_1, \ldots, x_n + y_n) \]

Commutativity of addition in \(\mathbb{F}^n\)

If $x,y \in \mathbb{F}^n $, then $x + y = y + x$.

Proof

Suppose \( x = (x_1, \ldots, x_n) \) and \( y = (y_1, \ldots, y_n) \). Then

\[ x + y = (x_1, \ldots, x_n) + (y_1, \ldots, y_n) \] \[ = (x_1 + y_1, \ldots, x_n + y_n) \] \[ = (y_1 + x_1, \ldots, y_n + x_n) \] \[ = (y_1, \ldots, y_n) + (x_1, \ldots, x_n) \] \[ = y + x, \]

where the second and fourth equalities above hold because of the definition of addition in \( \mathbb{F}^n \) and the third equality holds because of the usual commutativity of addition in \( \mathbb{F} \).

$\square$

Zero

Definition

Let 0 denote the list of length $n$-whose coordinates are all 0:

$$ 0 = (0, \ldots, 0) $$

Additive inverse in \( \mathbb{F}^n \)

Definition

For \( x \in \mathbb{F}^n \), the additive inverse of \( x \), denoted \(-x\), is the vector \(-x \in \mathbb{F}^n\) such that

\[x + (-x) = 0\]

Definition scalar multiplication in \( \mathbb{F}^n \)

Definition

The product of a number \( \lambda \) and a vector in \( \mathbb{F}^n \) is computed by multiplying each coordinate of the vector by \( \lambda \):

\[ \lambda(x_1, \ldots, x_n) = (\lambda x_1, \ldots, \lambda x_n) \]

here \( \lambda \in \mathbb{F} \) and \( (x_1, \ldots, x_n) \in \mathbb{F}^n \).

Addition and scalar multiplication

Definition

• An addition on a set \( V \) is a function that assigns an element \( u + v \in V \) to each pair of elements \( u, v \in V \).

• Scalar multiplication on a set \( V \) is a function that assigns an element \( \lambda v \in V \) to each \( \lambda \in F \) and each \( v \in V \).

Vector space

Definition

A vector space is a set \( V \) along with an addition on \( V \) and a scalar multiplication on \( V \) such that the following properties hold:

• Commutativity

  \[u + v = v + u \text{ for all } u, v \in V \]

• Associativity

  $$ (u + v) + w = u + (v + w) $$

  and

  $$ (ab)v = a(bv) \quad \forall u, v, w \in V $$

  and all \( a, b \in \mathbb{F}\)

• Additive identity
  there exists an element \( 0 \in V \) such that \( v + 0 = v \text{ for all } v \in V \)

• additive inverse
  for every \( v \in V \), there exists \( w \in V \) such that \( v + w = 0 \)

• Multiplicative identity

  \[1v = v \quad \forall v \in V \]

• Distributive properties

  $$ a(u + v) = au + av $$

  and

  $$ (a + b)v = av + bv \quad \forall a, b \in \mathbb{F} \ \text{ and } \forall u, v \in V. $$

Points and vectors

Definition

Elements of a vector space are called points or vectors.

Real vector space and complex vector space

Definitions

• A vector space over $\mathbb{R}$ is called a real vector space.

• A vector space over $\mathbb{C}$ is called a complex vector space.

Set of functions

Notation

If \( S \) is a set, then \( \mathbb{F}^S \) denotes the set of functions from \( S \) to \( \mathbb{F} \).

• For \( f, g \in \mathbb{F}^S \), the sum \( f + g \in \mathbb{F}^S \) is the function defined by

  \[ (f + g)(x) = f(x) + g(x) \]

  for all \( x \in S \)

• For \( \lambda \in \mathbb{F} \) and \( f \in \mathbb{F}^S \), the product \( \lambda f \in \mathbb{F}^S \) is the function defined by

  \[ (\lambda f)(x) = \lambda f(x) \]

  for all \( x \in S \).

Unique additive identity

Theorem

A vector space has a unique additive identity.

Proof

Suppose 0 and 0' are both additive identities for some vector space $V$. Then

$\square$

Unique additive inverse

Theorem

Every element in a vector space has a unique additive inverse.

Proof

Suppose \( v \) and \( w \) are both additive inverses of some element \( u \in V \). Then: \( v + u = 0 \) and \( w + u = 0 \)
Adding the additive inverse of \( v + u \) to both sides of the equation above gives \( 0 = w - v \), as desired.

$\square$

Notation

Let $v,W \in V$, then

• $-v$ denotes the additive inverse of $v$

• $w-v$ is defined to be $w + (-w)$

Numbers and vectors

Theorem

$0v = 0$ every $v \in V$.

Proof

$$ 0v = (0+0)v = 0v + 0v $$

Adding the additive inverse of $0v$ to both sides of the equation above gives $0v = 0$, as desired.

$\square$

Theorem

$a0 = 0$ for every $a \in \mathbb{F}$.

Proof

$$a0 = a (0+0) = a0 + a0 $$

Adding the additive inverse of $a0$ to both sides of the equation above gives $0 = a0$, as desired.

$\square$

Theorem

$(-1)v = -v$ for every $v \in V$.

Proof

$$ v + (-1)v = 1v + (-1)v = (1+(-1))v = 0v = 0$$

This equation says that $(-1)v$, when added to $v$, gives 0. Thus $(-1)v$ is the additive inverse of $v$, as desired.

$\square$

Subspaces

Definition

A subset \( W \) of a vector space \( V \) is called a subspace if:

• It contains the zero vector.

• It is closed under vector addition.

• It is closed under scalar multiplication.

Sum of subsets

Definition

Let \( U \) and \( W \) be subsets of a vector space \( V \). The sum of \( U \) and \( W \), denoted \( U + W \), is the set defined by:

\[ U + W = \{ u + w \mid u \in U, w \in W \} \]

Sum of subspaces is the smallest containing subspace

Theorem

Suppose \( U_1, \ldots, U_m \) are subspaces of \( V \). Then \( U_1 + \cdots + U_m \) is the smallest subspace of \( V \) containing \( U_1, \ldots, U_m \).

Proof

Let \( W \) be the smallest subspace of \( V \) containing \( U_1, \ldots, U_m \). We need to show that \( W \) is contained in \( U_1 + \cdots + U_m \).
Since \( W \) is a subspace, it is closed under addition and scalar multiplication. Therefore, if \( w \in W \), then for any \( u_i \in U_i \) and \( \lambda \in \mathbb{F} \), we have \( w + u_i \in W \) and \( \lambda w \in W \).
Now, consider any element \( z \in U_1 + \cdots + U_m \). By definition, \( z \) can be written as \( z = u_1 + \cdots + u_m \) for some \( u_i \in U_i \). Since each \( U_i \) is contained in \( W \), it follows that \( z \in W \).
Thus, we have shown that \( W \subseteq U_1 + \cdots + U_m \), which proves the theorem.

$\square$

Direct Sum

Definition

Suppose \( U_1, \ldots, U_m \) are subspaces of \( V \).

• The sum \( U_1 + \cdots + U_m \) is called a direct sum if each element of \( U_1 + \cdots + U_m \) can be written in only one way as a sum \( u_1 + \cdots + u_m \), where each \( u_j \) is in \( U_j \).

• If \( U_1 + \cdots + U_m \) is a direct sum, then \( U_1 \oplus \cdots \oplus U_m \) denotes \( U_1 + \cdots + U_m \), with the \(\oplus\) notation serving as an indication that this is a direct sum.

Condition for a direct sum

Theorem

Suppose \( U_1, \ldots, U_m \) are subspaces of \( V \). Then \( U_1 + \cdots + U_m \) is a direct sum if and only if the intersection \( U_i \cap U_j = \{0\} \) for all \( i \neq j \).

Proof

Suppose \( U_1, \ldots, U_m \) are subspaces of \( V \). We will show that \( U_1 + \cdots + U_m \) is a direct sum if and only if the intersection \( U_i \cap U_j = \{0\} \) for all \( i \neq j \).
(\( \Rightarrow \)) Assume \( U_1 + \cdots + U_m \) is a direct sum. Let \( v \in U_i \cap U_j \) for some \( i \neq j \). Then we can write \[ v = u_i + \cdots + u_m \] for some \( u_k \in U_k \). However, since \( v \in U_i \) and \( v \in U_j \), this representation is not unique, contradicting the assumption that the sum is direct. Thus, we must have \( v = 0 \), proving that \( U_i \cap U_j = \{0\} \).
(\( \Leftarrow \)) Now assume \( U_i \cap U_j = \{0\} \) for all \( i \neq j \). Let \( v \in U_1 + \cdots + U_m \). Then we can write \[ v = u_1 + \cdots + u_m \] for some \( u_k \in U_k \). Suppose there is another representation \[ v = w_1 + \cdots + w_m \] for some \( w_k \in U_k \). Then we have \[ u_1 + \cdots + u_m = w_1 + \cdots + w_m. \] Rearranging gives \[ (u_1 - w_1) + \cdots + (u_m - w_m) = 0. \] Since the \( U_i \) are pairwise disjoint, it follows that each \( u_k - w_k = 0 \), proving the uniqueness of the representation. Thus, \( U_1 + \cdots + U_m \) is a direct sum.

$\square$

Direct sum of two subspaces

Theorem

Suppose \( U \) and \( W \) are subspaces of \( V \). Then \( U \oplus W \) is a direct sum if and only if \( U \cap W = \{0\} \).

Proof

Suppose \( U \) and \( W \) are subspaces of \( V \). We will show that \( U \oplus W \) is a direct sum if and only if \( U \cap W = \{0\} \).
(\( \Rightarrow \)) Assume \( U \oplus W \) is a direct sum. Let \( v \in U \cap W \). Then we can write \[ v = u + w \] for some \( u \in U \) and \( w \in W \). However, since \( v \in U \) and \( v \in W \), this representation is not unique, contradicting the assumption that the sum is direct. Thus, we must have \( v = 0 \), proving that \( U \cap W = \{0\} \).
(\( \Leftarrow \)) Now assume \( U \cap W = \{0\} \). Let \( v \in U \oplus W \). Then we can write \[ v = u + w \] for some \( u \in U \) and \( w \in W \). Suppose there is another representation \[ v = w_1 + w_2 \] for some \( w_k \in W \). Then we have \[ u + w = w_1 + w_2. \] Rearranging gives \[ (u - w_1) + (w - w_2) = 0. \] Since \( U \cap W = \{0\} \), it follows that each \( u - w_1 = 0 \) and \( w - w_2 = 0 \), proving the uniqueness of the representation. Thus, \( U \oplus W \) is a direct sum.

$\square$