Chapter 2 Finite-Dimensional Vector Spaces

2A Span and Linear Independence

2.2 definition: linear combination

A linear combination of a list $𝑣_1, …, 𝑣_𝑚$ of vectors in $𝑉$ is a vector of the form

$$𝑎_1 𝑣_1 +⋯ + 𝑎_𝑚 𝑣_𝑚,$$

where $𝑎_1, …, 𝑎_𝑚 ∈ 𝐅$.

2.4 definition: span

The set of all linear combinations of a list of vectors $𝑣_1, …, 𝑣_𝑚$ in $𝑉$ is called the span of $𝑣_1, …, 𝑣_𝑚$, denoted by $\text{span}(𝑣_1, …, 𝑣_𝑚)$. In other words,

$$\text{span}(𝑣_1, \cdots, 𝑣_𝑚) = \{ a_1 v_1 + \cdots + a_m v_m : a_1, \cdots, a_m \in F \}.$$

The span of the empty list $()$ is defined to be $\{0\}$.

$\square$

2.6 span is the smallest containing subspace

The span of a list of vectors in $𝑉$ is the smallest subspace of $𝑉$ containing all vectors in the list.

Proof:

Let

$$\text{span}(𝑣_1, \cdots, 𝑣_𝑚) = \{ a_1 v_1 + \cdots + a_m v_m : a_1, \cdots, a_m \in F \}.$$

$v_i \in S$, because we can set $i = 1, j \neq i = 0$.

$$v + w = (a_1 v_1 + \cdots + a_m v_m) + (b_1 v_1 + \cdots + b_m v_m) \\ = (a_1+b_1) v_1 + \cdots + (a_m+b_m) v_m \in S$$

So addition is closed.

$$𝜆 v = 𝜆 (a_1 v_1 + \cdots + a_m v_m) \\ = (𝜆a_1) v_1 + \cdots + (𝜆a_m) v_m \in S$$

So scalar multiplication is closed.

Finally $0 \in S$ by setting $a_1 = \cdots = a_m = 0$.

Every sub-space of $𝑉$ that contains each $𝑣_𝑘$ contains span $(v_1, \cdots, v_m)$ . Thus $\text{span}(𝑣_1, \cdots, 𝑣_𝑚)$ is the smallest subspace of $𝑉$ containing all the vectors $𝑣_1, …, 𝑣_𝑚$.

$\square$

2.7 definition: spans

If $\text{span}(𝑣_1, \cdots, 𝑣_𝑚)$ equals $𝑉$, we say that the list $𝑣_1, …, 𝑣_𝑚$ spans $𝑉$.

2.9 definition: finite-dimensional vector space

A vector space is called finite-dimensional if some list of vectors in it spans the space.

2.10 definition: polynomial, $𝒫(𝐅)$

A function $𝑝 ∶ 𝐅 → 𝐅$ is called a polynomial with coefficients in $𝐅$ if there exist $𝑎_0, …, 𝑎_𝑚 ∈ 𝐅$ such that

$$p(z) = a_0 + a_1 z + \cdots + a_{m} z^{m}$$

for all $z \in F$.

$𝒫(𝐅)$ is the set of all polynomials with coefficients in $𝐅$.

$\square$

If a polynomial that is identically zero as a function on $𝐅$, then all coefficients are zero (will be proved in 4.8)

Conclusion: the coefficients of a polynomial are uniquely determined by the polynomial.

2.11 definition: degree of a polynomial, deg 𝑝

A polynomial $p \in 𝒫(𝐅)$ is said to have degree $𝑚$ if there exist scalars $a_1, \cdots, a_m \in 𝐅$ with $a_m \neq 0$, such that for every $𝑧 ∈ 𝐅$, we have

$$p(z) = a_0 + a_1 z + \cdots + a_{m} z^{m}.$$

The polynomial that is identically $0$ is said to have degree $−∞$.

The degree of a polynomial $𝑝$ is denoted by $\deg 𝑝$.

2.12 notation: $𝒫_𝑚(𝐅)$

For $𝑚$ a nonnegative integer, $𝒫_𝑚(𝐅)$ denotes the set of all polynomials with coefficients in $𝐅$ and degree at most $𝑚$.

2.13 definition: infinite-dimensional vector space

A vector space is called infinite-dimensional if it is not finite-dimensional.

2.14 example: $𝒫(𝐅)$ is infinite-dimensional.

Assume otherwise, then some list of polynomials can span $𝒫(𝐅)$, let $m$ be the highest degree of the polynomials, in the list, then $z^{m+1}$ is not in the span.

$\square$

Linear Independence

2.15 definition: linearly independent

A list $v_1, \cdots, v_m$ of vectors in $𝑉$ is called linearly independent if the only choice of $a_1, \cdots, a_m \in F$ that makes

$$a_1 v_1 + \cdots + a_m v_m = 0$$

is $a_1 = \cdots = a_m = 0$.

The empty list $( )$ is also declared to be linearly independent.

2.17 definition: linearly dependent

• A list of vectors in $𝑉$ is called linearly dependent if it is not linearly independent.
• A list $v_1, \cdots, v_m$ of vectors in $𝑉$ is linearly dependent if there exist $a_1 = \cdots = a_m \in F$, not all $0$, such that $a_1 v_1 + \cdots + a_m v_m = 0$.

2.19 linear dependence lemma

Suppose $v_1, \cdots, v_m$ is a linearly dependent list in $𝑉$. Then there exists $k \in \{1,2,\cdots,m\}$ such that

$$v_k \in \text{span}(𝑣_1, \cdots, 𝑣_{k-1}).$$

Furthermore, if $𝑘$ satisfies the condition above and the 𝑘th term is removed from $v_1, \cdots, v_m$, then the span of the remaining list equals $\text{span}(𝑣_1, \cdots, 𝑣_𝑚)$.

Proof:

We can find $a_1, \cdots, a_m \in F$, not all $0$ such that

$$a_1 v_1 + \cdots + a_m v_m = 0$$

Starting from $m$, let $k$ be the largest integer such that $a_k \neq 0$. Then

$$a_1 v_1 + \cdots + a_m v_m = a_1 v_1 + \cdots + a_k v_k = 0$$

That is

$$v_k = a_k^{-1} (a_1 v_1 + \cdots + a_{k-1} v_{k-1})$$