Basics of linear combinations線型結合せんけいけつごう and span張はる空間くうかん

date2026-07-14document_iddoc_98dfd2e7cd6a1b242c8d9e5430620a08description線型結合、張る空間、部分空間を具体的なベクトル計算から導入し、基底・階数・列空間へ進む前提を作る講義である。prerequisitesベクトルの基本演算 / 一次方程式の基本type講義content_typelecturestatusactive

mathlinear-algebraundergraduatelecture

1Introduction

A linear combination線型結合せんけいけつごう uses given vectors as ingredients and asks what directions can be reached by choosing coefficients. This is the bridge to basis基底きてい, rank階数かいすう, and column space列空間れつくうかん.

2Terms and definitions

Let $V$ be a vector space over a field $K$ . For vectors $v_{1}, \dots, v_{k}$ and scalars $c_{1}, \dots, c_{k} \in K$ ,

c_{1} v_{1} + \dots + c_{k} v_{k}

is a linear combination線型結合せんけいけつごう.

The span張はる空間くうかん of $v_{1}, \dots, v_{k}$ is the set of all such linear combinations線型結合せんけいけつごう:

span (v_{1}, \dots, v_{k}) = {c_{1} v_{1} + \dots + c_{k} v_{k} ∣ c_{1}, \dots, c_{k} \in K} .

A subspace部分空間ぶぶんくうかん is a set of vectors that contains $0$ and is closed under addition and scalar multiplication.

The span is the smallest subspace containing the listed vectors. To see that $span (v_{1}, \dots, v_{k})$ is a subspace, first choose $c_{1} = \dots = c_{k} = 0$ to get the zero vector. If

x = \sum_{i = 1}^{k} a_{i} v_{i}, y = \sum_{i = 1}^{k} b_{i} v_{i}

are in the span, then

x + y = \sum_{i = 1}^{k} (a_{i} + b_{i}) v_{i}

is again a linear combination of the same vectors. For a scalar $r \in K$ ,

r x = \sum_{i = 1}^{k} (r a_{i}) v_{i}

is also in the span. Thus the span contains $0$ and is closed under addition and scalar multiplication, so it is a subspace.

3Strategy

Start with two vectors and vary their coefficients係数けいすう to see what region can be reached. Then connect this viewpoint to the matrix method that arranges column vectors列れつベクトル as columns.

data/lecture/math/linear-algebra/vector-operations.lecture.n.md

4Intuitive explanation

Scalar multiples of one nonzero vector form a line through the origin. Combining two vectors pointing in different directions may produce the entire plane.

A linear combination線型結合せんけいけつごう investigates everything that can be built from the given material vectors. The span is not merely a scattered collection of points: it contains the zero vector and is closed under addition and scalar multiplication, so it is a subspace部分空間ぶぶんくうかん.

5Precise explanation

Let $v_{1} = (1, 0)^{T}$ and $v_{2} = (0, 1)^{T}$ . Then

a v_{1} + b v_{2} = (a, b)^{T},

so arbitrary choices of $a, b$ represent every vector in $R^{2}$ .

On the other hand, for $w_{1} = (1, 1)^{T}$ and $w_{2} = (2, 2)^{T}$ ,

a w_{1} + b w_{2} = (a + 2 b) (1, 1)^{T},

so the result is only a line. Having two vectors is not enough if their directions are redundant.

6Common misunderstandings

More vectors do not necessarily span a larger space; repeated directions matter.
In a linear combination, the coefficients are arbitrary. Do not look at only one fixed set of coefficients.
The span is determined by the directions and placement of the material vectors.

7Why span is a subspace部分空間ぶぶんくうかん

The zero vector is in the span by taking all coefficients equal to $0$ . If

x = \sum_{i = 1}^{k} a_{i} v_{i}, y = \sum_{i = 1}^{k} b_{i} v_{i},

then

x + y = \sum_{i = 1}^{k} (a_{i} + b_{i}) v_{i},

so the span is closed under addition. For a scalar $r$ ,

r x = \sum_{i = 1}^{k} (r a_{i}) v_{i},

so the span is closed under scalar multiplication.

8Worked examples

Let $v_{1} = (1, 0)^{T}$ and $v_{2} = (0, 1)^{T}$ . Then

a v_{1} + b v_{2} = (a, b)^{T},

so all of $R^{2}$ is reached.

By contrast, if $w_{1} = (1, 1)^{T}$ and $w_{2} = (2, 2)^{T}$ , then

a w_{1} + b w_{2} = (a + 2 b) (1, 1)^{T} .

The span is only the line through $(1, 1)^{T}$ . Two vectors do not necessarily span a plane if their directions are redundant.

9Theorem and proof: span is the smallest subspace部分空間ぶぶんくうかん containing the given vectors

Let $S = {v_{1}, \dots, v_{k}}$ . The set $span (S)$ is a subspace部分空間ぶぶんくうかん containing $S$ . If $U$ is any subspace部分空間ぶぶんくうかん containing $S$ , then $U$ is closed under linear combinations線型結合せんけいけつごう, so every $a_{1} v_{1} + \dots + a_{k} v_{k}$ lies in $U$ . Therefore

span (S) \subseteq U .

Thus the span is the smallest subspace部分空間ぶぶんくうかん containing the original vectors.

10Common misunderstandings

More vectors do not automatically mean a larger span; repeated directions add no new reach.
The coefficients in a linear combination線型結合せんけいけつごう are arbitrary, not fixed in advance.
A span is not just a scattered set of points. It is closed under the vector-space operations.

11Scope

The definition works in any vector space, including polynomial spaces and function spaces, as long as addition and scalar multiplication are defined.

12Final forms

[PARSE ERROR: Undefined("Command(\"boxed\")")] span (v_{1}, \dots, v_{k}) = {c_{1} v_{1} + \dots + c_{k} v_{k}}

[PARSE ERROR: Undefined("Command(\"boxed\")")] Aspanisthesetofallreachablelinearcombinations [PARSE ERROR: Undefined("RBrace")]

[PARSE ERROR: Undefined("Command(\"boxed\")")] span (v_{1}, \dots, v_{k}) isasubspace [PARSE ERROR: Undefined("RBrace")]

13Theorem and proof: the span is the smallest subspace

Let $S = {v_{1}, \dots, v_{k}}$ . Then

span (S) = {a_{1} v_{1} + \dots + a_{k} v_{k} ∣ a_{i} \in F}

is the smallest subspace部分空間ぶぶんくうかん containing $S$ .

First, $span (S)$ is closed under linear combinations. If

x = \sum_{i} a_{i} v_{i}, y = \sum_{i} b_{i} v_{i},

then for scalars $c, d$ ,

c x + d y = \sum_{i} (c a_{i} + d b_{i}) v_{i},

which is again a linear combination of vectors in $S$ . Thus it is a subspace.

Next, if $U$ is any subspace containing $S$ , then $U$ is closed under linear combinations, so it contains every linear combination of $v_{1}, \dots, v_{k}$ . Hence $span (S) \subseteq U$ .

14Exercise links

data/exercise/math/linear-algebra/vectors-and-linear-combinations.exercise.n.md