Independence of column vectors列れつベクトル and the bridge to rank階数かいすう

date2026-07-14document_iddoc_89aec023d343d7e3e5ebe5794315ae5bdescription列ベクトルの一次独立性を具体計算で判定し、pivot・列空間・階数へ接続する講義である。prerequisites線型結合と張る空間の基本 / 行基本変形の基本 / 階段形と簡約階段形type講義content_typelecturestatusactive

mathlinear-algebraundergraduatelecture

1Introduction

The key point is that linear independence一次独立性いちじどくりつせい of column vectors leads directly to pivots and rank階数かいすう. rank階数かいすう is not the number of nonzero entries. It is the number of independent column directions.

2Terms and definitions

Vectors $v_{1}, \dots, v_{k}$ are linearly independent一次独立いちじどくりつ if

c_{1} v_{1} + \dots + c_{k} v_{k} = 0

implies

c_{1} = \dots = c_{k} = 0.

The rank階数かいすう of a matrix is the maximum number of independent column vectors.

3Strategy

Arrange the column vectors列れつベクトル as a matrix行列ぎょうれつ and use elimination to identify the pivot主成分しゅせいぶん columns. The original columns in those pivot positions form a basis基底きてい for the column space列空間れつくうかん.

data/lecture/math/vector/row-and-column-vectors.lecture.n.md data/lecture/math/linear-algebra/row-echelon-and-reduced-row-echelon-forms.lecture.n.md

4Intuitive explanation

If two column vectors point in the same direction, one is a scalar multiple of the other and does not add a new direction. Rank階数かいすう counts the essential independent directions after such redundancy is removed.

5Precise explanation

Let $a_{1}, \dots, a_{n}$ be the columns of a matrix $A$ . The linear relation

c_{1} a_{1} + \dots + c_{n} a_{n} = 0

is equivalent to

A c = 0,

where $c = (c_{1}, \dots, c_{n})^{T}$ . Therefore column independence is tested by the homogeneous system $A c = 0$ .

Elementary row operations行基本変形ぎょうきほんへんけい preserve the solution set of this homogeneous system. Algebraically, a row operation is multiplication by an invertible matrix可逆行列かぎゃくぎょうれつ $E$ on the left, so $E A c = 0$ is equivalent to $A c = 0$ .

What is preserved is the linear relations among the columns. The visible column vectors themselves change from $a_{i}$ to $E a_{i}$ , so the column space列空間れつくうかん itself generally changes during row reduction. Therefore pivot columns found in an echelon form must be read back in the same positions of the original matrix.

6Row operations and pivot主成分しゅせいぶん columns

elementary row operations行基本変形ぎょうきほんへんけい preserve the solution set of $A c = 0$ . Algebraically, a row operation is multiplication by an invertible可逆かぎゃく matrix $E$ on the left, and

E A c = 0 ⟺ A c = 0.

However, row operations change the visible column vectors from $a_{i}$ to $E a_{i}$ . Therefore pivot主成分しゅせいぶん columns found in an echelon form must be read by their positions in the original matrix, not by taking the transformed columns themselves as the original column-space basis基底きてい.

7Worked example

Let

a_{1} = (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), a_{2} = (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), a_{3} = (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}) .

Then $a_{3} = a_{1} + a_{2}$ . Hence the three columns are linearly dependent, while $a_{1}$ and $a_{2}$ give two independent directions. The rank階数かいすう is $2$ .

8Theorem/proof checkpoint

The pivot主成分しゅせいぶん columns of the original matrix form a basis基底きてい of the column space列空間れつくうかん. They span because every non-pivot主成分しゅせいぶん column has a linear relation with earlier pivot主成分しゅせいぶん columns, detected by the echelon form. They are independent because a nontrivial relation among pivot主成分しゅせいぶん columns would produce a nontrivial solution to the corresponding homogeneous system, contradicting the pivot主成分しゅせいぶん positions.

9Common misunderstandings

rank階数かいすう is not a count of nonzero entries.
The transformed columns after row reduction are not generally the original column space列空間れつくうかん.
Having many columns does not guarantee high rank階数かいすう; dependent columns add no new direction.

10Scope

Independence of column vectors列れつベクトル is the finite-dimensional basis for rank階数かいすう. In an abstract vector space, vectors may not literally be arranged as columns, but the definition by linear combinations線型結合せんけいけつごう is the same.

11Final forms

[PARSE ERROR: Undefined("Command(\"boxed\")")] c_{1} a_{1} + \dots + c_{n} a_{n} = 0 ⟺ A c = 0

[PARSE ERROR: Undefined("Command(\"boxed\")")] rank (A) = maximumnumberofindependentcolumns

12Exercise links

data/exercise/math/linear-algebra/vector-spaces-bases-and-rank.exercise.n.md