Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 7

Linear independence

Definition

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. A family of vectors ${}v_{i}$, ${}i\in I$, (where ${}I$ denotes a finite index set) is called linearly independent if an equation of the form

$\sum _{i\in I}s_{i}v_{i}=0{\text{ with }}s_{i}\in K$

is only possible when ${}s_{i}=0$

for all ${}i$.

If a family is not linearly independent, then it is called linearly dependent. A linear combination ${}\sum _{i\in I}s_{i}v_{i}=0$ is called a representation of the null vector. It is called the trivial representation if all coefficients ${}s_{i}$ equal ${}0$ and, if at least one coefficient is not ${}0$, a nontrivial representation of the null vector. A family of vectors is linearly independent if and only if one can represent with the family the null vector only in the trivial way. This is equivalent with the property that no vector of the family can be expressed as a linear combination by the others.

Example

The standard vectors in ${}K^{n}$ are linearly independent. A representation

${}\sum _{i=1}^{n}s_{i}e_{i}=0\,$

just means

${}s_{1}{\begin{pmatrix}1\\0\\\vdots \\0\end{pmatrix}}+s_{2}{\begin{pmatrix}0\\1\\\vdots \\0\end{pmatrix}}+\cdots +s_{n}{\begin{pmatrix}0\\0\\\vdots \\1\end{pmatrix}}={\begin{pmatrix}0\\0\\\vdots \\0\end{pmatrix}}\,.$

The ${}i$-th row yields directly ${}s_{i}=0$.

Example

The three vectors

${\begin{pmatrix}3\\3\\3\end{pmatrix}},\,\,{\begin{pmatrix}0\\4\\5\end{pmatrix}},\,{\text{ and }}\,{\begin{pmatrix}4\\8\\9\end{pmatrix}}$

are linearly dependent. The equation

${}4{\begin{pmatrix}3\\3\\3\end{pmatrix}}+3{\begin{pmatrix}0\\4\\5\end{pmatrix}}-3{\begin{pmatrix}4\\8\\9\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\,$

is a nontrivial representation of the null vector.

Remark

The vectors ${}v_{1}={\begin{pmatrix}a_{11}\\\vdots \\a_{m1}\end{pmatrix}},\ldots ,v_{n}={\begin{pmatrix}a_{1n}\\\vdots \\a_{mn}\end{pmatrix}}\in K^{m}$ are linearly dependent if and only if the homogeneous linear system

${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&0\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&0\\\vdots &\vdots &\vdots \\a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}&=&0\end{matrix}}$

has a nontrivial solution.

For an infinite family of vectors, there is also a concept of linear independence.

Definition

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. Then a family of vectors ${}v_{i}$, ${}i\in I$, in ${}V$ is called linearly independent if an equation

$\sum _{i\in J}s_{i}v_{i}=0{\text{ with }}s_{i}\in K{\text{ for a finite subset }}J\subseteq I$

is only possible if ${}s_{i}=0$

for all ${}i$.

By this definition, linear independence of an arbitrary family is reduced to the finite case. Keep in mind that in a vector space, there are no infinite sums. An expression like ${}\sum _{n\in \mathbb {N} }s_{n}v_{n}=0$ does not have any sense.

Lemma

Let ${}K$ be a field, let ${}V$ be a ${}K$-vector space, and let ${}v_{i}$, ${}i\in I$,

be a family of vectors in ${}V$. Then the following statements hold.

1. If the family is linearly independent, then for each subset ${}J\subseteq I$, also the family ${}v_{i}$ , ${}i\in J$, is linearly independent.
2. The empty family is linearly independent.
3. If the family contains the null vector, then it is not linearly independent.
4. If a vector appears several times in the family, then the family is not linearly independent.
5. A single vector ${}v$ is linearly independent if and only if ${}v\neq 0$.
6. Two vectors ${}v$ and ${}u$ are linearly independent if and only if ${}u$ is not a scalar multiple of ${}v$ and vice versa.

Proof

See Exercise 7.14 .

$\Box$

Bases

Definition

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. Then a linearly independent generating system ${}v_{i}\in V$, ${}i\in I$,

of ${}V$ is called a basis of ${}V$.

Example

The standard vectors in ${}K^{n}$ form a basis. The linear independence was shown in Example 7.2 . To show that they also form a generating system, let

${}v={\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}}\in K^{n}\,$

be an arbitrary vector. Then we have immediately

${}v=\sum _{i=1}^{n}b_{i}e_{i}\,.$

Hence, we have a basis, which is called the standard basis of ${}K^{n}$.

Example

We consider the ${}K$-linear subspace ${}U\subset K^{n}$ given by

${}U={\left\{v\in K^{n}\mid \sum _{i=1}^{n}v_{i}=0\right\}}\,.$

A basis for ${}U$ is given by the ${}n-1$ vectors

$u_{1}=(1,-1,0,0,\ldots ,0),\,u_{2}=(0,1,-1,0,\ldots ,0),\,,\ldots ,u_{n-1}=(0,0,\ldots ,0,1,-1),\,.$

These vectors belong evidently to ${}U$. The linear independence can be checked in ${}K^{n}$. From an equation

${}\sum _{i=1}^{n-1}a_{i}u_{i}=0\,$

we can deduce step by step ${}a_{1}=0$, ${}a_{2}=0$, etc. That the system is a generating system follows from

${}{\begin{pmatrix}v_{1}\\v_{2}\\v_{3}\\\vdots \\v_{n-1}\\v_{n}\end{pmatrix}}=v_{1}{\begin{pmatrix}1\\-1\\0\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}){\begin{pmatrix}0\\1\\-1\\\vdots \\0\\0\end{pmatrix}}+(v_{1}+v_{2}+v_{3}){\begin{pmatrix}0\\0\\1\\-1\\\vdots \\0\end{pmatrix}}+\cdots +(v_{1}+v_{2}+v_{3}+\cdots +v_{n-1}){\begin{pmatrix}0\\0\\0\\\vdots \\1\\-1\end{pmatrix}}\,,$

where the equality in the last row rests on the condition

${}\sum _{i=1}^{n}v_{i}=0\,.$

For the complex numbers, the elements ${}1,{\mathrm {i} }$ form a real basis. In the space of all ${}m\times n$-matrices, that is, ${}\operatorname {Mat} _{m\times n}(K)$, those matrices where exactly one entry is ${}1$, and all other entries are ${}0$, form a basis, see Exercise 7.27 .

Example

In the polynomial ring ${}K[X]$ over a field ${}K$, the powers ${}X^{n}$, ${}n\in \mathbb {N}$, form a basis. By definition, every polynomial

$a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots +a_{2}X^{2}+a_{1}X+a_{0}$

is a linear combination of the powers ${}X^{0}=1,X^{1}=X,\ldots ,X^{n}$. Moreover, these powers are linearly independent. For if

${}a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots +a_{2}X^{2}+a_{1}X+a_{0}=0\,,$

then all coefficients equal ${}0$ (this is part of the concept of a polynomial).

Characterization of a basis

The following theorem gives an important characterization for a family of vectors to be basis.

Theorem

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. Let ${}v_{1},\ldots ,v_{n}\in V$

be a family of vectors. Then the following statements are equivalent.

1. The family is a basis of ${}V$.
2. The family is a minimal generating system; that is, as soon as we remove one vector ${}v_{i}$, the remaining family is not a generating system any more.
3. For every vector ${}u\in V$, there is exactly one representation

   ${}u=s_{1}v_{1}+\cdots +s_{n}v_{n}\,.$

4. The family is maximally linearly independent; that is, as soon as some vector is added, the family is not linearly independent any more.

Proof

Proof by ring closure. ${}(1)\Rightarrow (2)$. The family is a generating system. Let us remove a vector, say ${}v_{1}$, from the family. We have to show that the remaining family, that is ${}v_{2},\ldots ,v_{n}$, is not a generating system anymore. So suppose that it is still a generating system. Then, in particular, ${}v_{1}$ can be written as a linear combination of the remaining vectors, and we have

${}v_{1}=\sum _{i=2}^{n}s_{i}v_{i}\,.$

But then

${}v_{1}-\sum _{i=2}^{n}s_{i}v_{i}=0\,$

is a nontrivial representation of ${}0$, contradicting the linear independence of the family. ${}(2)\Rightarrow (3)$. Due to the condition, the family is a generating system, hence every vector can be represented as a linear combination. Suppose that for some ${}u\in V$, there is more than one representation, say

${}u=\sum _{i=1}^{n}s_{i}v_{i}=\sum _{i=1}^{n}t_{i}v_{i}\,,$

where at least one coefficient is different. Without loss of generality, we may assume ${}s_{1}\neq t_{1}$. Then we get the relation

${}{\left(s_{1}-t_{1}\right)}v_{1}=\sum _{i=2}^{n}{\left(t_{i}-s_{i}\right)}v_{i}\,$

Because of ${}s_{1}-t_{1}\neq 0$, we can divide by this number and obtain a representation of ${}v_{1}$ using the other vectors. In this situation, due to Exercise 6.25 , also the family without ${}v_{1}$ is a generating system of ${}V$, contradicting the minimality. ${}(3)\Rightarrow (4)$. Because of the unique representability, the zero vector has only the trivial representation. This means that the vectors are linearly independent. If we add a vector ${}u$, then it has a representation

${}u=\sum _{i=1}^{n}s_{i}v_{i}\,,$

and, therefore,

${}0=u-\sum _{i=1}^{n}s_{i}v_{i}\,$

is a non-trivial representation of ${}0$, so that the extended family ${}u,v_{1},\ldots ,v_{n}$ is not linearly independent. ${}(4)\Rightarrow (1)$. The family is linearly independent, we have to show that it is also a generating system. Let ${}u\in V$. Due to the condition, the family ${}u,v_{1},\ldots ,v_{n}$ is not linearly independent. This means that there exists a non-trivial representation

${}0=su+\sum _{i=1}^{n}s_{i}v_{i}\,.$

Here ${}s\neq 0$, because otherwise this would be a non-trivial representation of ${}0$ with the original family ${}v_{1},\ldots ,v_{n}$. Hence, we can write

${}u=-\sum _{i=1}^{n}{\frac {s_{i}}{s}}v_{i}\,,$

yielding a representation for ${}u$.

$\Box$

Remark

Let a basis ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ of a ${}K$-vector space ${}V$ be given. Due to Theorem 7.11   (3), this means that for every vector ${}u\in V$, there exists a unique representation (a linear combination)

${}u=s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}\,.$

Here, the uniquely determined elements ${}s_{i}\in K$ (scalars) are called the coordinates of ${}u$ with respect to the given basis. This means that for a given basis, there is a correspondence between vectors and coordinate tuples ${}(s_{1},s_{2},\ldots ,s_{n})\in K^{n}$. We say that a basis determines a linear coordinate system of ${}V$. To paraphrase, a basis gives, in particular, a bijective mapping

$\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\longmapsto s_{1}v_{1}+s_{2}v_{2}+\cdots +s_{n}v_{n}.$

The inverse mapping

${\left(\Psi _{\mathfrak {v}}\right)}^{-1}\colon V\longrightarrow K^{n}$

is also called the coordinate mapping.

Theorem

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space with a finite generating system. Then ${}V$ has a finite

basis.

Proof

Let ${}v_{i}$, ${}i\in I$, be a finite generating system of ${}V$ with a finite index set ${}I$. We argue with the characterization from Theorem 7.11   (2). If the family is minimal, then we have a basis. If not, then there exists some ${}k\in I$ such that the remaining family, where ${}v_{k}$ is removed, that is, ${}v_{i}$, ${}i\in I\setminus \{k\}$, is also a generating system. In this case, we can go on with this smaller index set. With this method, we arrive at a subset ${}J\subseteq I$ such that ${}v_{i}$, ${}i\in J$, is a minimal generating set, hence a basis.

$\Box$

Remark

In general, the Theorem of Hamel says that every vector space has a basis. The proof of this theorem uses strong set-theoretical methods, in particular the axiom of choice and the Lemma of Zorn. This is the reason why many statements about finite-dimensional vector spaces pass over to vector spaces of infinite dimension. In this course, we will concentrate on the finite-dimensional case.

Footnotes

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