properties of linear independence

Let $V$ be a vector space over a field $k$ . Below are some basic properties of linear independence.

1.
$S\\subseteq V$ is never linearly independent if $0\\in S$ .
Proof.
Since $1\\cdot 0=0$ . ∎
2.
If $S$ is linearly independent, so is any subset of $S$ . As a result, if $S$ and $T$ are linearly independent, so is $S\\cap T$ . In addition, $\\varnothing$ is linearly independent, its spanning set being the singleton consisting of the zero vector $0$ .
Proof.
If $r_{1}v_{1}+\\cdots r_{n}v_{n}=0$ , where $v_{i}\\in T$ , then $v_{i}\\in S$ , so $r_{i}=0$ for all $i=1,\\ldots,n$ . ∎
3.
If $S_{1}\\subseteq S_{2}\\subseteq\\cdots$ is a chain of linearly independent subsets of $V$ , so is their union.
Proof.
Let $S$ be the union. If $r_{1}v_{1}+\\cdots r_{n}v_{n}=0$ , then $v_{i}\\in S_{a(i)}$ , for each $i$ . Pick the largest $S_{a(i)}$ so that all $v_{i}$ ’s are in it. Since this set is linearly independent, $r_{i}=0$ for all $i$ .∎
4.
$S$ is a basis for $V$ iff $S$ is a maximal linear independent subset of $V$ . Here, maximal means that any proper superset of $S$ is linearly dependent.
Proof.
If $S$ is a basis for $V$ , then it is linearly independent and spans $V$ . If we take any vector $v\otin S$ , then $v$ can be expressed as a linear combination of elements in $S$ , so that $S\\cup\\{v\\}$ is no longer linearly independent, for the coefficient in front of $v$ is non-zero. Therefore, $S$ is maximal.
Conversely, suppose $S$ is a maximal linearly independent set in $V$ . Let $W$ be the span of $S$ . If $W\eq V$ , pick an element $v\\in V-W$ . Suppose $0=r_{1}v_{1}+\\cdots r_{n}v_{n}+rv$ , where $v_{i}\\in S$ , then $-rv=r_{1}v_{1}+\\cdots+r_{n}v_{n}$ . If $r\eq 0$ , then $v$ would be in the span of $S$ , contradicting the assumption. So $r=0$ , and as a result, $r_{i}=0$ , since $S$ is linearly independent. This shows that $S\\cup\\{v\\}$ is linearly independent, which is impossible since $S$ is assumed to be maximal. Therefore, $W=V$ .∎

Remark. All of the properties above can be generalized to modules over rings, except the last one, where the implication is only one-sided: basis implying maximal linear independence.

单词	PropertiesOfLinearIndependence
释义	properties of linear independence Let $V$ be a vector space over a field $k$ . Below are some basic properties of linear independence. 1. $S\\subseteq V$ is never linearly independent if $0\\in S$ . Proof. Since $1\\cdot 0=0$ . ∎ 2. If $S$ is linearly independent, so is any subset of $S$ . As a result, if $S$ and $T$ are linearly independent, so is $S\\cap T$ . In addition, $\\varnothing$ is linearly independent, its spanning set being the singleton consisting of the zero vector $0$ . Proof. If $r_{1}v_{1}+\\cdots r_{n}v_{n}=0$ , where $v_{i}\\in T$ , then $v_{i}\\in S$ , so $r_{i}=0$ for all $i=1,\\ldots,n$ . ∎ 3. If $S_{1}\\subseteq S_{2}\\subseteq\\cdots$ is a chain of linearly independent subsets of $V$ , so is their union. Proof. Let $S$ be the union. If $r_{1}v_{1}+\\cdots r_{n}v_{n}=0$ , then $v_{i}\\in S_{a(i)}$ , for each $i$ . Pick the largest $S_{a(i)}$ so that all $v_{i}$ ’s are in it. Since this set is linearly independent, $r_{i}=0$ for all $i$ .∎ 4. $S$ is a basis for $V$ iff $S$ is a maximal linear independent subset of $V$ . Here, maximal means that any proper superset of $S$ is linearly dependent. Proof. If $S$ is a basis for $V$ , then it is linearly independent and spans $V$ . If we take any vector $v\otin S$ , then $v$ can be expressed as a linear combination of elements in $S$ , so that $S\\cup\\{v\\}$ is no longer linearly independent, for the coefficient in front of $v$ is non-zero. Therefore, $S$ is maximal. Conversely, suppose $S$ is a maximal linearly independent set in $V$ . Let $W$ be the span of $S$ . If $W\eq V$ , pick an element $v\\in V-W$ . Suppose $0=r_{1}v_{1}+\\cdots r_{n}v_{n}+rv$ , where $v_{i}\\in S$ , then $-rv=r_{1}v_{1}+\\cdots+r_{n}v_{n}$ . If $r\eq 0$ , then $v$ would be in the span of $S$ , contradicting the assumption. So $r=0$ , and as a result, $r_{i}=0$ , since $S$ is linearly independent. This shows that $S\\cup\\{v\\}$ is linearly independent, which is impossible since $S$ is assumed to be maximal. Therefore, $W=V$ .∎ Remark. All of the properties above can be generalized to modules over rings, except the last one, where the implication is only one-sided: basis implying maximal linear independence.
随便看	Banach-Steinhaus theorem BanachTarskiParadox Banach-Tarski paradox Band band BaouendiTrevesApproximationTheorem Baouendi-Treves approximation theorem BarbualatsLemma Barbălat’s lemma BargmannFockSpace Bargmann-Fock space BargmannTransform Bargmann transform BarnesIntegralRepresentationOfTheHypergeometricFunction Barnes’ integral representation of the hypergeometric function BarnsleyFern Barnsley fern BaronisTheorem Baroni’s theorem barpartialOperator Barrel barrel BarycentricCoordinates barycentric coordinates BasalUnits

properties of linear independence

Proof.

Proof.

Proof.

Proof.