# Linear Algebra II

## September 11, 2014

### Lecture 6

Filed under: 2014 Fall — Y.K. Lau @ 8:00 PM

Today we completed our introduction on quotient vector spaces and introduced a couple of important concepts:

• Linear combination & Span,
• Spanning set,
• Minimal spanning set,
• Linear independence,
• Basis & dimension.

[This is also a checklist for your learning outcomes! You have to understand all these concepts and the theory related to them.]

Below is a flowchart to help you see how/what the theory is developed.

Remark 1. One may try to use set operations to make subspaces. Suppose ${U}$ and ${V}$ are subspaces of a vector space. Will

(i)   ${U\cup V}$,       (ii)   ${U\cap V}$,       (iii)   ${U\setminus V}$

be subspaces?
In general the answer is (i) NO,   (ii) YES,   (iii) NO.

Indeed, we have the following results:

Let ${U}$ and ${V}$ be subspaces of a vector space. Then ${U\cup V}$ is a subspace if and only if ${U\subset V}$ or ${V\subset U}$.

(Proof. See Assignment 1 Q. 3.)

${U\cap V}$ is a subspace.

(Proof. Exercise.)

${U\setminus V}$ is never a subspace.

(Proof. Exercise.)

[In view of this, you may appreciate more the definition of “Span”, which teaches you how to make subspaces.]

Remark 2. Recall the following two convention:

1. ${{\rm Span}(\emptyset)=\{\underline{0}\}}$ (the zero subspace).
2. Let ${S}$ be any subset (not necessarily finite) of a vector space. Then

$\displaystyle {\rm Span}(S)=\left\{a_1\underline{v}_1+a_2\underline{v}_2+\cdots +a_m\underline{v}_m: \ \begin{array}{l} a_1,\cdots, a_m\in {\mathbb R}, \\ \underline{v}_1,\cdots, \underline{v}_m\in S, \\ m\in {\mathbb N}\end{array}\right\}.$

Remark 3. We mentioned in lecture that linearly independence is a notion to characterize minimal spanning set. But how?! This is explained in the following

Claim: Let ${V}$ be a vector space. A linearly independent spanning set of ${V}$ is a minimal spanning set for ${V}$.

Proof. Suppose ${B=\{v_1,\cdots, v_n\}}$ is a linearly independent spanning set but not minimal.

That means we can find a proper subset ${C}$ of ${B}$ such that ${V={\rm Span}(C)}$.

As ${\displaystyle {} \ C\mathop{\subset}_{\neq} B \ {} }$, ${C}$ has less than ${n}$ elements.

i.e. The spanning set ${C}$ has less elements than the linearly independent set ${B}$.

This contradicts to the Fundamental Thm (in p. 305 of the textbook). So ${B}$ is minimal.

Finally we remark the often used equivalent definitions of linearly independence/dependence:

Let ${\underline{v}_1,\underline{v}_2,\cdots, \underline{v}_n}$ be vectors in a vector space. The following are equivalent:

1. ${\underline{v}_1,\underline{v}_2,\cdots, \underline{v}_n}$ are linearly independent.
2. The (vector) equation

${s_1\underline{v}_1+s_2\underline{v}_2+\cdots+s_n \underline{v}_n=\underline{0}}$

(regarding ${s_1,\cdots, s_n}$ as variables) has the trivial solution ${s_1=\cdots = s_n=0}$ only.

Below are various equivalent ways to define linearly dependent vectors.

Given vectors ${\underline{u}_1,\underline{u}_2,\cdots, \underline{u}_m}$, then the following are equivalent:

1. ${\underline{u}_1,\underline{u}_2,\cdots, \underline{u}_m}$ are linearly dependent.
2. ${\underline{u}_1,\underline{u}_2,\cdots, \underline{u}_m}$ are not linearly independent.
3. The equation ${s_1\underline{u}_1+s_2\underline{u}_2+\cdots+s_m \underline{u}_m=\underline{0}}$ has non-trivial solution(s).
4. There exist ${s_1,s_2,\cdots, s_m}$, not all zero, such that ${s_1\underline{u}_1+s_2\underline{u}_2+\cdots+s_m \underline{u}_m=\underline{0}}$.