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}}.




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