Review of basic concepts and facts in linear algebra

Matrix Analysis @ HITSZ

Instructor: Zijun Luo

Fall 2012

guideline

• Prerequisite: • Text: Matrix Analysis, Roger A. Horn, Charles R.

Johnson, Cambridge University Press, Reprint edition (February 23,1990). ISBN 0521386322, 575p

• Homework assignment: weekly on Friday• Exam: final exam (open book) on Oct., 2014 (te

mporary) • Content: vector space, norms, eigenvalues, unit

ary matrix, Hermitian matrix, matrix factorization, canonical form, nonnegative matrix

• QQ群： 314073497

Definition : Vector Space ( V, + ; F )A vector space (over a field F) consists of a set V along with 2 operations ‘+’ and ‘

’ s.t.(1) For the vector addition + :

v, w, u V a) v + w V ( Closure )b) v + w = w + v ( Commutativity )c) ( v + w ) + u = v + ( w + u ) ( Associativity )d) 0 V s.t. v + 0 = v ( Zero element )e) v V s.t. v v = 0 ( Inverse )

(2) For the scalar multiplication : v, w V and a, b F,a) a v V ( Closure )b) ( a + b ) v = a v + b v ( Distributivity )c) a ( v + w ) = a v + a wd) ( a b ) v = a ( b v ) = a b v ( Associativity )e) 1 v = v (Scalar identity of multiplication)

Expression of force, velocity, gradient,

Subspace A set U is a subspace of a vector space V if

• Every element of U is in V, and• U is a vector space.

Linear Combinations• Consider a set of vectors { v1,….,vn}

• and a set of scalars { a1, …, an}

• A linear combination of the vectors is• a1v1+a2v2+…+anvn

Remark: Vector space = Collection of linear combinations of vectors.

Definition : Span

Let S = { s1 , …, sn | sk ( V,+,R ) } be a set of n vectors in vector space V.

The span of S is the set of all linear combinations of the vectors in S, i.e.,

1

,n

k k k kk

span S c S c

s s R span 0with

Lemma : The span of any subset of a vector space is a subspace.

Proof:

Let S = { s1 , …, sn | sk ( V,+, ) }1 1

,n n

k k k kk k

u v span S

u s v sand

1

n

k k kk

a b au bv

w u v s1

n

k kk

w span S

s ,a b RQED

Note: span S is the smallest vector space containing all members of S.

Example:

Proof:

The problem is tantamount to showing that for all x, y R, unique a,b R s.t.

1 1

1 1

xa b

y

i.e.,a b x

a b y

has a unique solution for arbitrary x & y.

Since 1

2a x y 1

2b x y ,x y R QED

21 1,

1 1span

R

Example:

{ 1+x , 1x } is linearly independent.

Proof:

Let 21 1 0 0 0a x b x x x

→0

0

a b

a b

→

0

0

a

b

Otherwise they are linearly independent.

Example:

Let1 2 3

3 2 4

4 9 18

5 2 4

v v v then S = { v1 ,v2 , v3 } is L.D.

Note: v3-2v2=0

BasisDefinition : Basis

A basis of a vector space V is an ordered set of linearly independent (non-zero) vectors that spans V, i.e. any vector in V can be represented as a linear combination of the basis.

Example 1.2:

2 1,

4 1B

is a basis for R2

B is L.I. :

2 1 0

4 1 0a b

→

2 0

4 0

a b

a b

→0

0

a

b

B spans R2:

2 1

4 1

xa b

y

→

2

4

a b x

a b y

→ 1

22

a y x

b x y

1V

V

One of the most remarkable features of vector spaces is the notion of dimension.

We need one simple result that makes this happen, the basis theorem.

Norms are a way of putting a measure of distance on vector spaces.

The purpose is for the refined analysis of vector spaces from the viewpoint of many applications. It is also to all the comparison of various vectors on the basis of their length.

Norms derived from inner product

:

:

1.

2.

If , we can proof that it is a vector norm.

Firstly, we will proof the

Then, we proof the

With the triangle inequality, we can easily proof that

is a vector norm. we say it is derived from the inner product.

In fact, (i) is a norm derived from the standard inner product:

=

Distance: make sense

Basic concepts:

• vector space, subspace, span,

• linear combination, linearly independent, linear dependent, basis, dimension,

• vector norm, inner product.

Important principles:

• *A span is a subspace.

• *Zero vector is l.d. to all vectors; *Subset of a l.i. set is l.i.; *L.i. vectors can be added to form a basis; *Every basis has the same number of vectors; *Each vector has a unique basis-representation;

• *Every inner product has the Cauchy-Schwarz inequality. * Every inner product can be used to define a vector norm; * Every vector norm is a continuous function; *All vector norms are equivalent;

CONCLUSION

HOMEWORK