Linear Algebra. Session 5roquesol/Math_304_Fall... · Session 5. Vector Spaces Span: e ective description Let S be a subset of a vector space V If S = fv 1;v 2; ;v ngthen Span(S)

Linear Algebra. Session 5

Dr. Marco A Roque Sol

08/01/2017

Dr. Marco A Roque Sol Linear Algebra. Session 5

Vector Spaces

System of linear equations

a11x1 + a12x2 + · · ·+ a1nxn = b1a21x1 + a22x2 + · · ·+ a2nxn = b2

...am1x1 + am2x2 + · · ·+ amnxn = bm

Any solution (x1, x2, · · · , xn) is an element of Rn.

Theorem

The solution set of the system is a subspace of Rn if and only if allbi = 0


Vector Spaces

Theorem

The solution set of a system of linear equations in n variables is asubspace of Rn if and only if all equations a re homogeneous

Proof

“only if“: the zero vector 0 = (0, 0, ..., 0), which belongs to everysubspace, is a solution only if all equations are homogeneous.

”if”: a system of homogeneous linear equations is equivalent to amatrix equation Ax = 0, where A is the coefficient matrix of thesystem and all vectors are regarded as column vector A0 = 0⇒ 0is a solution ⇒ solution set is not empty.

If Ax = 0 and Ay = 0, then A(x + y) = Ax + Ay = 0 + 0⇒solution set is closed under addition.


Vector Spaces

If Ax = 0 then A(rx) = rAx = r0 = 0⇒, solution set is closedunder scaling.

Thus, the solution set is a subspace of Rn � .

Let V be a vector space and v1, v2, · · · , vn. Consider the set L ofall linear combinations r1v1 + r2v2 + · · ·+ rnvn where r1, r2, · · · , rn

Theorem

L is a subspace of V.


Vector Spaces

Proof

First of all, L is not empty. For example,0 = 0v1 + 0v2 + · · ·+ 0vnbelongs to L

The set L is closed under addition since(r1v1 + r2v2 + · · ·+ rnvn) + (s1v1 + s2v2 + · · ·+ snvn) =(r1 + s1)v1 + (r2 + s2)v2 + · · ·+ (rn + sn)vn

The set L is closed under scalar multiplication sincet(r1v1 + r2v2 + · · ·+ rnvn) = ((tr1)v1 + (tr2)v2 + · · ·+ (trn)vn)

Thus, L is a subspace of V � .


Vector Spaces

Span: implicit definition

Let S be a subset of a vector space V

Definition.

The span of the set S, denoted by Span(S), is the smallestsubspace of V that contains S. That is

Span(S)

For any subspace W ⊂ V one has thatS ⊂W⇒ Span(S) ⊂W

Remark The span of any set S ⊂ V is well defined (namely, it isthe intersection of all subspaces of V that contain S).


Vector Spaces

Span: effective description

Let S be a subset of a vector space V

If S = {v1, v2, · · · , vn} then Span(S) is the set of all linearcombinations r1v1 + r2v2 + · · ·+ rnvn where r1, r2, · · · , rn ∈ R.

If S is an infinite set then Span(S) is the set of all linearcombinations r1u1 + r2u2 + · · ·+ rkuk whereu1,u2, · · · ,uk ∈ S and r1, r2, · · · , rk ∈ R (1 ≤ k).

If S is the empty set then Span(S) = 0


Vector Spaces

Examples of subspaces of M2,2(R)

The Span of

S =

{(1 00 0

),

(0 00 1

)}

consists of all matrices of the form

Span(S) = a

(1 00 0

)+ b

(0 00 1

)=

(a 00 b

)

This is the subspace of diagonal matrices.


Vector Spaces

The Span of

S =

{(1 00 0

),

(0 00 1

),

(0 11 0

)}consists of all matrices of the form

Span(S) = a

(1 00 0

)+ b

(0 00 1

)+ c

(0 11 0

)=

(a cc b

)

This is the subspace of symmetric matrices (AT = A)


Vector Spaces

The Span of

S =

{(0 −11 0

)}


Span(S) = a

(0 −11 0

)=

(0 −aa 0

)

is the subspace of anti-symmetric matrices (AT = A)


Vector Spaces

The Span of

S =

{(1 00 0

),

(0 00 1

),

(0 10 0

)}


Span(S) = a

(1 00 0

)+b

(0 00 1

)+c

(0 10 0

)=

(a c0 b

)

is the subspace of upper triangular matrices (AT = A)


Vector Spaces

The Span of

S =

{(1 00 0

),

(0 00 1

),

(0 10 0

),

(0 01 0

)}


Span(S) = a

(1 00 0

)+b

(0 00 1

)+c

(0 10 0

)+d

(0 01 0

)=

(a bc d

)

is the entire space M2,2(R).


Vector Spaces

Spanning set

Definition. A subset S of a vector space V is called a spanningset for V if Span(S) = V

Examples

Vectors e1 = (1, 0, 0), e2 = (0, 1, 0). and e3 = (0, 0, 1) form aspanning set for R3 as

(x , y , z) = xe1 + ye2 + ze3

Matrices(1 00 0

) (0 00 1

) (0 10 0

)(0 01 0

)

form a spanning set for M2,2(R)


Vector Spaces

Linear Dependence and Independence .

A set of k vectors x(1), ..., x(k) is said to be linearly dependent ifthere exists a set of real or complex numbers c1, ...., ck , at leastone of which is nonzero, such that

c1x(1) + ...+ ckx(k) = 0

On the other hand, if the only set c1, ..., ck for which the aboveequation is satisfied is c1 = c2 = · · · = ck = 0,then the set ofvectors x(1), ..., x(k) is called linearly independent.


Vector Spaces.

Consider now a set of n vectors, each of which has n components ,

x(1) =

x11x21

...xn1

; x(2) =

x12x22

...xn2

; · · · x(n) =

x1nx2n

...xnn

the above equation can be written as .

x11c1 + x12c2 + . . .+ x1ncnx21c1 + x22c2 + . . .+ x2ncn

......

xn1c1 + xn2c2 + . . .+ xnncn

= 0


Vector Spaces

or equivalently

Xc = 0

If X is nonsingular (X−1 exists), then the only solution of is c = 0,but if X is singular (X−1 does not exist) there are nonzerosolutions.

Example 5.l

Determine wether the vectors are linearly indepent or not

x(1) =

12

− 1

; x(2) =

213

; x(3) =

− 41

− 11


Vector Spaces

Solution

To determine whether x (1), x (2), and x (3) are linearly dependent,we seek constants c1, c2, and c3 such that

c1x(1) + c2x(2) + c3x(3) = 0

written in the matrix form 1 2 42 1 1−1 3 −11

c1c2c3

= 0


Vector Spaces

Using elementary row operations on the augmented matrix1 2 −4

∣∣∣ 0

2 1 1∣∣∣ 0

−1 3 −11∣∣∣ 0

(a) Add (−2) times the first row to the second row, and add thefirst row to the third row.

1 2 −4∣∣∣ 0

0 −3 9∣∣∣ 0

0 5 −15∣∣∣ 0


Vector Spaces

(b) Divide the second row by (−3), then add (−5) times thesecond row to the third row.

1 2 −4∣∣∣ 0

0 1 −3∣∣∣ 0

0 0 0∣∣∣ 0

Thus we obtain the equivalent system(

c1 + 2c2 − 4c3 = 0c2 − 3c3 = 0

)= 0


Vector Spaces

Hence, we have 2 equations in 3 unknowns, so one of them, let’ssay c3 will be a free parameter (real number) α and the solution ofthe system is

c3 = α; c2 = 3c3 = 3α; c1 = −2c2 + 4c3 = −2c3 = −2α

c1c2c3

=

− 2α3αα

= α

− 231

Hence, there are infinitely solutions and the set of vectors islinearly dependent.


Vector Spaces

Example 5.2

let v1 = (1, 2, 0), v2 = (3, 1, 1), and v3 = (4,−7, 3). Determinewhether vectors v1, v2, v3 are linearly independent.

Solution

We have to check if there exist r1, r2, r3 ∈ R not all zero such thatr1v1 + r2v2 + r3v3 = 0.

This vector equation is equivalent to a system

r1 + 3r2 + 4r3 = 02r1 + r2 − 7r3 = 00r1 + r2 + 3r3 = 0


Vector Spaces

and the agmented matrix is :

[A|0] =

1 3 4 02 1 −7 00 1 3 0

The reduced row echelon form is

1 0 −5 00 1 3 00 0 0 0

1 3 −5 00 1 3 00 0 0 0

Hence, the third varible r3 is free and therefore the vectorsv1, v2, v3 are linearly dependent.


Vector Spaces

Example 5.3

Vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1) are linearlyindependent

Solution

xe1 + ye2 + ze3 = 0⇒ (x , y , z) = (0, 0, 0)⇒ x = y = z = 0Hence, the set of vectors is linearly dependent.

Example 5.4

The Matrices

E11 =

(1 00 0

)E12 =

(0 10 0

)E21 =

(0 01 0

)E22 =

(0 00 1

)are linearly independent.


Vector Spaces

Solution

aE11 + bE12 + cE21 + dE22 = 0⇒(a bc d

)= 0 =

(0 00 0

)⇒

a = b = c = d = 0. Hence, the set of matrices is linearlydependent.

Example 5.5

Polynomials 1, x , x2, ..., xn are linearly independent

Solutiona0 + a1x + · · ·+ anx

n = 0 = 0 + 0x + · · ·+ 0xn ⇒ a0 = a1 =· · · = an = 0. Hence, the set of polynomial is linearly dependent.


Vector Spaces

Example 5.6

The set of polynomials {p1 = 1, p2 = x − 1, p3 = (x − 1)2} islinearly independent

Solution

a1p1 + a2p2 + a3p3 = a1x + a2(x − 1) + a3(x − 1)2 =(a1 − a2 + a3) + (a2 − 2a3)x + a3x

2 = 0 = 0 + 0x + 0x2 ⇒a1 − a2 + a3 = 0; a2 − 2a3 = 0; a3 = 0⇒ a1 = a2 = a3 = 0

Hence, the set of polynomial is linearly dependent.


Vector Spaces

Example 5.7

Let v1 = (2, 5) and v2 = (1, 3). Show that {v1, v2} is a spanningset for R2.v1 = (2, 5) and v2 = (1, 3)

Solution

Take any vector w = (a, b) ∈ R2. We have to check that thereexist r1, r2 ∈ R such that r1v1 + r2v2 = w ⇐⇒{

2r1 + r2 = a5r1 + 3r2 = b

Coefficient matrix A:

A =

(2 15 3

)⇒ det(A) = 1 6= 0


Vector Spaces

Since the matrix A is invertible, the system has a unique solutionfor any a and b. Thus, Span(v1, v2) = R2

Example 5.8

Let v1 = (2, 5) and v2 = (1, 3). Show that {v1, v2} is a spanningset for R2.

Alternative Solution

First let us show that vectors e1 = (1, 0) and e2 = (0, 1) belong toSpan(v1, v2)

r1v1 + r2v2 = e1 ⇐⇒{

2r1 + r2 = 15r1 + 3r2 = 0

⇐⇒{

r1 = 3r2 = −5


Vector Spaces

r1v1 + r2v2 = e2 ⇐⇒{

2r1 + r2 = 05r1 + 3r2 = 1

⇐⇒{

r1 = −1r2 = 2

Thus 3v1 − 5v2 = e1 and −v1 + 2v2 = e2.Then, for any vector w = (a, b) ∈ R2, we w = ae1 + be2 =a(3v1 − 5v2) + b(−v1 + 2v2) = (3a− b)v1 + (−5a + 2b)v2

Remarks on the alternative solution (Example 5.4) of Example 5.3

Notice that R2 is spanned by vectors e1 = (1, 0) and e2 = (0, 1)since (a, b) = ae1 + be2

This is why we have checked that vectors e1 and e2 belong toSpan(v1, v2). Then,e1, e2 ∈ Span(v1, v2)⇒ Span(e1, e2) ⊂Span(v1, v2)⇒ R2 ⊂ Span(v1, v2)⇒ Span(v1, v2) = R2


Vector Spaces

Theorem

The following conditions are equivalent:

i) Vectors v1, v2, · · · , vk are linearly dependent.ii) One of vector from v1, v2, · · · , vk is a linear combination of theother k − 1 vectors.

proof

POC !!!! �


Documents

Linear Algebra. Session 5roquesol/Math_304_Fall... · Session 5. Vector Spaces Span: e ective description Let S be a subset of a vector space V If S = fv 1;v 2; ;v ngthen Span(S)