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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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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 � .
Dr. Marco A Roque Sol Linear Algebra. Session 5
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).
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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)
Dr. Marco A Roque Sol Linear Algebra. Session 5
Vector Spaces
The Span of
S =
{(0 −11 0
)}
consists of all matrices of the form
Span(S) = a
(0 −11 0
)=
(0 −aa 0
)
is the subspace of anti-symmetric matrices (AT = A)
Dr. Marco A Roque Sol Linear Algebra. Session 5
Vector Spaces
The Span of
S =
{(1 00 0
),
(0 00 1
),
(0 10 0
)}
consists of all matrices of the form
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)
Dr. Marco A Roque Sol Linear Algebra. Session 5
Vector Spaces
The Span of
S =
{(1 00 0
),
(0 00 1
),
(0 10 0
),
(0 01 0
)}
consists of all matrices of the form
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).
Dr. Marco A Roque Sol Linear Algebra. Session 5
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)
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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.
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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
Dr. Marco A Roque Sol Linear Algebra. Session 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
Dr. Marco A Roque Sol Linear Algebra. Session 5
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 !!!! �
Dr. Marco A Roque Sol Linear Algebra. Session 5