Download • When does a vector equation have a solution? • When does a

MTH 309 Y
Recall:
LECTURE 5.1
TUE 2014.02.11
Vector equations are equivalent to systems of linear equations:
1
2
3
⇥
+
2
4
2
vector
equation
⇥
10
=
3
⇥
2
3
+4
2
1
1
2
2
= 10
= 3
system of
linear equations
Upshot. A vector equation can have either:
• no solutions
• exactly one solution
• infinitely many solutions
Next:
• When does a vector equation have a solution?
• When does a vector equation have a unique solution?
MTH 309 Y
LECTURE 5.2
TUE 2014.02.11
Recall:
A vector w ∈ R� is a linear combination of v1� � � � � v� ∈ R� if there
exist scalars �1� � � � � �� such that
�1v1 + �2v2 + · · · + ��v� = w
Example.
Let
⎡
⎤
1
v1 = ⎣ 0 ⎦ �
0
⎡
⎤
0
v2 = ⎣ 1 ⎦ �
0
⎡
⎤
1
w=⎣1⎦
3
Express w as a linear combination of v1� v2 or show that this is not
possible.
MTH 309 Y
LECTURE 5.3
TUE 2014.02.11
Geometric picture of the last example
w=
v1
⎡
⎤
0
v2 = ⎣ 1 ⎦
0
⎡
v2
⎤
1
v1 = ⎣ 0 ⎦
0
3
⇤
⌅
1
1⇥
3
�1v1 + �2v2
⎡
⎤
�1
�1v1 + �2v2 = ⎣ �2 ⎦
0
MTH 309 Y
LECTURE 5.4
TUE 2014.02.11
Definition. If v1� v2� � � � � v� ∈ R� then
⎧
⎨
⎫
⎬
the set of
Span(v1� v2� � � � � v�) =
all linear combinations
⎩
⎭
�1v1 + �2v2 + · · · + ��v�
Example.
v1 =
⇤
⌅
1
0 ⇥�
0
v2 =
⇤
⌅
0
1⇥
0
MTH 309 Y
LECTURE 5.5
TUE 2014.02.11
Proposition. A vector w is in Span(v1� � � � � v�) iff the vector equation
�1v1 + · · · + ��v� = w
has a solution.
MTH 309 Y
LECTURE 5.6
TUE 2014.02.11
Geometricofinterpretation
of Span
Geometric interpretation
Span
v
u
Span(u)
= {�u | � ∈ R}
Geometric interpretation
of Span
Span(u⇥
v) = { =1u{+u 2|v |
Span(u)
v
Span(u⇥ v) = {
1u
⇥ }2
1R
R}
u
+
2v
Span(u� v) = {�1u + �2v | �1� �2 ∈ R}
|
1⇥ 2
R}
MTH 309 Y
LECTURE 5.7
TUE 2014.02.11
Proposition. For any vectors v1� � � � � v� ∈ R� the zero vector
0 ∈ R� is in Span(v1� � � � � v�).
Geometric interpretation of Span
v
Geometric interpretation of Span
u
Span(u⇥
v) = { =1u{+u 2|v |
Span(u)
v
⇥ }2
1R
R } Span(u⇥ v) = {
1u
u
+
2v
|
1⇥ 2
R}
MTH 309 Y
LECTURE 5.8
TUE 2014.02.11
Linear independence
Definition. A homogenous vector equation is a vector equation
of the form
�1v1 + · · · + ��v� = 0
Definition. Let v1� � � � � v� ∈ R�. The set {v1� � � � � v�} is linearly
independent if the homogenous equation
�1v1 + · · · + ��v� = 0
has only one, trivial solution �1 = 0� � � � � �� = 0.
Otherwise this set is linearly dependent.
MTH 309 Y
LECTURE 5.9
TUE 2014.02.11
Theorem. Let v1� � � � � v� ∈ R�. The set {v1� � � � � v�} is linearly
independent then the equation
�1v1 + · · · + ��v� = w
has only one solution for each w ∈ Span(v1� � � � � v�).
If the set is linearly dependent then this equation has infinitely
many solutions for each w ∈ Span(v1� � � � � v�)
MTH 309 Y
Example.
Let
LECTURE 5.10
⎡
⎤
1
v1 = ⎣ 2 ⎦ �
−2
⎡
⎤
3
v2 = ⎣ 5 ⎦ �
4
TUE 2014.02.11
⎡
⎤
1
v3 = ⎣ 3 ⎦
−12
Check if the set {v1� v2� v3} is linearly independent.
MTH 309 Y
LECTURE 5.11
TUE 2014.02.11
Some properties of linearly (in)dependent sets
1) A set of one vector {v1} is linearly dependent iff v1 = 0.
2) A set of two vectors {v1� v2} is linearly dependent iff one vector
is a scalar multiple of the other vector.
MTH 309 Y
LECTURE 5.12
TUE 2014.02.11
3) If {v1� � � � � v�} is a set of � vectors in R� and � > � then this set
is linearly dependent.
MTH 309 Y
LECTURE 5.13
TUE 2014.02.11
Upshot: How to find the number of solutions of a vector equation
vector equation
�1 v 1 + · · · + �� v � = w
NO
zero
solutions
is w in
Span(v1 � � � � � v� )
?
YES
only one
solution
YES
is the set
{v1 � � � � � v� }
linearly
indep.
?
NO
infinitely many
solution