Vectors in Spaces | Engineering Mathematics - Civil Engineering (CE) PDF Download

 Page 1


Chapter 3. Vector spaces
1/22
Page 2


Chapter 3. Vector spaces
1/22
Linear combinations
Suppose thatv
1
,v
2
,...,v
n
andv are vectors inR
m
.
De?nition 3.1 – Linear combination
We say that v is a linear combination of v
1
,v
2
,...,v
n
, if there exist
scalars x
1
,x
2
,...,x
n
such thatv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
.
Geometrically, the linear combinations of a nonzero vector form a line.
The linear combinations of two nonzero vectors form a plane, unless
the two vectors are collinear, in which case they form a line.
Theorem 3.2 – Expressing a vector as a linear combination
Let B denote the matrix whose columns are the vectorsv
1
,v
2
,...,v
n
.
Expressingv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
as a linear combination of the
given vectors is then equivalent to solving the linear system Bx =v.
2/22
Page 3


Chapter 3. Vector spaces
1/22
Linear combinations
Suppose thatv
1
,v
2
,...,v
n
andv are vectors inR
m
.
De?nition 3.1 – Linear combination
We say that v is a linear combination of v
1
,v
2
,...,v
n
, if there exist
scalars x
1
,x
2
,...,x
n
such thatv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
.
Geometrically, the linear combinations of a nonzero vector form a line.
The linear combinations of two nonzero vectors form a plane, unless
the two vectors are collinear, in which case they form a line.
Theorem 3.2 – Expressing a vector as a linear combination
Let B denote the matrix whose columns are the vectorsv
1
,v
2
,...,v
n
.
Expressingv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
as a linear combination of the
given vectors is then equivalent to solving the linear system Bx =v.
2/22
Linear combinations: Example
We expressv as a linear combination ofv
1
,v
2
,v
3
in the case that
v
1
=
?
?
1
1
2
?
?
, v
2
=
?
?
1
2
4
?
?
, v
3
=
?
?
3
1
2
?
?
, v =
?
?
3
2
4
?
?
.
Let B denote the matrix whose columns arev
1
,v
2
,v
3
and consider
the linear system Bx =v. Using row reduction, we then get
[Bv] =
?
?
1 1 3 3
1 2 1 2
2 4 2 4
?
?
-?
?
?
1 0 5 4
0 1 -2 -1
0 0 0 0
?
?
.
In particular, the system has in?nitely many solutions given by
x
1
= 4-5x
3
, x
2
=-1+2x
3
.
Let us merely pick one solution, say, the one with x
3
= 0. Then we
get x
1
= 4 and x
2
=-1, sov =x
1
v
1
+x
2
v
2
+x
3
v
3
= 4v
1
-v
2
.
3/22
Page 4


Chapter 3. Vector spaces
1/22
Linear combinations
Suppose thatv
1
,v
2
,...,v
n
andv are vectors inR
m
.
De?nition 3.1 – Linear combination
We say that v is a linear combination of v
1
,v
2
,...,v
n
, if there exist
scalars x
1
,x
2
,...,x
n
such thatv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
.
Geometrically, the linear combinations of a nonzero vector form a line.
The linear combinations of two nonzero vectors form a plane, unless
the two vectors are collinear, in which case they form a line.
Theorem 3.2 – Expressing a vector as a linear combination
Let B denote the matrix whose columns are the vectorsv
1
,v
2
,...,v
n
.
Expressingv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
as a linear combination of the
given vectors is then equivalent to solving the linear system Bx =v.
2/22
Linear combinations: Example
We expressv as a linear combination ofv
1
,v
2
,v
3
in the case that
v
1
=
?
?
1
1
2
?
?
, v
2
=
?
?
1
2
4
?
?
, v
3
=
?
?
3
1
2
?
?
, v =
?
?
3
2
4
?
?
.
Let B denote the matrix whose columns arev
1
,v
2
,v
3
and consider
the linear system Bx =v. Using row reduction, we then get
[Bv] =
?
?
1 1 3 3
1 2 1 2
2 4 2 4
?
?
-?
?
?
1 0 5 4
0 1 -2 -1
0 0 0 0
?
?
.
In particular, the system has in?nitely many solutions given by
x
1
= 4-5x
3
, x
2
=-1+2x
3
.
Let us merely pick one solution, say, the one with x
3
= 0. Then we
get x
1
= 4 and x
2
=-1, sov =x
1
v
1
+x
2
v
2
+x
3
v
3
= 4v
1
-v
2
.
3/22
Linear independence
Suppose thatv
1
,v
2
,...,v
n
are vectors inR
m
.
De?nition 3.3 – Linear independence
We say thatv
1
,v
2
,...,v
n
are linearly independent, if none of them is
a linear combination of the others. This is actually equivalent to saying
that x
1
v
1
+x
2
v
2
+...+x
n
v
n
= 0 implies x
1
=x
2
=... =x
n
= 0.
Theorem 3.4 – Checking linear independence in R
m
The following statements are equivalent for each m×n matrix B.
1 The columns of B are linearly independent.
2 The system Bx = 0 has only the trivial solutionx = 0.
3 The reduced row echelon form of B has a pivot in every column.
4/22
Page 5


Chapter 3. Vector spaces
1/22
Linear combinations
Suppose thatv
1
,v
2
,...,v
n
andv are vectors inR
m
.
De?nition 3.1 – Linear combination
We say that v is a linear combination of v
1
,v
2
,...,v
n
, if there exist
scalars x
1
,x
2
,...,x
n
such thatv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
.
Geometrically, the linear combinations of a nonzero vector form a line.
The linear combinations of two nonzero vectors form a plane, unless
the two vectors are collinear, in which case they form a line.
Theorem 3.2 – Expressing a vector as a linear combination
Let B denote the matrix whose columns are the vectorsv
1
,v
2
,...,v
n
.
Expressingv =x
1
v
1
+x
2
v
2
+...+x
n
v
n
as a linear combination of the
given vectors is then equivalent to solving the linear system Bx =v.
2/22
Linear combinations: Example
We expressv as a linear combination ofv
1
,v
2
,v
3
in the case that
v
1
=
?
?
1
1
2
?
?
, v
2
=
?
?
1
2
4
?
?
, v
3
=
?
?
3
1
2
?
?
, v =
?
?
3
2
4
?
?
.
Let B denote the matrix whose columns arev
1
,v
2
,v
3
and consider
the linear system Bx =v. Using row reduction, we then get
[Bv] =
?
?
1 1 3 3
1 2 1 2
2 4 2 4
?
?
-?
?
?
1 0 5 4
0 1 -2 -1
0 0 0 0
?
?
.
In particular, the system has in?nitely many solutions given by
x
1
= 4-5x
3
, x
2
=-1+2x
3
.
Let us merely pick one solution, say, the one with x
3
= 0. Then we
get x
1
= 4 and x
2
=-1, sov =x
1
v
1
+x
2
v
2
+x
3
v
3
= 4v
1
-v
2
.
3/22
Linear independence
Suppose thatv
1
,v
2
,...,v
n
are vectors inR
m
.
De?nition 3.3 – Linear independence
We say thatv
1
,v
2
,...,v
n
are linearly independent, if none of them is
a linear combination of the others. This is actually equivalent to saying
that x
1
v
1
+x
2
v
2
+...+x
n
v
n
= 0 implies x
1
=x
2
=... =x
n
= 0.
Theorem 3.4 – Checking linear independence in R
m
The following statements are equivalent for each m×n matrix B.
1 The columns of B are linearly independent.
2 The system Bx = 0 has only the trivial solutionx = 0.
3 The reduced row echelon form of B has a pivot in every column.
4/22
Linear independence: Example
We checkv
1
,v
2
,v
3
,v
4
for linear independence in the case that
v
1
=
?
?
1
1
2
?
?
, v
2
=
?
?
1
2
3
?
?
, v
3
=
?
?
4
5
9
?
?
, v
4
=
?
?
3
1
2
?
?
.
Letting B be the matrix whose columns are these vectors, we get
B =
?
?
1 1 4 3
1 2 5 1
2 3 9 2
?
?
-?
?
?
1 0 3 0
0 1 1 0
0 0 0 1
?
?
.
Since the third column does not contain a pivot, we conclude that the
given vectors are not linearly independent.
On the other hand, the 1st, 2nd and 4th columns contain pivots, so
the vectorsv
1
,v
2
,v
4
are linearly independent. As forv
3
, this can be
expressed as a linear combination of the other three vectors.
5/22