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Lecture 3
A fundamental problem in linear algebra is solving vector equations for an unknown
vector
...
3
4
6
Here the unknowns are the scalars x1 and x2
...
3
4
6
In some sense, the vector b is a combination of the vectors v1 and v2
...
Definition 3
...
, vp be vectors in Rn
...
, vp if there exists scalars x1 , x2 ,
...
The scalars in a linear combination are called the coefficients of the linear combination
...
Therefore, we can say that b is a linear combination of v1 , v2 , v3 with coefficients x1 = 3,
x2 = 4, and x3 = −2
...
2
The linear combination problem
The linear combination problem is the following:
21
Vector Equations
Problem: Given vectors v1 ,
...
, vp ?
For example, say you are given the vectors
1
1
2
v1 = 2 , v2 = 1 , v3 = 1
1
0
2
and also
0
b = 1
...
1)
For obvious reasons, equation (3
...
To gain some intuition with the linear combination problem, let’s do an example
by inspection
...
3
...
Are b1 and b2 linear combinations of v1 , v2 ?
Solution
...
On the other hand, by inspection
we have that
−3
0
−3
−3v1 + 7v2 = 0 + 0 = 0 = b2
0
7
7
and thus yes, b2 is a linear combination of v1 , v2 , v3
...
Going forward, we are going to need a systematic way to solve
the linear combination problem that does not rely on pure inspection
...
Consider again the vectors
1
1
2
0
v1 = 2 , v2 = 1 , v3 = 1 , b = 1
...
2)
22
Lecture 3
First, let’s expand the left-hand side of equation (3
...
x1
0
2x3
x1 + 2x3
We want equation (3
...
In
other words, set
x1 + x2 + 2x3
0
2x1 + x2 + x3 = 1
...
3)
x1 + 2x3 = −2
...
We know how to do this
...
3) is
1 1 2 0
[A b] = 2 1 1 1
1 0 2 −2
Notice that the 1st column of A is just v1 , the second column is v2 , and the third column
is v3 , in other words, the augment matrix is
[A b] = v1 v2 v3 b
Applying the row reduction algorithm, the solution is
x1 = 0, x2 = 2, x3 = −1
and thus these coefficients solve the linear combination problem
...
, vp in only one (or unique) way
...
We summarize the previous discussion with the following:
The problem of determining if a given vector b is a linear combination of the vectors
v1 , v2 ,
...
23
Vector Equations
Applying the existence and uniqueness Theorem 2
...
If the linear system is inconsistent then b is not a linear combination of v1 , v2 ,
...
e
...
, xp such that x1 v1 + x2 v2 + · · · + xp vp = b
...
If the linear system is consistent and the solution is unique then b can be written as a
linear combination of v1 , v2 ,
...
3
...
, vp in infinitely many ways
...
4
...
Form the augmented matrix:
v1 v2
The RREF of the augmented matrix is
1 2 7
b = −2 5 4
−5 6 −3
1 0 3
0 1 2
0 0 0
and therefore the solution is x1 = 3 and x2 = 2
...
Therefore, the above linear combination is the only way to
write b as a linear combination of v1 and v2
...
5
...
The augmented matrix of the corresponding linear system is
1 0 2 1
0 1 1 0
...
0 0 0 −1
The last row is inconsistent, and therefore the linear system does not have a solution
...
Example 3
...
Is the vector b = (8, 8, 12) a linear combination of the vectors
2
4
6
v1 = 1 , v2 = 2 , v3 = 4?
3
6
9
Solution
...
3 6 9 12
0 0 0 0
The system is consistent and therefore b is a linear combination of v1 , v2 , v3
...
In terms of the parameter t, the
solution set is
x1 = −8 − 2t
x2 = t
x3 = 4
Choosing any t gives scalars that can be used to write b as a linear combination of v1 , v2 , v3
...
Given vectors
v1 , v2 ,
...
, vp in an obvious way
...
, vp :
0 = 0v1 + 0v2 + · · · + 0vp
...
, vp , for example,
v2 = 0v1 + (1)v2 + 0v3 + · · · + 0vp
...
, vp ,
for example,
xv2 = 0v1 + xv2 + 0v3 + · · · + 0vp
...
, xp , we see that there are infinitely many vectors b
that can be written as a linear combination of v1 , v2 ,
...
The “space” of all the possible
linear combinations of v1 , v2 ,
...
3
...
, vp }, we have been considering the problem of whether
or not a given vector b is a linear combination of {v1 , v2 ,
...
We now take another
point of view and instead consider the idea of generating all vectors that are a linear
combination of {v1 , v2 ,
...
So how do we generate a vector that is guaranteed to be
a linear combination of {v1 , v2 ,
...
3
6
9
12
Thus, by construction, the vector b = (8, 8, 12) is a linear combination of {v1 , v2 , v3 }
...
Definition 3
...
, vp be vectors
...
, vp is called the span of v1 , v2 ,
...
, vp }
...
If b is
a linear combination of v1 , v2 ,
...
, vp },
and we write this as
b ∈ span{v1 , v2 ,
...
26
Lecture 3
By definition, writing that b ∈ span{v1 , v2 ,
...
, xp
such that
x1 v1 + x2 v2 + · · · + xp vp = b
...
, vp } is an infinite set of vectors, it is not necessarily true that
it is the whole space Rn
...
, vp } is just a collection of infinitely many vectors but it has some
geometric structure
...
, vp }
...
1
...
1: The span of a single non-zero vector in R2
...
That is, span{v1 , v2 } = R2
...
In R3 , the span of two vectors v1 , v2 ∈ R3 that are not multiples
of each other is a plane through the origin containing v1 and v2 , see Figure 3
...
In R3 , the
4
3
span{v,w}
2
1
z 0
−1
−2
−3
−4
−4
−3
−2
−1
0
y
1
2
3
4
3
2
1
0
x
−1
−2
−3
−4
Figure 3
...
span of a single vector is a line through the origin, and the span of three vectors that do not
depend on each other (we will make this precise soon) is all of R3
...
8
...
By definition, b is in the span of v1 and v2 if there exists scalars x1 and x2 such
that
x1 v1 + x2 v2 = b,
that is, if b can be written as a linear combination of v1 and v2
...
Using row reduction, the augmented matrix is consistent and there is only one solution (see
Example 3
...
Therefore, yes, b ∈ span{v1 , v2 } and the linear combination is unique
...
9
...
From Example 3
...
Example 3
...
Is the vector b = (8, 8, 12) in the span of the vectors v1 = (2, 1, 3), v2 =
(4, 2, 6), v3 = (6, 4, 9)?
Solution
...
6, we have that
1 2 3 4
REF
v1 v2 v3 b −−→ 0 0 1 4
...
In this case, the solution set
contains d = 1 free parameters and therefore, it is possible to write b as a linear combination
of v1 , v2 , v3 in infinitely many ways
...
11
...
(a) The vector b = (1, 2, 3) is in the span of the set of vectors
2
4
−1
3 , −7 , −5
...
(c) Suppose that the augmented matrix v1 v2 v3 b has an inconsistent row
...
(d) The span of the vectors {v1 , v2 , v3 } (at least one of which is nonzero) contains only the
vectors v1 , v2 , v3 and the zero vector 0
...
, vp
and the problem of writing b as a linear combination of v1 , v2 ,
...
4
...
Definition 4
...
...
,
...
...
...
...
...
...
...
...
...
...
...
am3 · · · amn
xn
am1 x1 + am2 x2 + · · · + amn xn
{z
} | {z }
A
x
For the product Ax to be well-defined, the number of columns of A must equal the number
of components of x
...
2
...
31
The Matrix Equation Ax = b
(a)
A = 1 −1 3 0 ,
2
−4
x=
−3
8
(b)
3 3 −2
,
A=
4 −4 −1
1
x= 0
−1
(c)
−1 1
0
4
1 −2
A=
3 −3 3 ,
0 −2 −3
−1
x= 2
−2
Solution
...
Theorem 4
...
(a) For any vectors u, v in Rn it holds that
A(u + v) = Au + Av
...
Example 4
...
For the given data, verify that the properties of Theorem 4
...
, v=
, u=
A=
−1
3
2 1
4
...
...
...
...
...
=
...
...
1)
33
The Matrix Equation Ax = b
There is an important way to decompose matrix-vector multiplication involving a linear
combination
...
, vn denote the columns of A and consider the
following linear combination:
x1 a11
x2 a12
xn a1n
x1 a21 x2 a22
xn a2n
x1 v1 + x2 v2 + · · · + xn vn =
...
...
...
x1 am1 + x2 am2 + · · · + xn amn
(4
...
1) and (4
...
, xn ) then
Ax = x1 v1 + x2 v2 + · · · + xn vn
...
Example 4
...
Given
−1 1
0
4
1 −2
A=
3 −3 3 ,
0 −2 −3
−1
x = 2 ,
−2
compute Ax in two ways: (1) using the original Definition 4
...
4
...
If A is a m × n matrix then we must have x ∈ Rn and the output
vector Ax is in Rm
...
(⋆)
Equation (⋆) is a matrix equation where the unknown variable is x
...
For example,
34
Lecture 4
suppose that
1 0
−3
A=
, b=
...
Since b does not have equal entries then
the equation Ax = b has no solution
...
As we have seen, the vector
Ax is a linear combination of the columns of A with the coefficients given by the components
of x
...
In Lecture 2, we showed that the linear combination problem can be
a
solved by solving
system of linear equations
...
From now on, a system of linear equations such as
a11 x1
a21 x1
a31 x1
+ a12 x2
+ a22 x2
+ a32 x2
...
+ a13 x3
+ a23 x3
+ a33 x3
...
am1 x1 + am2 x2 + am3 x3
+ · · · + a1n xn
+ · · · + a2n xn
+ · · · + a3n xn
...
...
+ · · · + amn xn
= b1
= b2
= b3
...
= bm
will be written in the compact form
Ax = b
where A is the coefficient matrix of the linear system, b is the output vector, and x is the
unknown vector to be solved for
...
Theorem 4
...
The following statements are equivalent:
(a) The equation Ax = b has a solution
...
(c) The linear system represented by the augmented matrix A b is consistent
...
7
...
A= 1
−3 −7 −6
12
Solution
...
−3 −7 −6 12
0 0 −12 0
Here r = rank(A) = 3 and therefore d = 0, i
...
, no free parameters
...
Thus, the solution to the matrix
equation is unique (no free parameters) and is given by
−11
x= 3
0
Let’s verify that Ax = b:
1
3 −4 −11
−11 + 9 + 0
−2
5
2 3 = −11 + 15 + 0 = 4 = b
Ax = 1
−3 −7 −6
0
33 − 21
12
In other words, b is a linear combination of the columns of A:
1
3
−4
−2
−11 1 + 3 5 + 0 2 = 4
−3
7
−6
12
36
Lecture 4
Example 4
...
Solve, if possible, the matrix equation Ax = b if
3
1 2
...
Row reducing the augmented matrix A b we get
1 2 3
1 2 3
−2R1 +R2
−−−−−→
...
In other words, b is not a linear combination of the columns of A
...
9
...
0 3 6
−1
Solution
...
The
linear system Ax = b has m = 2 equations and n = 3 unknowns
...
The
therefore
augmented matrix A b is
1 −1 2 2
A b =
...
Thus, the solution set to the linear system is
x1 = 53 − 4t
x2 = − 13 − 2t
x3 = t
where t is an arbitrary number
...
Equivalently, b can be written as a linear combination of
the columns of A in infinitely many ways
...
−1
Recall from Definition 3
...
, vp , which we denoted
by span{v1 , v2 ,
...
, vp
...
10
...
The vector b is in span{v1 , v2 } if we can find scalars x1 , x2 such that
x1 v1 + x2 v2 = b
...
then we need to solve the matrix equation Ax = b
...
5
−2 6 4 ∼ 0 1 1
...
5
x=
1
...
5v1 + 1
...
, vp are vectors in Rn and it happens to be true that span{v1 , v2 ,
...
, vp } spans all of Rn
...
6,
we have the following
...
11: Let A ∈ Mm×n be a matrix with columns v1 , v2 ,
...
The following are equivalent:
(a)
(b)
(c)
(d)
span{v1 , v2 ,
...
, vn
...
The rank of A is m
...
12
...
From Theorem 4
...
The RREF of A is
1
2 −1
1 0 0
−3 −4 2 ∼ 0 1 0
5
2
3
0 0 1
which does indeed have r = 3 leading entries
...
Therefore, the vectors v1 , v2 , v3 span R3 :
span{v1 , v2 , v3 } = R3
...
After this lecture you should know the following:
• how to multiply a matrix A with a vector x
• that the product Ax is a linear combination of the columns of A
• how to solve the matrix equation Ax = b if A and b are known
• how to determine if a set of vectors {v1 , v2 ,
...
6)
• when the columns of a matrix A ∈ Mm×n span all of Rm (Theorem 4
...
3
39
The Matrix Equation Ax = b
40
Lecture 5
Lecture 5
Homogeneous and Nonhomogeneous
Systems
5
...
Definition 5
...
A homogeneous system Ax = 0 always has at least one solution, namely, the zero solution
because A0 = 0
...
The zero solution
x = 0 is called the trivial solution and any non-zero solution is called a nontrivial
solution
...
5), we know that a
consistent linear system will have either one solution or infinitely many solutions
...
Recall that the number of parameters in the solution set is d = n − r, where r is the rank
of the coefficient matrix A and n is the number of unknowns
...
2
...
The linear system will have a nontrivial solution if the solution set has at least one
free parameter
...
The rank of the coefficient matrix is r = 2 and thus there will be
d = 3 − 2 = 1 free parameter in the solution set
...
Therefore, the general solution of the
linear system is
x1 = 2t
x2 = 3t
x3 = t
The general solution can be written in vector notation as
2
x = 3 t
1
2
Or more compactly if we let v = 3 then x = vt
...
In other words, the
1
solution set of the linear system is the span of the vector v:
span{v}
...
Hence, to solve a homogeneous system, we can row
reduce the coefficient matrix A only and then set all rows equal to zero when performing
back substitution
...
3
...
2 5 5 2 9
42
Lecture 5
Solution
...
The second row of rref(A) gives the equation
x2 + x3 + x5 = 0
...
From the first row we obtain the equation
x1 + x4 + 2x5 = 0
The unknown x5 has already been assigned, so we must now choose either x1 or x4 to be a
parameter
...
In other words, any solution x is in the span of
v1 , v2 , v3 :
x ∈ span{v1 , v2 , v3 }
...
3 holds in general and is summarized in
the following theorem
...
4: Consider the homogenous linear system Ax = 0, where A ∈ Mm×n and
0 ∈ Rm
...
1
...
2
...
, vd such
that any solution x of the linear system can be written as
x = t1 v1 + t2 v2 + · · · + tp vd
...
, vd :
x ∈ span{v1 , v2 ,
...
A solution x to a homogeneous system written in the form
x = t1 v1 + t2 v2 + · · · + tp vd
is said to be in parametric vector form
...
2
Nonhomogeneous systems
As we have seen, a homogeneous system Ax = 0 is always consistent
...
A
natural question arises: What is the relationship between the solution set of the homogeneous
system Ax = 0 and that of the nonhomogeneous system Ax = b when it is consistent? To
answer this question, suppose that p is a solution to the nonhomogeneous system Ax = b,
that is, Ap = b
...
Now let q = p + v
...
Therefore, Aq = b
...
We have
therefore proved the following theorem
...
5: Suppose that the linear system Ax = b is consistent and let p be a
solution
...
Another way of stating Theorem 5
...
The proof is a simple computation:
Av = A(q − p) = Aq − Ap = b − b = 0
...
, vd span the solution
set of the homogeneous system Ax = 0
...
We saw in Lecture 3 that we can interpret the span of a set of vectors as a plane containing
the zero vector 0
...
Therefore, the solution set of Ax = b is a shift of the span{v1 , v2 ,
...
This is illustrated in Figure 5
...
p + span{v}
p + tv
p
b
b
b
b
b
span{v}
tv
v
0
Figure 5
...
Example 5
...
Write the general solution, in parametric vector form, of the linear system
3x1 + x2 − 9x3 = 2
x1 + x2 − 5x3 = 0
2x1 + x2 − 7x3 = 1
...
The RREF of the augmented
3 1 −9
1 1 −5
2 1 −7
matrix is:
2
1 0 −2 1
0 ∼ 0 1 −3 −1
1
0 0 0
0
The system is consistent and the rank of the coefficient matrix is r = 2
...
Letting x3 = t be the parameter, from the
second row of the RREF we have
x2 = 3t − 1
And from the first row of the RREF we have
x1 = 2t + 1
Therefore, the general solution of the system in parametric vector form is
1
2t + 1
2
x = 3t − 1 = −1 +t 3
0
t
1
| {z } |{z}
p
v
45