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Rice University **We aren't endorsed by this school

Course

MATH 355

Subject

Mathematics

Date

Sep 24, 2023

Pages

2

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Math 355: Linear Algebra
Monday August 28
1. Let
v
1
=
1
0
1
,
v
2
=
1
1
2
,
v
3
=
3
1
4
, and
v
4
=
−
1
0
−
2
.
Determine via RR if
w
=
5
−
1
2
is in the span of
{
v
1
;
v
2
;
v
3
;
v
4
}
.
We set up and row reduce our augmented matrix:
1
1
3
−
1
5
0
1
1
0
−
1
1
2
4
−
2
2
−→
1
1
3
−
1
5
0
1
1
0
−
1
0
0
0
1
1
.
A solution to this system of equations gives a linear combination of
v
1
, v
2
, v
3
and
v
4
to make
w
. Because
the matrix has no inconsistencies (ie: row of zeros and a nonzero element on the right side), there exists
a solution to the system which gives a linear combination of
v
1
, v
2
, v
3
and
v
4
to make
w
. Therefore,
the vector
w
is in the span of
{
v
1
;
v
2
;
v
3
;
v
4
}
.
2. Now, without doing any additional calculations
, determine if
u
=
301
.
34
−
234
.
8
−
17
.
1
is in Span
{
v
1
;
v
2
;
v
3
;
v
4
}
.
Is there anything special about
u
? What can you say more generally?
For a vector to NOT be in the span of the provided list, there must be some inconsistency in the
augmented matrix. Note that the result of row reduction of the coefficient matrix does not change.
Therefore, we still have no row of all zeros, an inconsistency is not possible. Thus, all vectors in
R
3
,
(which includes
u
), are in Span
{
v
1
;
v
2
;
v
3
;
v
4
}
.
Takeaway: A system of equation is inconsistent if and only if in the RREF for the augmented matrix,
we see a row of zero coefficients followed by a non-zero entry in the augmented column.
3. What happened with the row reduction involving
v
1
, v
2
, v
3
, v
4
that allowed us to make the conclusion
that we did? Let's try to find what the most general statement might be:
Given a list of
n
vectors
{
v
1
;
v
2
;
. . . v
n
}
in
R
m
, we can deduce that the span of
{
v
1
;
v
2
;
. . . v
n
}
is
as long as
when we row reduce
.
Blank 1 :
R
m
Blank 2 :
m
pivots or the number of pivots is the same as the number of rows
Blank 3 : the augmented matrix of
{
v
1
;
v
2
;
. . . v
n
}
Takeaway: When the number of pivots is the same as the number of rows in the RREF of the augment
matrix, there is always a solution to the system of equations.
Check your statement with a TA.
4. Let
u
1
=
1
0
1
,
u
2
=
1
1
2
,
u
3
=
4
1
5
. Determine via RR if
w
=
1
2
3
is in Span
{
u
1
;
u
2
;
u
3
}
.
We augment and row reduce:
1
1
4
1
0
1
1
2
1
2
5
3
−→
1
1
4
1
0
1
1
2
0
0
0
0
.
There are no inconsistencies,
hence
w
is in the span.
1

5. Without doing any additional calculations
, can you tell if every vector in
R
3
is in the span of
{
u
1
;
u
2
;
u
3
}
?
False. The vector
0
0
1
is not in the span.
Please tell your TA when you get to this point, and wait until after the mini-lecture before you start
the next page.
6. We now have a row-reduction condition that tells us exactly when a list of
n
vectors
{
v
1
;
v
2
;
. . . v
n
}
in
R
m
spans
R
m
. Are there conditions on
m
and
n
that tell us that this row-reduction condition MUST
occur? That this row-reduction condition CANNOT occur?
If
n < m
, then
{
v
1
, dots, v
n
}
CANNOT span
m
. This is because there will always be at least one row
of all zero coefficients when row reduced. Indeed, the number of pivot is always smaller than both
m
and
n
.
Since
n < m
, there can only be at most
n
pivots.
Since we have
m
rows and
m
is strictly
greater than
n
, there must be rows that do not contain any pivot. These rows must contain only zeros.
In other words, the row reduction condition:
number of pivots = number of rows
cannot occur. In particular, if the
k
th
row is the row of all zeros in the row reduced augmented matrix,
the vector
0
.
.
.
1
.
.
.
0
where the 1 is in the
k
th
row will not be in the span of the list.
Note well that there is NO condition on
n
that forces the row reduction condition
number of pivots = number of rows
to be true. For example, consider the list of
{
v
1
, v
2
, . . . , v
n
}
where every vector is the zero vector. Then
any nonzero vector is not in the span of our list regardless of the relation of
m
and
n
.
Takeaway: If there are more rows than columns, then the augmented matrix can be inconsistent.
Takeaway: If
k < m
, then any set of
k
vectors in
R
m
cannot span
R
m
.
7. Consider again the vectors
v
1
=
1
0
1
,
v
2
=
1
1
2
,
v
3
=
3
1
4
, and
v
4
=
−
1
0
−
2
from the first
problem on the previous page.
Let
S
denote the set
a
b
0
:
a, b
∈
R
. (If you want, you can think of
S
as the xy-plane inside
R
3
.)
Are the following statements true?
•
Every vector in
S
is in Span
{
v
1
;
v
2
;
v
3
;
v
4
}
.
This is true! There will be no inconsistencies in the augmented matrix for any
a, b
•
The list
{
v
1
;
v
2
;
v
3
;
v
4
}
spans
S
.
This is false! Remember the definition of spans. We say that a list of vectors spans a space
V
if
and only if
V
= Span
{
v
1
, v
2
, v
3
, v
4
}
. Here, Span
{
v
1
, v
2
, v
3
, v
4
}
=
R
3
, and
S
̸
=
R
3
. In summary,
just because every vector in a set is included in the span of a list of vectors, it does not always
make that set the span of the list of vectors, and hence to say that the list spans the set may be
incorrect.
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