Section III. Reduced Echelon Form
59
2.5 Lemma In an echelon form matrix, no nonzero row is a linear combination
of the other nonzero rows.
Proof Let R be an echelon form matrix and consider its non-~ rows. First
0
observe that if we have a row written as
Section II. Linear Geometry
45
nish
~ +~
uv
start
~
v
~
u
Proof (Well use some algebraic properties of dot product that we have not yet
checked, for instance that ~ (a + ~ ) = ~ a + ~ ~ and that ~ ~ = ~ ~ . See
u~b
u~ ub
uv vu
Exercise 18.) Since all the
Section II. Linear Geometry
43
by our own terms labeling subsets of Rk of the forms cfw_ ~ + t~ | t 2 R and
p
v
cfw_ ~ + t~ + sw | t, s 2 R as lines and planes doesnt make them act like the
p
v
~
lines and planes of our past experience. Rather, we must
Chapter One. Linear Systems
48
(b) Find the angle between the diagonal of the unit cube in R3 and one of the
axes.
(c) Find the angle between the diagonal of the unit cube in Rn and one of the
axes.
(d) What is the limit, as n goes to 1, of the angle betw
Section III. Reduced Echelon Form
III
51
Reduced Echelon Form
After developing the mechanics of Gausss Method, we observed that it can be
done in more than one way. For example, from this matrix
!
22
43
we could derive any of these three echelon form matr
Chapter One. Linear Systems
56
1.15 The proof of Lemma 1.5 contains a reference to the i 6= j condition on the row
combination operation.
(a) Write down a 22 matrix with nonzero entries, and show that the -1 1 + 1
operation is not reversed by 1 1 + 1 .
(b
Section III. Reduced Echelon Form
53
set when it is empty, of course. Example 1.1 and 1.2 show how in a single
element solution set case the single element is in the column of constants. The
next example shows how to read the parametrization of an innite
Section II. Linear Geometry
41
example, this
0
1
01
01
1
1
2
B0C
B1C
B2C
BC
BC
BC
L = cfw_ B C + t B C + s B C | t, s 2 R
@-1A
@0A
@0A
-2
-1
-2
is a degenerate plane because it is actually a line, since the vectors are multiples
of each other and we can
Section II. Linear Geometry
39
The vector that in the description is associated with the parameter t
!
!
!
2
3
1
=
-1
1
2
is the one shown in the picture as having its whole body in the line it is a
direction vector for the line. Note that points on the l
Section I. Solving Linear Systems
29
pattern too. This system illustrates.
x
+ z + w = -1
2x - y
+ w= 3
x + y + 3z + 2w = 1
-21 +2
x
!
-1 +3
+ z + w = -1
-y - 2z - w = 5
y + 2z + w = 2
It has no solutions because the nal two equations conict. But the asso
Section I. Solving Linear Systems
27
can express the leading variable in terms of free ones. We must verify that this
then also holds for the next equation up, the (m - (t + 1)-th equation. For that,
take each variable that leads in a lower equation x`m ,
Section I. Solving Linear Systems
31
of coecients is nonsingular the system has a unique solution for any constants
on the right side: for instance, Gausss Method shows that this system
x + 2y = a
3x + 4y = b
has the unique solution x = b - 2a and y = (3a
Section I. Solving Linear Systems
(a) 2x + y - z = 1
4x - y
=3
(d)
33
(b) x
-z
=1
y + 2z - w = 3
x + 2y + 3z - w = 7
(c)
x- y+ z
=0
y
+w=0
3x - 2y + 3z + w = 0
-y
-w=0
a + 2b + 3c + d - e = 1
3a - b + c + d + e = 3
X 3.16 For the system
2x - y
- w= 3
y +
Section II. Linear Geometry
37
so that we represent the one over and two up arrows shown above in this way.
!
1
2
We often draw the arrow as starting at the origin, and we then say it is in the
canonical position (or natural position or standard position
Section II. Linear Geometry
II
35
Linear Geometry
If you have seen the elements of vectors then this section is an optional
review. However, later work will refer to this material so if this is not a
review then it is not optional.
In the rst section we h