Exercise for the break
Which of the following geometric shapes can be the image of a square under a
linear mapping
from
R
2
{\displaystyle {}\mathbb {R} ^{2}}
to
R
2
{\displaystyle {}\mathbb {R} ^{2}}
?
Exercises
Let
K
{\displaystyle {}K}
be a field, and let
V
{\displaystyle {}V}
and
W
{\displaystyle {}W}
be
K
{\displaystyle {}K}
-vector spaces. Let
φ
:
V
⟶
W
{\displaystyle \varphi \colon V\longrightarrow W}
be a linear map. Prove the following facts.
For a linear subspace
S
⊆
V
{\displaystyle {}S\subseteq V}
,
also the image
φ
(
S
)
{\displaystyle {}\varphi (S)}
is a linear subspace of
W
{\displaystyle {}W}
.
In particular, the image
Im
φ
=
φ
(
V
)
{\displaystyle {}\operatorname {Im} \varphi =\varphi (V)\,}
of the map is a subspace of
W
{\displaystyle {}W}
.
For a linear subspace
T
⊆
W
{\displaystyle {}T\subseteq W}
,
also the preimage
φ
−
1
(
T
)
{\displaystyle {}\varphi ^{-1}(T)}
is a linear subspace of
V
{\displaystyle {}V}
.
In particular,
φ
−
1
(
0
)
{\displaystyle {}\varphi ^{-1}(0)}
is a subspace of
V
{\displaystyle {}V}
.
Determine the kernel of the linear map
φ
:
R
4
⟶
R
2
,
{\displaystyle \varphi \colon \mathbb {R} ^{4}\longrightarrow \mathbb {R} ^{2},}
given by the matrix
M
=
(
2
3
0
−
1
4
2
2
5
)
.
{\displaystyle {}M={\begin{pmatrix}2&3&0&-1\\4&2&2&5\end{pmatrix}}\,.}
Determine the kernel of the linear map
R
4
⟶
R
3
,
(
x
y
z
w
)
⟼
(
2
1
5
2
3
−
2
7
−
1
2
−
1
−
4
3
)
(
x
y
z
w
)
.
{\displaystyle \mathbb {R} ^{4}\longrightarrow \mathbb {R} ^{3},{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}\longmapsto {\begin{pmatrix}2&1&5&2\\3&-2&7&-1\\2&-1&-4&3\end{pmatrix}}{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}.}
We consider the
linear mapping
φ
:
R
2
→
R
2
{\displaystyle {}\varphi \colon \mathbb {R} ^{2}\rightarrow \mathbb {R} ^{2}}
given by the
matrix
(
8
4
5
9
)
{\displaystyle {}{\begin{pmatrix}8&4\\5&9\end{pmatrix}}}
.
Determine the
image
of the line defined by the equation
5
x
−
7
y
=
0
.
{\displaystyle {}5x-7y=0\,.}
Determine the
preimage
of the line given by the equation
6
x
−
11
y
=
0
.
{\displaystyle {}6x-11y=0\,.}
Let
K
{\displaystyle {}K}
be a
field ,
and let
V
{\displaystyle {}V}
denote a
finite-dimensional
K
{\displaystyle {}K}
-vector space .
Let
U
⊆
V
{\displaystyle {}U\subseteq V}
be a linear subspace. Show that there exists a
K
{\displaystyle {}K}
-vector space
W
{\displaystyle {}W}
and a surjective
K
{\displaystyle {}K}
-linear mapping
φ
:
V
⟶
W
{\displaystyle \varphi \colon V\longrightarrow W}
such that
U
=
kern
φ
{\displaystyle {}U=\operatorname {kern} \varphi }
holds.
Let a
linear mapping
φ
:
R
3
→
R
2
{\displaystyle {}\varphi \colon \mathbb {R} ^{3}\rightarrow \mathbb {R} ^{2}}
as in
Example 10.11
be given, in order to represent three-dimensional shapes in the plane. Imagine the
preimage
for a point
P
∈
R
2
{\displaystyle {}P\in \mathbb {R} ^{2}}
!
How do the corresponding line equations look like? What points
Q
∈
R
3
{\displaystyle {}Q\in \mathbb {R} ^{3}}
have the same image point as the corner
(
1
,
1
,
1
)
{\displaystyle {}(1,1,1)}
of the unit cube?
Let
A
=
(
3
1
5
−
1
0
2
)
{\displaystyle {}A={\begin{pmatrix}3&1&5\\-1&0&2\end{pmatrix}}}
,
and let
f
:
R
3
⟶
R
2
,
x
⟼
A
⋅
x
{\displaystyle f\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{2},x\longmapsto A\cdot x}
denote the corresponding
linear mapping .
Determine a
basis
and the
dimension
of
kern
f
{\displaystyle {}\operatorname {kern} f}
, and of
Im
f
{\displaystyle {}\operatorname {Im} f}
.
Find a
linear subspace
V
⊆
R
3
{\displaystyle {}V\subseteq \mathbb {R} ^{3}}
such that
R
3
=
V
⊕
kern
f
{\displaystyle {}\mathbb {R} ^{3}=V\oplus \operatorname {kern} f\,}
holds.
Does there also exist a linear subspace
U
⊂
R
2
{\displaystyle {}U\subset \mathbb {R} ^{2}}
,
U
≠
0
{\displaystyle {}U\neq 0}
,
such that
R
2
=
U
⊕
Im
f
{\displaystyle {}\mathbb {R} ^{2}=U\oplus \operatorname {Im} f}
?
Let
V
{\displaystyle {}V}
be a
finite-dimensional
K
{\displaystyle {}K}
-vector space ,
and
f
:
V
→
V
{\displaystyle {}f\colon V\rightarrow V}
be an
endomorphism .
Show that the following statements are equivalent.
kern
f
=
kern
(
f
∘
f
)
{\displaystyle {}\operatorname {kern} f=\operatorname {kern} {\left(f\circ f\right)}}
,
kern
f
∩
Im
f
=
{
0
}
{\displaystyle {}\operatorname {kern} f\cap \operatorname {Im} f=\{0\}}
,
V
=
kern
f
⊕
Im
f
{\displaystyle {}V=\operatorname {kern} f\oplus \operatorname {Im} f}
,
Im
f
=
Im
(
f
∘
f
)
{\displaystyle {}\operatorname {Im} f=\operatorname {Im} {\left(f\circ f\right)}}
.
We consider the
mapping
R
⟶
R
,
x
⟼
x
2
−
1.
{\displaystyle \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{2}-1.}
Show that for this mapping, neither the image of a
linear subspace
U
⊆
R
{\displaystyle {}U\subseteq \mathbb {R} }
must be a linear subspace, nor the preimage of a linear subspace must be a linear subspace.
Let
K
{\displaystyle {}K}
be a
field ,
and let
V
{\displaystyle {}V}
be a
K
{\displaystyle {}K}
-vector space .
For a vector
v
∈
V
{\displaystyle {}v\in V}
,
the
mapping
V
⟶
V
,
x
⟼
x
+
v
,
{\displaystyle V\longrightarrow V,x\longmapsto x+v,}
is called the
translation by the vector
v
{\displaystyle {}v}
.
A translation is in general not a linear mapping, as
0
{\displaystyle {}0}
is not sent to
0
{\displaystyle {}0}
. It is, however, an
affine-linear mapping ,
we will come back later to this concept.
Let
F
⊆
R
n
{\displaystyle {}F\subseteq \mathbb {R} ^{n}}
be a "geometric shape“, for example, a circle or a rectangle in the plane. Let
θ
w
:
R
n
⟶
R
n
,
x
⟼
x
+
w
,
{\displaystyle \theta _{w}\colon \mathbb {R} ^{n}\longrightarrow \mathbb {R} ^{n},x\longmapsto x+w,}
be the
translation
with the vector
w
{\displaystyle {}w}
, and let
F
′
=
θ
w
(
F
)
{\displaystyle {}F'=\theta _{w}(F)}
denote the translated shape. Let
φ
:
R
n
⟶
R
m
{\displaystyle \varphi \colon \mathbb {R} ^{n}\longrightarrow \mathbb {R} ^{m}}
be a
linear mapping .
Show that the
image
φ
(
F
′
)
{\displaystyle {}\varphi (F')}
arises from the image
φ
(
F
)
{\displaystyle {}\varphi (F)}
by a translation.
How does the
graph
of a
linear mapping
f
:
R
⟶
R
,
{\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} ,}
g
:
R
⟶
R
2
,
{\displaystyle g\colon \mathbb {R} \longrightarrow \mathbb {R} ^{2},}
h
:
R
2
⟶
R
{\displaystyle h\colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} }
look like? How can you see in a sketch of the graph the kernel of the map?
Give an example for a
linear mapping
φ
:
R
2
⟶
R
2
{\displaystyle \varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2}}
that is not
injective
but such that its
restriction
Q
2
⟶
R
2
{\displaystyle \mathbb {Q} ^{2}\longrightarrow \mathbb {R} ^{2}}
is injective.
Let
φ
:
R
2
→
R
2
{\displaystyle {}\varphi \colon \mathbb {R} ^{2}\rightarrow \mathbb {R} ^{2}}
be the
linear mapping
determined by the matrix
M
=
(
5
1
2
3
)
{\displaystyle {}M={\begin{pmatrix}5&1\\2&3\end{pmatrix}}}
(with respect to the standard basis).
Determine the describing matrix of
φ
{\displaystyle {}\varphi }
with respect to the basis
(
1
4
)
{\displaystyle {}{\begin{pmatrix}1\\4\end{pmatrix}}}
and
(
4
2
)
{\displaystyle {}{\begin{pmatrix}4\\2\end{pmatrix}}}
.
The telephone companies
A
,
B
{\displaystyle {}A,B}
and
C
{\displaystyle {}C}
compete for a market, where the market customers in a year
j
{\displaystyle {}j}
are given by the customers-tuple
K
j
=
(
a
j
,
b
j
,
c
j
)
{\displaystyle {}K_{j}=(a_{j},b_{j},c_{j})}
(where
a
j
{\displaystyle {}a_{j}}
is the number of customers of
A
{\displaystyle {}A}
in the year
j
{\displaystyle {}j}
etc.). There are customers passing from one provider to another one during a year.
The customers of
A
{\displaystyle {}A}
remain for
80
%
{\displaystyle {}80\%}
with
A
{\displaystyle {}A}
, while
10
%
{\displaystyle {}10\%}
of them goes to
B
{\displaystyle {}B}
, and the same percentage goes to
C
{\displaystyle {}C}
.
The customers of
B
{\displaystyle {}B}
remain for
70
%
{\displaystyle {}70\%}
with
B
{\displaystyle {}B}
, while
10
%
{\displaystyle {}10\%}
of them goes to
A
{\displaystyle {}A}
, and
20
%
{\displaystyle {}20\%}
goes to
C
{\displaystyle {}C}
.
The customers of
C
{\displaystyle {}C}
remain for
50
%
{\displaystyle {}50\%}
with
C
{\displaystyle {}C}
, while
20
%
{\displaystyle {}20\%}
of them goes to
A
{\displaystyle {}A}
, and
30
%
{\displaystyle {}30\%}
goes to
B
{\displaystyle {}B}
.
a) Determine the linear map (i.e. the matrix) that expresses the customers-tuple
K
j
+
1
{\displaystyle {}K_{j+1}}
with respect to
K
j
{\displaystyle {}K_{j}}
.
b) Which customers-tuple arises from the customers-tuple
(
12000
,
10000
,
8000
)
{\displaystyle {}(12000,10000,8000)}
within one year?
c) Which customers-tuple arises from the customers-tuple
(
10000
,
0
,
0
)
{\displaystyle {}(10000,0,0)}
in four years?
The newspapers
A
,
B
{\displaystyle {}A,B}
and
C
{\displaystyle {}C}
sell subscriptions, and they compete in a local market with
100000
{\displaystyle {}100000}
customers. Within a year, one can observe the following movements.
The subscribers of
A
{\displaystyle {}A}
stick with a percentage of
80
%
{\displaystyle {}80\%}
to
A
{\displaystyle {}A}
,
10
%
{\displaystyle {}10\%}
switch to
B
{\displaystyle {}B}
,
5
%
{\displaystyle {}5\%}
switch to
C
{\displaystyle {}C}
and
5
%
{\displaystyle {}5\%}
become nonreaders.
The subscribers of
B
{\displaystyle {}B}
stick with a percentage of
60
%
{\displaystyle {}60\%}
to
B
{\displaystyle {}B}
,
10
%
{\displaystyle {}10\%}
switch to
A
{\displaystyle {}A}
,
20
%
{\displaystyle {}20\%}
switch to
C
{\displaystyle {}C}
and
10
%
{\displaystyle {}10\%}
become nonreaders.
The subscribers of
C
{\displaystyle {}C}
stick with a percentage of
70
%
{\displaystyle {}70\%}
to
C
{\displaystyle {}C}
, nobody switches to
A
{\displaystyle {}A}
,
10
%
{\displaystyle {}10\%}
switch to
B
{\displaystyle {}B}
and
20
%
{\displaystyle {}20\%}
become nonreaders.
Among the nonreaders,
10
%
{\displaystyle {}10\%}
subscribe to
A
,
B
{\displaystyle {}A,B}
or
C
{\displaystyle {}C}
, the rest remains nonreaders.
a) Establish the matrix that describes the movement of customers within a year.
b) In a certain year, each of the three newspapers has
20000
{\displaystyle {}20000}
subscribers and there are
40000
{\displaystyle {}40000}
nonreaders. How does the distribution look like after a year?
c) The three newspapers expand to another city, where there are no newspapers at all so far, but also
100000
{\displaystyle {}100000}
potential customers. How many subscribers does each newspaper have
(and how many nonreaders)
after three years, if the same movements hold in the new city?
We consider the linear map
φ
:
K
3
⟶
K
2
,
(
x
y
z
)
⟼
(
1
2
5
4
1
1
)
(
x
y
z
)
.
{\displaystyle \varphi \colon K^{3}\longrightarrow K^{2},{\begin{pmatrix}x\\y\\z\end{pmatrix}}\longmapsto {\begin{pmatrix}1&2&5\\4&1&1\end{pmatrix}}{\begin{pmatrix}x\\y\\z\end{pmatrix}}.}
Let
U
⊆
K
3
{\displaystyle {}U\subseteq K^{3}}
be the subspace of
K
3
{\displaystyle {}K^{3}}
, defined by the linear equation
2
x
+
3
y
+
4
z
=
0
{\displaystyle {}2x+3y+4z=0}
,
and let
ψ
{\displaystyle {}\psi }
be the restriction of
φ
{\displaystyle {}\varphi }
on
U
{\displaystyle {}U}
. On
U
{\displaystyle {}U}
, there are given vectors of the form
u
=
(
0
,
1
,
a
)
,
v
=
(
1
,
0
,
b
)
and
w
=
(
1
,
c
,
0
)
.
{\displaystyle u=(0,1,a),\,v=(1,0,b){\text{ and }}w=(1,c,0).}
Compute the "change of basis" matrix between the bases
b
1
=
v
,
w
,
b
2
=
u
,
w
and
b
3
=
u
,
v
{\displaystyle {\mathfrak {b}}_{1}=v,w,\,{\mathfrak {b}}_{2}=u,w{\text{ and }}{\mathfrak {b}}_{3}=u,v}
of
U
{\displaystyle {}U}
, and the transformation matrix of
ψ
{\displaystyle {}\psi }
with respect to these three bases
(and the standard basis of
K
2
{\displaystyle {}K^{2}}
).
We consider the families of vectors
u
=
(
1
0
1
0
)
,
(
0
1
1
0
)
,
(
0
0
1
1
)
,
(
1
0
0
1
)
and
v
=
(
1
0
1
)
,
(
0
1
1
)
,
(
0
0
1
)
{\displaystyle {\mathfrak {u}}={\begin{pmatrix}1\\0\\1\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\\1\\0\end{pmatrix}},\,{\begin{pmatrix}0\\0\\1\\1\end{pmatrix}},\,{\begin{pmatrix}1\\0\\0\\1\end{pmatrix}}\,\,{\text{ and }}\,\,{\mathfrak {v}}={\begin{pmatrix}1\\0\\1\end{pmatrix}},\,{\begin{pmatrix}0\\1\\1\end{pmatrix}},\,{\begin{pmatrix}0\\0\\1\end{pmatrix}}}
in
R
4
{\displaystyle {}\mathbb {R} ^{4}}
and in
R
3
{\displaystyle {}\mathbb {R} ^{3}}
. We denote the
standard bases
by
e
4
{\displaystyle {}{\mathfrak {e}}_{4}}
and
e
3
{\displaystyle {}{\mathfrak {e}}_{3}}
. We consider the
linear mapping
f
:
R
4
⟶
R
3
{\displaystyle f\colon \mathbb {R} ^{4}\longrightarrow \mathbb {R} ^{3}}
given by the
matrix
A
=
(
−
2
3
2
3
−
3
5
0
1
−
1
2
−
2
−
2
)
{\displaystyle {}A={\begin{pmatrix}-2&3&2&3\\-3&5&0&1\\-1&2&-2&-2\end{pmatrix}}\,}
with respect to the standard bases. Determine the
describing matrices
of
f
{\displaystyle {}f}
with respect to the bases
a)
e
4
{\displaystyle {}{\mathfrak {e}}_{4}}
and
v
{\displaystyle {}{\mathfrak {v}}}
,
b)
u
{\displaystyle {}{\mathfrak {u}}}
and
e
3
{\displaystyle {}{\mathfrak {e}}_{3}}
,
c)
u
{\displaystyle {}{\mathfrak {u}}}
and
v
{\displaystyle {}{\mathfrak {v}}}
.
Prove
Theorem 9.7
using
Theorem 11.5
.
Let
φ
:
V
⟶
V
{\displaystyle \varphi \colon V\longrightarrow V}
be an
endomorphism
on a
finite-dimensional
K
{\displaystyle {}K}
-vector space
V
{\displaystyle {}V}
. Show that
φ
{\displaystyle {}\varphi }
is a
homothety
if and only if the
describing matrix
is independent of the chosen
basis .
Prove
Corollary 8.10
using
Corollary 11.9
and
Exercise 10.23
.
Prove
Lemma 9.5
with the help of
Lemma 11.10
and
Example 10.13
.
The following exercise uses the concept of an isomorphism for groups, it also needs the concept of cardinality. The result might be somehow confusing.
Let
V
{\displaystyle {}V}
and
W
{\displaystyle {}W}
be
finite-dimensional
real vector spaces ,
both not the zero-space. Are these spaces
isomorphic
as
commutative groups ?
Hand-in-exercises
Sketch the
image
of the pictured circles under the
linear mapping
given by the
matrix
(
2
0
0
3
)
{\displaystyle {}{\begin{pmatrix}2&0\\0&3\end{pmatrix}}}
from
R
2
{\displaystyle {}\mathbb {R} ^{2}}
to itself.
Determine the image and the kernel of the linear map
f
:
R
4
⟶
R
4
,
(
x
1
x
2
x
3
x
4
)
⟼
(
1
3
4
−
1
2
5
7
−
1
−
1
2
3
−
2
−
2
0
0
−
2
)
⋅
(
x
1
x
2
x
3
x
4
)
.
{\displaystyle f\colon \mathbb {R} ^{4}\longrightarrow \mathbb {R} ^{4},{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}}\longmapsto {\begin{pmatrix}1&3&4&-1\\2&5&7&-1\\-1&2&3&-2\\-2&0&0&-2\end{pmatrix}}\cdot {\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}}.}
Let
E
⊆
R
3
{\displaystyle {}E\subseteq \mathbb {R} ^{3}}
be the plane defined by the linear equation
5
x
+
7
y
−
4
z
=
0
{\displaystyle {}5x+7y-4z=0}
.
Determine a linear map
φ
:
R
2
⟶
R
3
,
{\displaystyle \varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{3},}
such that the image of
φ
{\displaystyle {}\varphi }
is equal to
E
{\displaystyle {}E}
.
On the real vector space
G
=
R
4
{\displaystyle {}G=\mathbb {R} ^{4}}
of mulled wines, we consider the two linear maps
π
:
G
⟶
R
,
(
z
n
r
s
)
⟼
8
z
+
9
n
+
5
r
+
s
,
{\displaystyle \pi \colon G\longrightarrow \mathbb {R} ,{\begin{pmatrix}z\\n\\r\\s\end{pmatrix}}\longmapsto 8z+9n+5r+s,}
and
κ
:
G
⟶
R
,
(
z
n
r
s
)
⟼
2
z
+
n
+
4
r
+
8
s
.
{\displaystyle \kappa \colon G\longrightarrow \mathbb {R} ,{\begin{pmatrix}z\\n\\r\\s\end{pmatrix}}\longmapsto 2z+n+4r+8s.}
We consider
π
{\displaystyle {}\pi }
as the price function, and
κ
{\displaystyle {}\kappa }
as the caloric function. Determine a basis for
ker
(
π
)
{\displaystyle {}\operatorname {ker} {\left(\pi \right)}}
, one for
ker
(
κ
)
{\displaystyle {}\operatorname {ker} {\left(\kappa \right)}}
and one for
ker
(
π
×
κ
)
{\displaystyle {}\operatorname {ker} {\left(\pi \times \kappa \right)}}
.[ 1]
An animal population consists of babies (first year), freshers (second year), rockers (third year), mature ones (fourth year), and veterans (fifth year), these animals can not become older. The total stock of these animals in a given year
j
{\displaystyle {}j}
is given by a
5
{\displaystyle {}5}
-tuple
B
j
=
(
b
1
,
j
,
b
2
,
j
,
b
3
,
j
,
b
4
,
j
,
b
5
,
j
)
{\displaystyle {}B_{j}=(b_{1,j},b_{2,j},b_{3,j},b_{4,j},b_{5,j})}
.
During a year,
7
/
8
{\displaystyle {}7/8}
of the babies become freshers,
9
/
10
{\displaystyle {}9/10}
of the freshers become rockers,
5
/
6
{\displaystyle {}5/6}
of the rockers become mature ones, and
2
/
3
{\displaystyle {}2/3}
of the mature ones reach the fifth year.
Babies and freshers can not reproduce yet, then they reach sexual maturity, and
10
{\displaystyle {}10}
rockers generate
5
{\displaystyle {}5}
new pets, and
10
{\displaystyle {}10}
of the mature ones generate
8
{\displaystyle {}8}
new babies, and the babies are born one year later.
a) Determine the linear map (i.e., the matrix) that expresses the total stock
B
j
+
1
{\displaystyle {}B_{j+1}}
with respect to the stock
B
j
{\displaystyle {}B_{j}}
.
b) What will happen to the stock
(
200
,
150
,
100
,
100
,
50
)
{\displaystyle {}(200,150,100,100,50)}
in the next year?
c) What will happen to the stock
(
0
,
0
,
100
,
0
,
0
)
{\displaystyle {}(0,0,100,0,0)}
in five years?
Let
z
∈
C
{\displaystyle {}z\in \mathbb {C} }
be a complex number and let
C
⟶
C
,
w
⟼
z
w
,
{\displaystyle \mathbb {C} \longrightarrow \mathbb {C} ,w\longmapsto zw,}
be the multiplication map, which is a
C
{\displaystyle {}\mathbb {C} }
-linear map. How does the matrix related to this map with respect to the real basis
1
{\displaystyle {}1}
and
i
{\displaystyle {}{\mathrm {i} }}
look like? Let
z
1
{\displaystyle {}z_{1}}
and
z
2
{\displaystyle {}z_{2}}
be complex numbers with corresponding real matrices
M
1
{\displaystyle {}M_{1}}
and
M
2
{\displaystyle {}M_{2}}
.
Prove that the matrix product
M
2
∘
M
1
{\displaystyle {}M_{2}\circ M_{1}}
is the real matrix corresponding to
z
1
z
2
{\displaystyle {}z_{1}z_{2}}
.
Footnotes
↑ Do not mind that there may exist negative numbers. In a mulled wine, of course the ingredients do not enter with a negative coefficient. But if you would like to consider, for example, in how many ways you can change a particular recipe, without changing the total price or the total amount of energy, then the negative entries make sense.