Linear Algebra 2270
Homework 3 preparation for the quiz on: 06/10/2015
Axioms of real numbers R :
RA1) x
+ y
= y
+ x
RA2)
( x
+ y
) + z
= y
+ ( x
+ z
)
RA3) there is a natural element of addition denoted by 0, such that for any x
∈
R : x
+
0
= x
RA4) there is an additive inverse
− x such that x
+ (− x
) =
0
RA5) x
⋅ y
= y
⋅ x
RA6)
( x
⋅ y
) ⋅ z
= y
⋅ ( x
⋅ z
)
RA7) there is a natural element of multiplication denoted by 1, such that for any x
∈
R : x
⋅
1
= x
RA8) If x
≠
0, there is a multiplicative inverse
1 x
RA9) x
⋅ ( y
+ z
) = x
⋅ y
+ x
⋅ z such that x
⋅ ( 1 x
) =
1
Axioms of a vector space V :
For arbitrary u, v, w
∈
V and for arbitrary c, d
∈
R we have
V1) u
+ v
= v
+ u
V2)
( u
+ v
) + w
= u
+ ( v
+ w
)
V3) there exists a natural element of addition 0
∈
R , which satisfies u
+
0
= u
V4) there exists an additive inverse
(− u
)
, such that u
+ (− u
) =
0
V5) c
⋅ ( u
+ v
) = c
⋅ u
+ c
⋅ v
V6)
( c
+ d
) ⋅ u
= c
⋅ u
+ d
⋅ u
V7) c
⋅ ( d
⋅ u
) = ( c
⋅ d
) ⋅ u
V8) 1
⋅ u
= u
Problems:
1. Row reduce the following augmented matrices and write the solution set in the parametric form , i.e. find vectors q
0
, q
1
, . . . , q k
∈
R n such that the solution set is
{ u
= q
0
+ a
1 q
1
+
. . .
+ a k q k
∶ a
1
, . . . , a k
∈
R
}
⎢⎢
⎡⎢
⎢⎢
1
−
2 0 0
−
1
∣
2
1
−
2 1
−
1 4
∣
1
2
−
4 0 1 5
∣ −
1
⎥⎥
⎤⎥
, ⎢⎢
⎢⎢
⎡⎢
0
1
0
1
−
0
1
2
0
−
6
1
∣
∣ −
1
1 2 1
−
1 4
∣
5
3
⎥⎥
⎤⎥
(1)
2. Check if the following vectors are linearly independent or not. If they are dependent, find a linear combination that is equal to 0 , but has not all coefficients equal to 0. If you decide that the vectors are linearly independent, prove it. In the case of (a) and (b), draw the considered vectors.
(a) v
1
= [
1
2
]
, v
2
= [−
10
−
20
]
(b) v
1
= [
1
2
]
, v
2
= [
1
1
]
1

(c) v
1
=
⎢⎢
⎡⎢
⎢⎢ 2
⎦
1
0
⎤⎥
⎥⎥
⎥⎥
, v
2
=
⎢⎢
⎡⎢
⎢⎢ 1
⎦
1
0
⎤⎥
⎥⎥
⎥⎥
, v
3
=
⎢⎢
⎡⎢
−
1
⎤⎥
0
1
⎥⎥
⎦
(d) v
1
=
⎢⎢
⎡⎢
⎢⎢ 2
⎦
1
1
⎤⎥
⎥⎥
⎥⎥
, v
2
=
⎢⎢
⎡⎢
⎢⎢ 1
⎦
1
0
⎤⎥
⎥⎥
⎥⎥
, v
3
=
⎢⎢
⎡⎢
⎢⎢ 8
⎦
5
3
⎤⎥
⎥⎥
⎥⎥
3. Given the definition of “
+
” and “
⋅
” for the space of functions V
=
R
R
, prove that V1) – V8) are satisfied. You can use the axioms of real numbers as well as the definitions of 0 and
− f : for any x
∈
R 0
( x
) =
0 for any f
∈
R
R
,
− f is such a function that for any x
∈
R
(− f
)( x
) = − f
( x
)
4. Decide whether W is a subspace of V . Prove your claim. In the case of V
=
R
2
, plot the set W .
(a) V
=
R
2
, W
= {[
0 x
] ∈
R
2 ∶ x
∈
R
}
(b) V
=
R
2
, W
= {[
3 x
+
1
] ∈
R
2 ∶ x
∈
R
}
(c) V
=
R
2
, W
= {[ x
1 x
2
] ∈
R
2 ∶ x
1
, x
2
∈
R , x
1
+ x
2
=
0
}
(d) V
=
R
2
, W
= {[ x
1 x
2
] ∈
R
2 ∶ x
1
, x
2
∈
R , x
1
+ x
2
= −
1
}
(e) V
=
R
2
, W
= {
0
}
( W is a set containing only one element, 0 ).
(f) V
=
R
2
, W
= {
0 ,
[
1
0
]}
( W is a set with only two elements).
(g) V
=
R
3
, W
= {
⎡⎢
⎢⎢
⎢⎢ x
1 x
2
(h) V
=
R
3
, W
= {
⎣ x
3
⎡⎢
⎢⎢
⎢⎢ x
1 x x
2
3
⎤⎥
⎥⎥
⎤⎥
⎥⎥
∈
∈
R
R
3
3 ∶
∶ x x
1
1
, x
, x
2
2
, x
, x
3
3
∈
∈
R
R
, x
, x
1
1
+ x
= −
2
1
−
+
2 x x
2
3
}
=
0
}
(i) V
=
R
R
, W is a set of polynomials up to order 2. In other words
W
= { f
∈
R
R ∶ there are three numbers a
0
, a
1
, a
2
∈
R such that for all x
∈
R , f
( x
) = a
0
+ a
1 x
+ a
2 x
2 }
2