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A series of exercises and solutions related to vector spaces and subspaces in linear algebra. It covers concepts such as vector addition, scalar multiplication, and the properties of subspaces. Detailed explanations and step-by-step solutions, making it a valuable resource for students studying linear algebra.
Typology: Exams
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A. B. V ={∈ x , x − 1 > , where x is areal number } Law 1 If X and Y are any two vectors in V then (^) X + Y ∈ V X= ¿ 3 , 3 − 1 ≥¿ 3 , 2 >¿ Y=¿ 4 , 4 − 1 ≥¿ 4 , 3 >¿ X+Y= ¿ 3 , 2 >+¿ 4 , 3 ≥¿ 3 + 4 , 2 + 3 ≥¿ 7 , 5 >¿ ¿ 7 , 5 >∉ V because 5 ≠ 7 − 1 Since V does not satisfy law 1 by a counterexample V is not a vector space. C.1. Let R= a − b 2 a + 4 b 1 3 a + 5 b a , b ∈ R {[^ ]^ }
[ 3 b − a a ] [ [ ] [ ] [ [ ] [
] R does not contain the zero vector R =^
Because all vectors in set R must contain a 1 in the third coordinate, but there is a zero in the third coordinate of the zero vector. To check if R is closed under vector addition let A ∈ R and B ∈ R such that [
] [
]
[
If R was closed under vector addition, we could add A+B and get another sector in set R. When added, I get
. A+B is not an element of R because the third 2 45 coordinate in set R must be 1, but when added A+B, I get 2 in the third coordinate. Therefore, R is not closed under vector addition. To prove that R is closed under − 2 scalar multiplication I will use the already-found vector A, A =^
and multiply by − 2 the scalar r=3. So, rA =^3 ∗^
rA = [
. 3 102 rA is not in set R because every vector in set R must have a 1 in coordinate three, but rA has a 3 in coordinate three. Therefor (^) rA ∉ R so it is not closed under scalar multiplication. C.2. Let M={ a b a , b ∈ R } ] ] ] [ ] ]
[ ] [ ] 7 + C 2 1 + C^3 −^3 =^0 − 5 − 1 1 0 7 1 − 3 0 0 − 6 − 24 0 − 5 − 1 1 0 σG = σ g 1 f (^) 1 = 3 f (^) 1 − g 1 g 1 σG = σ g 1 σ f (^) 1 σ 3 f (^) 1 − g 1 σ g 1 = using commutativity of real number multiplication switch σ and 3 σG = σ^ g^1 σ^ f^^1 [ 3 σ f (^) 1 − g 1 σ g 1 ] Let σ^ g 1 = g 3 and σ^ f^ 1 = f^ 3 Substituting in for σ g ∧ σ f we^ get^ σG =^ g^3 f^^3 ∈^ M^ This means^ that subset^ M (^1 1) [ 3 f (^) 3 − g 3 g 3 ] is closed under scalar multiplication because it follows the required pattern. Subset M supports all three conditions to be a subspace, so M is a subspace of the vector space of two by two matrices. D. D = (
) D.1. Let set S be the set of column vectors from matrix D S ={ V (^) 1 , V (^) 2 , V (^) 3 } C 1 V (^) 1 + C 2 V (^) 2 + C 3 V (^) 3 = 0 [
] [
] [
] [
] [
− 5 − (^1 1 0) ] [
]
[ ] [ ]
[
] 0 4 0 [
(^0 4 16 0) ] [
] Since there is a row of zeros there is a free variable C 3 Let C 3 = t C 2 =− 4 t C 1 = t Let t= c 1 = 1 c 2 =− 4 C 3 = 1 [
] [
] [
] [
] − 5 [
]
[
]
[
]
[
] 1 − 4 + 3 0 7 − 4 − 3 =^0 − 5 + 4 + 1 0
[
]
2
1 1
3 3 2 3 1 1 0 − 1 6 0 1 4 11 0 0 0 6
Row 3 is a contradiction 0 ≠ 1. There is no solution to the equation therefore vector 2 u¿^3 − 2 is not in Col(D)
[ ] [ ] [
[ 0 1 4 ] x 2 = [
] ] D.3. First, find the reduced row echelon form 1 3 D= [
] 1 1 3 0 − 6 − 24 0 4 16
Next set the reduced row echelon form equal to zero 1 0 − 1 [ x 1 ]
(^0 0 0) x 3 0 x 1 − x 3 = 0 x 2 + 4 x 3 = 0 Let x 3 = t^ then x 1 = t^ ,^ x 2 =−^4 t t − 4 t t
(^1) ] t (^) the null space for matrix D is
− 4 t Where^ t^ is^ a^ real^ number 1 [ [ [ [
] ] ]
[
]
