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Linear-Algebra-Subspaces-1, guaranteed and verified 100% Pass

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Linear-Algebra-Subspaces-1, guaranteed and verified 100% PassLinear-Algebra-Subspaces-1, guaranteed and verified 100% PassLinear-Algebra-Subspaces-1, guaranteed and verified 100% PassLinear-Algebra-Subspaces-1, guaranteed and verified 100% PassLinear-Algebra-Subspaces-1, guaranteed and verified 100% PassLinear-Algebra-Subspaces-1, guaranteed and verified 100% PassLinear-Algebra-Subspaces-1, guaranteed and verified 100% Pass

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Institution
Math
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Math









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Institution
Math
Course
Math

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Uploaded on
January 1, 2025
Number of pages
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Written in
2024/2025
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Class notes
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Auroux, denis
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1


Subspaces


Def. A subset 𝑊 of a vector space 𝑉 is called a subspace of 𝑽 if 𝑊 is a vector space
with the operations of addition and scalar multiplication defined in 𝑉.


For any vector space 𝑉, 𝑉 and {0} are subspaces of 𝑉. {0} is called the zero
subspace of 𝑉.


Notice that vector space axioms 1,2,5,6,7, and 8 hold for all vectors in 𝑉 so they
hold for any subset 𝑊 of 𝑉. Thus to show that 𝑊, a subset of 𝑉, is a subspace of 𝑉
we only need to show:
1. 𝑣 + 𝑤 ∈ 𝑊 whenever 𝑣, 𝑤 ∈ 𝑊 (i.e. 𝑊 is closed under addition).
2. 𝑐𝑤 ∈ 𝑊 whenever 𝑐 ∈ ℝ and 𝑤 ∈ 𝑊 (i.e. 𝑊 is closed under scalar mult.)
3. The zero vector of 𝑉 is in 𝑊.
4. Every vector 𝑤 ∈ 𝑊 has an additive inverse in 𝑊.


In fact, we actually only need to show conditions 1 and 2 hold since if 𝑤 ∈ 𝑊 then
−𝑤 ∈ 𝑊 (by condition 2) and 𝑤 + (−𝑤 ) = 0 ∈ 𝑊 (by condition 1).


Ex. Show that 𝑊 = {< 𝑥, 𝑦, 𝑧 >∈ ℝ3 | 𝑧 = 3𝑥 + 𝑦} is a subspace of the vector
space ℝ3 with the usual vector addition and scalar multiplication.


1. Given 𝑤1 , 𝑤2 ∈ 𝑊, where 𝑤1 =< 𝑥1 , 𝑦1 , 3𝑥1 + 𝑦1 >, 𝑤2 =< 𝑥2 , 𝑦2 , 3𝑥2 + 𝑦2 >
then 𝑤1 + 𝑤2 =< 𝑥1 , 𝑦1 , 3𝑥1 + 𝑦1 > +< 𝑥2 , 𝑦2 , 3𝑥2 + 𝑦2 >
=< 𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 , 3𝑥1 + 3𝑥2 + 𝑦1 + 𝑦2 >
=< (𝑥1 + 𝑥2 ), (𝑦1 + 𝑦2 ), 3(𝑥1 + 𝑥2 ) + (𝑦1 + 𝑦2 ) >∈ 𝑊.

, 2


2. If 𝑤 ∈ 𝑊 and 𝑐 ∈ ℝ then
𝑐𝑤 = 𝑐 < 𝑥, 𝑦, 3𝑥 + 𝑦 >
=< 𝑐𝑥, 𝑐𝑦, 𝑐 (3𝑥 + 𝑦) >
=< 𝑐𝑥, 𝑐𝑦, 3(𝑐𝑥) + (𝑐𝑦) >∈ 𝑊 .


Thus 𝑊 is a subspace of ℝ3 .



Ex. Show that 𝑊 = {< 𝑥, 𝑦, 𝑧 >∈ ℝ3 | 𝑧 = 𝑥 + 2𝑦 + 4} is not a subspace of the
vector space ℝ3 with the usual vector addition and scalar multiplication.


Notice that 𝑊 actually violates both conditions we would need for it to be a
subspace of ℝ3 (although it only needs to violate one condition to fail to be a
subspace).
1. If 𝑣 =< 1,2,9 > and 𝑤 =< 2,1,8 > then 𝑣, 𝑤 ∈ 𝑊. However,
𝑣 + 𝑤 =< 1,2,9 > +< 2,1,8 >=< 3, 3, 17 >.
But < 3, 3, 17 > doesn’t satisfy 𝑧 = 𝑥 + 2𝑦 + 4 so 𝑣 + 𝑤 ∉ 𝑊.


2. Notice that 2𝑤 = 2 < 2, 1, 8 >=< 4, 2, 16 >.
But < 4, 2, 16 > doesn’t satisfy 𝑧 = 𝑥 + 2𝑦 + 4 so 2𝑤 ∉ 𝑊.
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