Let H= [] : 7x² + 8y² ≤1 scalar-to show that H is not a subspace of R2. H is not a subspace of R2 because the two vectors 1 0 √√8 (Use a comma to separate vectors as needed.) H is not a subspace of R2 because the scalar 4 and the vector which represents the set of points on and inside an ellipse in the xy-plane. Find two specific examples-two vectors, and a vector and a 1 0 √7 show that H is not closed under addition. show that H is not closed under scalar multiplication.
Let H= [] : 7x² + 8y² ≤1 scalar-to show that H is not a subspace of R2. H is not a subspace of R2 because the two vectors 1 0 √√8 (Use a comma to separate vectors as needed.) H is not a subspace of R2 because the scalar 4 and the vector which represents the set of points on and inside an ellipse in the xy-plane. Find two specific examples-two vectors, and a vector and a 1 0 √7 show that H is not closed under addition. show that H is not closed under scalar multiplication.
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter4: Vector Spaces
Section4.3: Subspaces Of Vector Spaces
Problem 49E
Related questions
Question
![ܪܘ
question
which represents the set of points on and inside an ellipse in the xy-plane. Find two specific examples-two vectors, and a vector and a
1
0
√7
show that H is not closed under addition.
show that H is not closed under scalar multiplication.
Let H=
0
: 7x² + 8y²
scalar-to show that H is not a subspace of R².
H is not a subspace of R2 because the two vectors
0
(Use a comma to separate vectors as needed.)
H is not a subspace of R2 because the scalar 4 and the vector
√√8
18](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc068f3d-cd1b-431f-8a49-4a351b92465c%2Faae9c651-e4ae-4422-a3ff-c02bba1d42a3%2Ftpnr9oc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:ܪܘ
question
which represents the set of points on and inside an ellipse in the xy-plane. Find two specific examples-two vectors, and a vector and a
1
0
√7
show that H is not closed under addition.
show that H is not closed under scalar multiplication.
Let H=
0
: 7x² + 8y²
scalar-to show that H is not a subspace of R².
H is not a subspace of R2 because the two vectors
0
(Use a comma to separate vectors as needed.)
H is not a subspace of R2 because the scalar 4 and the vector
√√8
18
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