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i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H The following is NOT subspace. The set of all polynomials of at most second degree, p(t) = ao + a₁t + a₂t² with rational coefficients. Part 1 of 3 Explain WHY it fails to be subspaces. To do this, all you need to do is come up with ONE specific example that makes one of the statements above false. Since most of you are new to this, this question will "guide" you through the steps. i. Is the 0 vector in the set? If so, what is 0 (that is, what are the values for all the coefficients, ao, a1, a2? If not, why is not in the set? Formats B IUX, X² A Edit - Insert

i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H The following is NOT subspace. The set of all polynomials of at most second degree, p(t) = ao + a₁t + a₂t² with rational coefficients. Part 1 of 3 Explain WHY it fails to be subspaces. To do this, all you need to do is come up with ONE specific example that makes one of the statements above false. Since most of you are new to this, this question will "guide" you through the steps. i. Is the 0 vector in the set? If so, what is 0 (that is, what are the values for all the coefficients, ao, a1, a2? If not, why is not in the set? Formats B IUX, X² A Edit - Insert

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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To be a subspace H of V, the following must all be true i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H The following is NOT subspace. The set of all polynomials of at most second degree, p(t) = ao + a₁t + a₂t² with rational coefficients. Explain WHY it fails to be subspaces. To do this, all you need to do is come up with ONE specific example that makes one of the statements above false. Since most of you are new to this, this question will "guide" you through the steps. i. Is the 0 vector in the set? If so, what is 0 (that is, what are the values for all the coefficients, ao, a1, a2? If not, why is o not in the set? Edit Insert Formats Part 1 of 3 BIUX₂ x² 4
Transcribed Image Text:To be a subspace H of V, the following must all be true i. 0 (the zero vector) is in H ii. It is closed under addition. i.e. if u, v are in H, then u + v are in H iii. It is closed under scalar multiplication. i.e if u is in H, then for any constant c, cu is in H The following is NOT subspace. The set of all polynomials of at most second degree, p(t) = ao + a₁t + a₂t² with rational coefficients. Explain WHY it fails to be subspaces. To do this, all you need to do is come up with ONE specific example that makes one of the statements above false. Since most of you are new to this, this question will "guide" you through the steps. i. Is the 0 vector in the set? If so, what is 0 (that is, what are the values for all the coefficients, ao, a1, a2? If not, why is o not in the set? Edit Insert Formats Part 1 of 3 BIUX₂ x² 4
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