GOAL Check whether or not a subset of R^n is a subspace....

GOAL Check whether or not a subset of R^n is a subspace. Apply the concept of linear independence (in terms of Definition 3.2.3, Theorem 3.2.7, and Theorem 3.2.8). Apply the concept of a basis, both in terms of Definition 3.2.3 and in terms of Theorem 3.2.10. Which of the sets W in Exercises 1 through 3 are subspaces of R^3? 1. W = {[x y z] : x + y + z = 1} 2. W = {[x y z] : x ? y ? z} 3. W = {[x + 2y + 3z 4x + 5y + 6z 7x + 8y + 9z] : x, y, z arbitrary constants} 4. Consider the vectors v1, v2, ..., vm in R^n. Is span(v1, ..., vm) necessarily a subspace of R^n? Justify your answer.

Transcript

00:01 So in this question, we're giving four different scenarios, four different sets, and we want to check whether these give subspaces.

00:10 And all right, so the first set, so this is sort of question one, is all the vectors x, y, z, such that x plus y plus z equals one.

00:24 And i want to claim that this is not a subspace of r3.

00:28 And i can just show you something that's breaks the subspace rule.

00:34 But in general, let's say you had two vectors from this set, an x, y, and a z, and i just add it to another vector in this set, in a bc.

00:46 Well, then you're going to get x plus a, y plus b, z plus c.

00:52 But if you go to check this condition, right? like, is it true that the sum of the entries equals one? well, you're going to get x plus a plus y plus b plus z plus c equals two, right? because x, y, and z is one, a, b, and c is one.

01:16 So i add that together, i get two.

01:18 So this does not live, this element, right, this combination, linear combination is not in w, right? so w is not closed under some, so it can't be a subspace.

01:28 All right, so question 2, your w is the vectors x, y, and z, such that x is less than nir equal to y is less than equal to z.

01:42 And again, i want to claim that this is not by subspace.

01:46 For example, consider the vector 1, 2, 3, which is certainly in this, then if i scale this by negative 1, i get the vector negative 1, negative 2, negative 3.

02:04 But negative 1 is not less than equal to negative 2, which is not less than equal to negative 3.

02:10 So this element is not back inside of the space.

02:13 It's not closed under scalar multiplication...

Question

Please give Ace some feedback