Vector Space Axioms Definition Let V be a set on which two...

Vector Space Axioms Definition Let V be a set on which two operations, called vector addition and vector scalar multiplication, have been defined. If u and v are in V, the sum of u and v is denoted by u + v, and if k is a scalar, the scalar multiple of u is denoted by ku. If the following axioms true for all u, v and w in V and for all scalars k and l, then V is called a vector space and its objects in V are called vectors. 1) u + v is in V 2) u + v = v + u 3) (u + v) + w = u + (v + w) 4) 0 + v = v 5) v + (-v) = 0 6) ku is in V 7) k(u + v) = ku + kv 8) (k + l)u = ku + lu 9) k(lu) = (kl)(u) 10) 1v = v QUESTION Consider P4 be a set of all the point on a plane through the origin in R4. The general equation of a plane through the origin in R4 as follow aw + bx + cy + dz = 0 where a, b, c, and d are fixed constant and at least one is not zero. Show that P4 with the standard addition and scalar multiplication is a vector space.

Transcript

00:01 In this question, we have been given that p4 is a set of all the points on a plane through the origin in the space r4.

00:11 The general equation of the plane is given as ax plus by plus cz plus sorry equation is given by aw plus bx plus cy plus d times of z equals to zero.

00:30 Abcd are fixed constant and at least one is not zero.

00:35 We need to show that p4 with a standard addition and the scalar multiplication is a vector space.

00:45 It is a vector space.

00:48 So to p4 for p4 to be a vector space, it must satisfy all the 10 axioms of the vector space.

00:56 So what i will do, i will take some two elements in w first.

01:00 So let w, x, y, z calling this as v1 and i am taking another element.

01:08 It is w1, x1, y1, z1.

01:11 Similarly, we can take another element w2, x2, y2 and z2.

01:16 I am calling it as v1, v2 and they are the elements of this p4 space.

01:23 Correct.

01:24 They belongs to i am calling this p4.

01:31 Correct.

01:32 They belongs to p4.

01:34 Then they will satisfy the given equation of the plane.

01:37 Aw1 plus bx1 plus c times of y1, z times of d times of z1 equals to zero.

01:47 Similarly, aw2 plus bx2 plus cy2 plus dz2.

01:54 This will also be equal to zero.

01:57 Correct.

01:57 If i consider first axiom which is v1 plus v2, i will get that w1 plus w2 comma x1 plus x2 comma y1 plus y2 and then i will get here z1 plus z2.

02:14 This also satisfies this equation a times w1 plus w2.

02:20 This belongs to p4 since a times w1 plus w2 plus b times x1 plus x2 c times y1 plus y2 plus d times z1 plus z2.

02:34 This will also be zero using these two conditions.

02:37 Correct.

02:39 The next property.

02:40 So, this was closer.

02:42 Then we are going to check for commutative.

02:46 So, v1 plus v2 comes out to be w1 plus w2.

02:52 Correct.

02:53 Comma x1 plus x2 comma y1 plus y2 comma z1 plus z2.

03:03 Since addition r is a vector space.

03:07 So, from that and it is also vector space.

03:10 So, using that property of commutating, we can write this as w2 plus w1 comma x2 plus x1 comma y2 plus y1 comma z2 plus z1...

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