(please solve within 15 minutes I will give thumbs up thanks) Let V = R2. For (vị , v2), (W1, w2) E V and k ER define vector addition by :(vi, v2) O (w1, w2) := (v + w-5, v2 + w2)and scalar multiplication by :ko (vi, v2) := (kvi - 5k + 5, kvz).It can be shown that (V, B. O) is a real vector space (all 10 axioms are true).Computer the following:the sum:(3, 3) (3,-4) =(the scalar multiple:-5 0 (3, 3) =(the zero vector:Oy =(the additive inverse of (x, y):e(x, y) =Verify that the following Axioms are true:• Axiom 4.• Axiom 5.• Axiom 7.* For these three axioms, you must submit the complete solutions in MOODLE. *

The objective of the question is computing the values.

Answered: Let V = R2. For (vị , v2), (W1, w2) E V…

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter4: Vector Spaces

Section4.1: Vector In R^n

Problem 61E: Illustrate properties 110 of Theorem 4.2 for u=(2,1,3,6), v=(1,4,0,1), w=(3,0,2,0), c=5, and d=2....

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Consider the vectors u=(6,2,4) and v=(1,2,0) from Example 10. Without using Theorem 5.9, show that among all the scalar multiples cv of the vector v, the projection of u onto v is the closest to u that is, show that d(u,projvu) is a minimum.
Illustrate properties 110 of Theorem 4.2 for u=(2,1,3,6), v=(1,4,0,1), w=(3,0,2,0), c=5, and d=2. THEOREM 4.2Properties of Vector Addition and Scalar Multiplication in Rn. Let u,v, and w be vectors in Rn, and let c and d be scalars. 1. u+v is vector in Rn. Closure under addition 2. u+v=v+u Commutative property of addition 3. (u+v)+w=u+(v+w) Associative property of addition 4. u+0=u Additive identity property 5. u+(u)=0 Additive inverse property 6. cu is a vector in Rn. Closure under scalar multiplication 7. c(u+v)=cu+cv Distributive property 8. (c+d)u=cu+du Distributive property 9. c(du)=(cd)u Associative property of multiplication 10. 1(u)=u Multiplicative identity property
Prove that in a given vector space V, the additive inverse of a vector is unique.
Prove that if A is similar to B and A is diagonalizable, then B is diagonalizable.
Find a basis for R2 that includes the vector (2,2).
Find a basis B for R3 such that the matrix for the linear transformation T:R3R3, T(x,y,z)=(2x2z,2y2z,3x3z), relative to B is diagonal.
Consider the matrices below. X=[1201],Y=[1032],Z=[3412],W=[3241] Find scalars a,b, and c such that W=aX+bY+cZ. Show that there do not exist scalars a and b such that Z=aX+bY. Show that if aX+bY+cZ=0, then a=b=c=0.
Rather than use the standard definitions of addition and scalar multiplication in R3, let these two operations be defined as shown below. (a) (x1,y1,z1)+(x2,y2,z2)=(x1+x2,y1+y2,z1+z2) c(x,y,z)=(cx,cy,0) (b) (x1,y1,z1)+(x2,y2,z2)=(0,0,0) c(x,y,z)=(cx,cy,cz) (c) (x1,y1,z1)+(x2,y2,z2)=(x1+x2+1,y1+y2+1,z1+z2+1) c(x,y,z)=(cx,cy,cz) (d) (x1,y1,z1)+(x2,y2,z2)=(x1+x2+1,y1+y2+1,z1+z2+1) c(x,y,z)=(cx+c1,cy+c1,cz+c1) With each of these new definitions, is R3 a vector space? Justify your answers.

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