Can you please solve me Q.2 part (b) what is the addetive inverse and explain wh

Question

Can you please solve me Q.2 part (b) what is the addetive inverse and explain why "lnear combinations" stay in the sace?

ttp://ccom.ucsd.edu/~idilts/teaching/102-Winter16/ com ucsd.edu GMatrix Algebra From a .. emath litb.ac.in LDU Decomposition of... .P Piazza Careers × File Edit Goto Favorites Help 1. Which of the following are subspaces of Ro? Explain why or why not. (a) All sequences that include infinitely many zeroes. (b) All sequences (zi, r2, ..) with r, 0 from some point onward. (The j might vary from sequence to sequence. (c) All arithmetic progressions: 2j+1 - rj is the same for allj (d) All geometric progressions (x1,kx1, k2x1, ...) allowing all k and r1 2. Explanation question similar to 2.1.5 in the book: (part c from the book is good too, but you don't have to do it. Notice that my question requires more explanation than the problem from the book. (a) Suppose addition in R2 adds an extra 1 to each component, so that (3,1)+(5,0) is (9,2) instead of (8,1). If scalar multiplication is unchanged, which rules of a vector space are broken? Explain why they are broken with sentences and an example. (b) Explain why the set of all positive real numbers, with x + y and cr redefined to equal the usual ry and r", is a vector space. You don't need to explicitly check each of the 8 properties, but do explain why "linear combinations" stay in the space. Also, explain: What is the "zero vector"? For an vector x, what is the "additive inverse", -r? 11:53 AM I'm Cortana. Ask me anything 0 1/17/2016

Explanation / Answer

By this new definition,

i) commutativity is preserved

ii) associativity is preserved

iii) Identity is not 0 as

(x,y) +(0,0) = (x+1, y+1) not equal to (0,0)+(x,y)

Hence identity element would be (-1,-1)

so that (x,y) + (-1,-1) = (x,y) = (-1,-1)+(x,y)

iv) Inverse of (x,y) would be (-x-2, -y-2)

Hence still a vector space with identity and inverse changed

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x+y = xy and cx = xc

Here Addition is closed and sclar multiplication is closed

x+y = y+x and x+y+z = xyz = x+(y+z)

Hence commutative and associative

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Identity element is 1, since x+1= x = 1+x

Inverse of x is 1/x since x+1/x =1 = 1/x+1

For any vector a and two scalars c and d

c(da) = c(ad) = adc = (cd)a

c(a+b) =(a+b)c

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