please help Rather than use the standard definitions of addi

please help

Rather than use the standard definitions of addition and scalar multiplication in R^1, suppose these two operations are defined as follows. (x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1 + x_2, y_1 + y_2, z_1 + z_2) c(x, y, z) = (cx, cy, 0) (x_1, y_1, z_1) + (x_2, y_2, z_2) = (0, 0, 0) x(x, y, z) = (cx, cy, cz) (x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1 + x_2 + 1, y_1 + y_2 + 1, Z_1 + Z_2 + 1) c(x, y, z) = (cx, cy, cz) (x_1, y_1, z_1) + (x_2, y_2, z_2) = (x_1 + x_2 + 1, y_1 + y_2 + 1, z_1 + z_2 + 1) c(x, y, z) = (cx + c - 1, cy + c - 1, cz + c - 1) With these new definition, is R\' a vector space? Justify your answers Prove in full detail that M_22, with the standard operations is a

Solution

Let u = (x₁ , y₁, z₁) ; v = (x₂ , y₂ , z₂)

(a) Now with the addition defined, u+v is in R³, hence it is closed under addition

(0,0,0) is a zero vector such that u+0 = u = 0+u

For every vector u, we have (-x₁, -y₁, -z₁) such that u+(-u) = 0

Also cu is in R³, hence it is closed under scalar multioplication

So R³ is a vector space with this definition

(b) Here as (0,0,0) is in R³, so it is closed under addition

But here we can not find a zero vector which adds to u and gives u because every time , we will get (0,0,0) and hence R³ is not a vector space with these operations

(c) As u+v is in R³, hence R³ is closed under addition

Consider (-1,-1,-1); then (x₁, y₁, z₁) + (-1,-1,-1) = (x₁-1+1, y₁-1+1, z₁-1+1) = (x₁, y₁, z₁)

Hence (-1,-1,-1) is zero vector

Now define v = (-2-x₁, -2-y₁, -2-z₁)

Then u+v = (-1,-1,-1)

So v is additive inverse of u

Also scalar multiplication is defined in R³

Hence R³ is vector space under these operations

(d) As we checked for addition in previouis step is well defined

For scalar multiplication , cu is in R³ and hence it is well defined

Now also c=1, cu = 1(x,y,z) = (x+1-1, y+1-1, z+1-1) = (x,y,z)

Hence 1 is multiplicative identity

So R³ is vector space under these operations