Linear Algebra subspace and nontrivialy proof help

[frogyspond]

I know

 

[Steven G]

I know I am not not fully understanding the nontrivial, subset information any help to beef up or changes would be greatly appreciated. I tried to dut and paste from my word document but my equations did not copy right I hope I rewrote it correctly. Thank you again for any help.

Give an example of a subset of R² that is a nontrivial subspace of R^2, Showing all work.

subset (x, 2x) in subset V x is all real numbers

1. The subset in this examples of V is all real numbers in the horizontal and vertical positions in the first and third quadrants in the Cartesian plane.

1. Proof that the zero vector is an element

0 = 0+0’ P(0’) holds
= 0’+0 commutativity of addition
= 0’ P(0) holds
(0*X₁) + (0*X₂) = O

1. Proof of closure under vector addition

If the X and Y are any two vectors in V, Then X + Y ∈ V.

x = av1 + 2av2 is a real number
y = bv1 + 2bv2 is a real number
x+y = abv + 2avbv

The addition of the original vertices to the new vertices will be in subset V.

1. Proof of closure under scalar multiplication

If any real number is multiplied by any vector in the subset of V, Then aX ∈ V
Variable c are all real numbers

The new vertices is twice the first vertices and with the first vertices in the subset the new vertices is in subset V.

I agree that we are only in quadrant 1 and 3.
We must show that the 0 vector is in V. Now what type of objects are in V? It is very important that you can answer this question! The objects are in R^2, so they are in the form (a,b). So our zero is not 0 as you wrote but rather I claim it is 0 = (0,0)=(0,2*0) and it is in V. Now for any (x,2x) in V, we have (x,2x) + (0,0) = x+0, 2x+0)=(x,2x) and (0,0) is our 0 in V.
Is V closed? Let, (a,2a) and (b,2b) be in V. Then (a,2a) + (b,2b)= (a+b, 2(a+b)) and is in V.
Let k be any real number and X be in V. It is easy to show that kX is in V (Show it!!). Hence V is a subspace of R^2.