Squaring a vector and subspace axioms

[Firepanda]

If a problem I'm doing asks to find

V² where V is a vector

is it simply the dot product of the vector, or the cross product?

The question: Which of the following sets of vectors v = {v₁,...,v_n} in Rⁿ are subspaces of Rⁿ (n>=3)

iii) All v such that V₂=V₁²

He proved it by saying it's not closed under addition (axiom of a subspace)

By (1,1,0) + (-1,1,0) = (0,2,0)

And that concludes his proof, but I'm not seeing what he's proved there at all.

Personally I would have done let a = (a₁,...,a_n) and b = (b₁,...,b_n)

then a+b = (a₁+b₁,...,a_n+b_n)

and so (a+b)²=(a₁+b₁,...,a_n+b_n)²=(a₁+b₁,...,a_n+b_n).(a₁+b₁,...,a_n+b_n)

= a₁²+2a₁b₁+b₁²+.. (dot product)

Which is a scalar field not a vector field, so it's not closed under addition.

Am I wrong?

 

[Dick]

They aren't asking you to square the vector. The v's are the components of the vector (v1,v2,v3). The question is take the subset of these vectors such that v2=v1^2. Like (2,4,1) and the first two vectors in your example.

 

[Firepanda]

They aren't asking you to square the vector.
The v's are the components of the vector (v1,v2,v3). The question is take the subset of these vectors such that v2=v1^2. Like (2,4,1) and the first two vectors in your example.

I was assuming v was a matrix of vectors, and each v₁ etc was a vector. hence '...following sets of vectors v = {v1,...,vn}'

I still don't understand the rest, especially his proof

 

[Dick]

It's not supposed to be a matrix. If V is the subset of R^3 such that v2=v1^2, then (1,1,0) (since 1^1=1) and (-1,1,0) (since (-1)^2=1) are in V. Their sum (0,2,0) is not in V since 0^2 is not equal to 2.

 

[Firepanda]

It's not supposed to be a matrix. If V is the subset of R^3 such that v2=v1^2, then (1,1,0) (since 1^1=1) and (-1,1,0) (since (-1)^2=1) are in V. Their sum (0,2,0) is not in V since 0^2 is not equal to 2.

ok thanks a lot I get it now :)