Squaring a vector and subspace axioms

In summary, the question is asking which of the following sets of vectors in Rn are subspaces of Rn (n>=3)? The answer is that none of the vectors are subspaces of Rn, because if v2=v1^2 then (1,1,0) and (-1,1,0) are not in V.
  • #1
Firepanda
430
0
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,...,vn} 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,...,an) and b = (b1,...,bn)

then a+b = (a1+b1,...,an+bn)

and so (a+b)2=(a1+b1,...,an+bn)2=(a1+b1,...,an+bn).(a1+b1,...,an+bn)

= a12+2a1b1+b12+.. (dot product)

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

Am I wrong?
 
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  • #2
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.
 
  • #3
Dick said:
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
 
  • #4
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.
 
  • #5
Dick said:
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 :)
 

FAQ: Squaring a vector and subspace axioms

What does it mean to square a vector?

Squaring a vector refers to multiplying the vector by itself. This is done by multiplying each component of the vector by itself and then adding all the resulting values together.

Why is squaring a vector important in mathematics?

Squaring a vector allows us to calculate the magnitude or length of the vector, which is an important concept in many mathematical and scientific applications. It also helps us understand the relationship between vectors and their components.

How do you square a vector mathematically?

To square a vector, you multiply each component of the vector by itself and then add all the resulting values together. For example, if a vector is represented as (a,b,c), its squared value would be (a^2 + b^2 + c^2).

What are the axioms of a subspace?

The axioms of a subspace are the set of rules that define a vector space within a larger vector space. These include closure under addition and scalar multiplication, existence of a zero vector, and existence of additive and multiplicative inverses.

How do the subspace axioms relate to squaring a vector?

The subspace axioms are important in understanding the properties of vectors, including squaring a vector. For example, the existence of a zero vector allows us to square a vector and still maintain its length as 0. The closure under addition and scalar multiplication also play a role in squaring a vector and maintaining its properties.

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