# Proving Vector Space Properties for a Set of Scalar Multiples of [1,3,2] in R3

• hedgie
In summary: Thanks dmoc I didn't notice the careless typo on the vector and thanks for the recommendation on writing that closure holds when I believe I have proven it. So the would this be proper proof for closure under addition:a<1,3,2> + b<1,3,2> = <a1+b1, a3+b3, a2+b2>And this proves closure under addition holds. As well what I typed for the scalar portion was OK?Yes, that is correct.
hedgie

## Homework Statement

Prove that the set of all scalar multiples of the vector [1,3,2] in R3 forms a vector space with th usual operations on 3-vectors.

## Homework Equations

I am struggling to get anywhere on with this on paper. I know intuitively it and since its an intro course its a elementary problem, but am not getting to an actual proof. Of note: I am in transition to proofs in an undergraduate math degree...hence the struggle (I am reading some books on this as well).

## The Attempt at a Solution

I have decent/real written attempt here.

I assume that I need to prove scalar multiplication as well as addition are closed, and then of course the other eight properties. To do so would I just take
Scalar = k
k[1,2,3] = [k, 3k, 2k]
and any general vector in R3 [a,b,c] + [1,3,2] = [a + 1, b + 3, c + 3]

Then go on to the other properties...or am I way off? I feel like this is not even close to correct or sufficient, and obviously not formal enough.

hedgie said:

## Homework Statement

Prove that the set of all scalar multiples of the vector [1,3,2] in R3 forms a vector space with th usual operations on 3-vectors.

## Homework Equations

I am struggling to get anywhere on with this on paper. I know intuitively it and since its an intro course its a elementary problem, but am not getting to an actual proof. Of note: I am in transition to proofs in an undergraduate math degree...hence the struggle (I am reading some books on this as well).

## The Attempt at a Solution

I have decent/real written attempt here.

I assume that I need to prove scalar multiplication as well as addition are closed, and then of course the other eight properties. To do so would I just take
Scalar = k
k[1,2,3] = [k, 3k, 2k]
and any general vector in R3 [a,b,c] + [1,3,2] = [a + 1, b + 3, c + 3]
Your general vector will look like k<1, 3, 2>.

To add two vectors that belong to this set, each of them has to be a scalar multiple of <1, 3, 2>. For example, a<1, 3, 2> and b<1, 3, 2>.
hedgie said:
Then go on to the other properties...or am I way off? I feel like this is not even close to correct or sufficient, and obviously not formal enough.

It sounds like you have the basic idea.

[a + 1, b + 3, c + 3]

You'll probably catch this, but the c + 3 should really be a c + 2. A better way to write closure under addition would be the way Mark44 wrote it.

Also, I recommend adding a confirmation statement such as "Closure under addition holds!" at the end of every specific property proof.

Thank you both! First, I meant to say "I don't have a decent/real written attempt here." I edited it but it must not have saved the 'don't' after the edit. Thanks dmoc I didn't notice the careless typo on the vector and thanks for the recommendation on writing that closure holds when I believe I have proven it.

So the would this be proper proof for closure under addition:
a<1,3,2> + b<1,3,2> = <a1+b1, a3+b3, a2+b2>
And this proves closure under addition holds.

As well what I typed for the scalar portion was OK?

Also am I missing something when it asks "in R3 forms a vector space with the usual operations on 3-vectors."? Do I have to prove something with the three basic row operations as well? Or is it referring to vectors with three variables?

I understand graphically, I believe, I am proving the vector <1,3,2> goes through the origin in either direction for any scalar...is that the proper way to think of it?

Thanks!

For closure under addition, I would suggest

-------------------------

Let a<1,3,2>, b<1,3,2> $$\in$$ R3,

a<1,3,2> + b<1,3,2> = <a1+b1, a3+b3, a2+b2>

<a1+b1, a3+b3, a2+b2> $$\in$$ R3

---------------------------

I don't know how strict your teacher/TA/grader will grade your proofs. Mine encouraged me to explicitly say that something is in the set of R3. By doing that, we then acknowledge that the end result <a1+b1, a3+b3, a2+b2> is in the set R3. This is what we're trying to prove, anyway.

Also am I missing something when it asks "in R3 forms a vector space with the usual operations on 3-vectors."? Do I have to prove something with the three basic row operations as well? Or is it referring to vectors with three variables?

This should just refer to vectors with 3 variables (or "elements") which are going to be in R^3. You shouldn't have to mention anything related to the 3 row operations here.

Last edited:
hedgie said:
Thank you both! First, I meant to say "I don't have a decent/real written attempt here." I edited it but it must not have saved the 'don't' after the edit. Thanks dmoc I didn't notice the careless typo on the vector and thanks for the recommendation on writing that closure holds when I believe I have proven it.

So the would this be proper proof for closure under addition:
a<1,3,2> + b<1,3,2> = <a1+b1, a3+b3, a2+b2>
And this proves closure under addition holds.
To make it crystal clear that the sum of two vectors is also in the set, continue another step, like so.
a<1,3,2> + b<1,3,2> = <a1+b1, a3+b3, a2+b2> = (a + b)<1, 3, 2>.

Also, the notation a1, b1, etc. might not be recognized as meaning a*1, b*1. We normally write expressions like 2x, 3b, and so on with the constant being written first, rather than as x2 or b3.

hedgie said:
As well what I typed for the scalar portion was OK?

Also am I missing something when it asks "in R3 forms a vector space with the usual operations on 3-vectors."? Do I have to prove something with the three basic row operations as well? Or is it referring to vectors with three variables?

I understand graphically, I believe, I am proving the vector <1,3,2> goes through the origin in either direction for any scalar...is that the proper way to think of it?

Thanks!

Thank you both for your time and help. It is greatly appreciated.

That was sloppy on my part to write it as 1a, etc... Thanks for pointing it out. Thanks also for the pointing out I should clarify it is in the set.

Thanks dmoc for your input on noting that I should note that it is in R^3.

Thanks again. This forum is terrific.

## 1. What is a vector space?

A vector space is a mathematical structure that consists of a set of objects called vectors, which can be multiplied by numbers called scalars. This results in another vector that satisfies certain properties, such as closure under addition and scalar multiplication.

## 2. What does it mean for a vector space to be simple?

A simple vector space is one that does not contain any proper subspaces, meaning that there are no smaller vector spaces within the larger vector space. In other words, the only subspaces of a simple vector space are the zero subspace and the entire vector space itself.

## 3. How can I prove that a vector space is simple?

To prove that a vector space is simple, you need to show that the only subspaces it contains are the zero subspace and the entire vector space itself. This can be done by assuming the existence of a non-trivial subspace and then showing that it must equal the entire vector space or the zero subspace.

## 4. What are some examples of simple vector spaces?

One example of a simple vector space is the set of all real numbers, with addition and scalar multiplication defined as usual. Another example is the set of all 2D vectors with real number coordinates, using the standard vector addition and scalar multiplication operations.

## 5. Why is it important to study simple vector spaces?

Studying simple vector spaces can help us understand the fundamental properties and structures of vector spaces. It also has practical applications in areas such as linear algebra, computer graphics, and physics. Additionally, simple vector spaces serve as building blocks for more complex vector spaces, making them an important concept in advanced mathematics.

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