Understanding Subspaces: Criteria and Examples (R3, Integers, R2)

[Rijad Hadzic]

Homework Statement

Homework Equations

The Attempt at a Solution

So for a subspace 2 criteria have to be met:

its closed under addition, and scalar multiplication.

Now I have the question

"Which of the following subsets of R3 are subspaces? The set of all vectors of the form (a,b,c) where a,b, and c are:

integers"

This subset is closed under addition, but if I multiply by a scalar, I get (ka,kb,kc), which seems like it is closed under scalar multiplication, but if that scalar is negative wouldn't I get: (-ka, -kb, -kc)? which means its actually not closed..

But then I have another question that says

"Consider the sets of vectors of the following form. Determine whether the sets are subspaces of R2 or R3. :"

(a,0)

Again passes addition, but I don't understand why it passes scalar multiplication as well?

if I multiply it by k, I get (ka,0)

which is of the same form as (a,0)

but if k is a negative scalar, wouldn't I get (-ka, 0), which means its not a subspace?

but my book is telling me that it is indeed a subspace.

Does anyone know what I'm missing here?

 

[Rijad Hadzic]

Ok, so for (a,0), maybe its because both (ka,0) and (-ka,0) are in R2? Because R2 isn't only limited to the positive axis?

If that's the reason, then why doesn't this imply to integers as well, since integers include both positive and negative whole numbers and zero..?

 

[member 587159]

Ok, so for (a,0), maybe its because both (ka,0) and (-ka,0) are in R2? Because R2 isn't only limited to the positive axis?

If that's the reason, then why doesn't this imply to integers as well, since integers include both positive and negative whole numbers and zero..?

It helps to write down the sets in set builder notation

For the first question, you have to determine whether the set $A := \{(a,b,c) \mid a,b,c \in \mathbb{Z} \}$ is a subspace of $\mathbb{R^3}$

As you noticed, this is not true. For example, take $(1,0,0) \in A$ and $\sqrt{2} \in \mathbb{R}$. Then $\sqrt{2}(1,0,0) = (\sqrt{2},0,0) \not\in A$. Hence, the set is not closed under scalar multiplication.

For the second question, you then have to determine whether the set $B :=\{(a,0) \mid a \in \mathbb{R} \}$ is a subspace of $\mathbb{R^2}$

You already showed it is closed under addition. Now, pick any scalar $k \in \mathbb{R}$, then $k(a,0) = (\underbrace{ka}_{\in \mathbb{R}},0) \in B$

One last thing: there is another thing you have to verify when you have to check whether a set is a subspace. You must verify whether the set is non empty, which is usually done by quicly observing that the zero vector is element of the set (or not)

 

[Rijad Hadzic]

I thought though that as long as the first two conditions are met (closure under addition and scalar multiplication) that the zero vector will be in the set each time?

 

[member 587159]

I thought though that as long as the first two conditions are met (closure under addition and scalar multiplication) that the zero vector will be in the set each time?

Nope, take the empty set. It is closed under scalar multiplication and addition but it does not contain the zero vector. When one writes a set, you are a priori not sure whether this set is the empty set in disguise (usually seeing this is rather trivial though) or something else.

 

[Rijad Hadzic]

Nope, take the empty set. It is closed under scalar multiplication and addition but it does not contain the zero vector. When one writes a set, you are a priori not sure whether this set is the empty set in disguise (usually seeing this is rather trivial though) or something else.

That makes sense. I find it odd that we didn't touch upon the empty set at all during class. Thanks for spreading your knowledge!