Is the following subset a subspace?

[Ryker]

Homework Statement

Determine if the following subset of R_n is a subspace: all vectors <a₁, a₂, ... , a_n>, such that a₁ = 1.

Homework Equations

The Attempt at a Solution

I'm going through the Linear Algebra: An Introductory Approach by Curtis and found this thing. I can't quite get around the fact on why this isn't a subspace. Can someone explain perhaps? How do I even start doing this? I tried setting up two different vectors, the first one <k, k₂a₂, ... , k_na_n> and the second one <l, l₂a₂, ... , l_na_n>, and then adding them up. But I don't even know what I'm supposed to be looking for.

edit: Or do k and l need to be 1 here? Because I guess then it's not hard seeing that multiplying such a vector by a scalar would result in a vector whose first coordinate isn't one, and that adding two, whose first coordinates are 1, amounts to the first one being 2. Hence, this would not be a subspace, because these resultant vectors would lie outside of our conditions.

Still, even if this is the case, this one is really basic, and I'm still having trouble getting my head around how to tackle more complicated cases.

 

[Inferior89]

There is one important property of a subspace that you need to have that you don't in this example, or rather there is a specific vector that you need to have in the subset for it to be a subspace.

 

[silvermane]

Homework Statement

Determine if the following subset of R_n is a subspace: all vectors <a₁, a₂, ... , a_n>, such that a₁ = 1.

Homework Equations

The Attempt at a Solution

I'm going through the Linear Algebra: An Introductory Approach by Curtis and found this thing. I can't quite get around the fact on why this isn't a subspace. Can someone explain perhaps? How do I even start doing this? I tried setting up two different vectors, the first one <k, k₂a₂, ... , k_na_n> and the second one <l, l₂a₂, ... , l_na_n>, and then adding them up. But I don't even know what I'm supposed to be looking for.

edit: Or do k and l need to be 1 here? Because I guess then it's not hard seeing that multiplying such a vector by a scalar would result in a vector whose first coordinate isn't one, and that adding two, whose first coordinates are 1, amounts to the first one being 2. Hence, this would not be a subspace, because these resultant vectors would lie outside of our conditions.

Still, even if this is the case, this one is really basic, and I'm still having trouble getting my head around how to tackle more complicated cases.

Well, you have to satisfy some properties:

1.) does it have an identity element?
2.) is it a subspace under linear superposition?
3.) does it have a "zero" element?
4.) does it have a multiplicitive inverse?
etc...

If ONE of those properties is NOT satisfied, then the set of vectors is not a subspace. Check through and see which property doesn't apply for the given set of vectors. Just reply if you need more clarification. :)

Hint: I've stated one that doesn't apply to your subspace! ;)

 

[silvermane]

With respect to your edit, try to simplify your reasoning. Start with commonly broken rules, such as if the space has a "zero" vector. As in your example above, its really easy to see that you will never have a vector (0,0,0,...,0) in your space (since you need a 1 where a1 is).
Then work to harder reasonings, such as linear superposition. Hope this tip helps you!

 

[Ryker]

3.) does it have a "zero" element?

Hint: I've stated one that doesn't apply to your subspace! ;)

This one? So basically a subspace has to satisfy everything a field does, plus those extra two conditions I listed (a + b and ka are elements of subspace)?

edit: OK, now that I think about it, those two "extra" conditions are basically addition and multiplication (except you multiply by a scalar, which isn't a part of the vector space, here), so are they already included in conditions for a field? This book isn't really user friendly, to be honest.

 

[silvermane]

This one? So basically a subspace has to satisfy everything a field does, plus those extra two conditions I listed (a + b and ka are elements of subspace)?

edit: OK, now that I think about it, those two "extra" conditions are basically addition and multiplication (except you multiply by a scalar, which isn't a part of the vector space, here), so are they already included in conditions for a field? This book isn't really user friendly, to be honest.

Yes, you are correct. :)

Explicitly stated, there's the rule of linear superposition where we have x,y (two vectors in a given subspace) and for the subspace to be a subspace, they must satisfy ax + by to be in the subspace as well.

Then we have 8 properties that all vector spaces satisfy:
1.) associativity of addition: u + (v+w) = (u+v) + w
2.) commutativity of addition: v+w = w+v
3.) identity element of addition ("zero" vector of addition): 0 such that v+0=v
4.) Inverse element of addition: for every v, there exists a w such that v+w=0
5.) distributivity of scalar multiplication (with respect to vector addition): a(v+w) = av+aw
6.) distributivity of scalar multiplication (with respect to field addition): (a+b)v = av+bv
7.) compativility of scalar multiplication (with respect to field multiplication): a(bv) = (ab)v
8.) identity element of multiplication (scalar): 1v=v where 1 is the appropriate identity in F.

I suggest knowing all these properties when you decide if something is a vector space or not. Most of all, understand why these are properties of vector spaces. It's hard to get at first, but after a good amount of practice, hopefully it comes easier over time. :)

 

[Ryker]

Explicitly stated, there's the rule of linear superposition where we have x,y (two vectors in a given subspace) and for the subspace to be a subspace, they must satisfy ax + by to be in the subspace as well.

And that must be true for any a and b, right? I guess this then basically seems those two conditions I mentioned above put together into one if I'm not mistaken.

Then we have 8 properties that all vector spaces satisfy:
1.) associativity of addition: u + (v+w) = (u+v) + w
2.) commutativity of addition: v+w = w+v
3.) identity element of addition ("zero" vector of addition): 0 such that v+0=v
4.) Inverse element of addition: for every v, there exists a w such that v+w=0
5.) distributivity of scalar multiplication (with respect to vector addition): a(v+w) = av+aw
6.) distributivity of scalar multiplication (with respect to field addition): (a+b)v = av+bv
7.) compativility of scalar multiplication (with respect to field multiplication): a(bv) = (ab)v
8.) identity element of multiplication (scalar): 1v=v where 1 is the appropriate identity in F.

I suggest knowing all these properties when you decide if something is a vector space or not. Most of all, understand why these are properties of vector spaces. It's hard to get at first, but after a good amount of practice, hopefully it comes easier over time. :)

Thanks, I see this is all mentioned in the book, as well, but first vector spaces were defined, and then in the next chapter subspaces are covered. And well, again, it is mentioned that if those extra conditions I mentioned are to be satisfied, those 8 you mentioned need to be, as well. But the examples and everything that's done in the book is kind of confusing, though I don't know if it's just me. With the other Maths class I'm taking, Honours Calculus, we only have lecture notes posted online and they are so much clearer when compared to this that it's ridiculous. And in this course, we haven't actually covered subspaces and vector spaces yet, but we have done fields and stuff. But since vectors and subspaces are in the beginning chapters of the book, whereas some of the stuff we covered is in subsequent ones, I figured I'd start at the beginning and work my way through it. But yeah, it's hell. Sometimes I'm just looking at stuff, and it makes no sense whatsoever to me, and it gets me frustrated really quickly as I'm not used to finding a math-related thing so incomprehensible.

 

[HallsofIvy]

This one? So basically a subspace has to satisfy everything a field does, plus those extra two conditions I listed (a + b and ka are elements of subspace)?

Probably just a typo but a subspace does NOT have to "satisfy everything a field does". It has to satisfy everything a vector space does. And it does not have tow "extra" conditions because a vector space is certainly "closed" under addition and scalar multiplication. The difference with a "subspace" is that since it is already a subset of a vector space, you already know that things like "addition is commutative", "every vector has an additive inverse" are true. The only things you need to show for a subspace is that it is closed under addition and scalar multiplication (that a+ b and ka are in the subset as long as a and b are).

 

[Ryker]

Ah, OK, thanks for the clarification. But then the book also says that the set of polynomial functions on R, where a polynomial function is such that for some fixed set of real numbers a₀, a₁, ..., a_n

$$f(x) = a_{0} + a_{1}x + ... + a_{n}x^{n}$$

is a subspace, as well.

But how can that be if it's a fixed set of real numbers? I mean, just multiplying the polynomial by a scalar changes the first element which doesn't have an x beside it. Ugh, I'm missing something here pretty badly.

 

[Inferior89]

Edit: I might be wrong what I am saying below because it says fixed a_i.. For example x^2 + 1 should not be a subspace because it doesn't include zero. Right?
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Check the criterias for subspace.

Does it have an identity element? a_0 = 1 and the rest of a_i = 0. Ok

Do you still stay in the subset if you multiply with a scalar. Of course, if you multiply a polynomial with a scalar it is still a polynomial. Ok.

Does it include the zero element. Set all a_i to 0. Ok.

Thus it is a subspace.

 

[Ryker]

Ah, OK, I thought a₀ must always stay the same, akin to the original question and a₁ in that. I did see that you always stay within the set of polynomials, but that fixed part confused me.