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Is V a vector space? and Show that W is not a subspace

  1. Mar 12, 2012 #1
    Hi all

    I hope you guys can help me. I am soo confused with this question: I would really liked a complete answer to this, I have an upcoming exam and I know these two will be on the exam.

    1. Let V be the set of all diagonal 2x2 matrices i.e. V = {[a 0; 0 b] | a, b are real numbers} with addition defined as A ⊕ B = AB, normal scalar multiplication. Prove all 10 axioms. Is V a vector space? If it is not a vector space, which axiom(s) fail?

    2. Let W = {(x,y) ∈ ℝ^2 | x^2 + y^2 <= 1}. Show that W is not a subspace of ℝ^2.

    The 10 axioms are:

    1. If u and v are objects in V, then u+v is in V
    2. u+v = v + u
    3. u+(v+w) = (u+v)+w
    4. There is an object 0 in V, called a zero vector for V, such that 0+u = u+0 = u for all u in V
    5. For each u in V, there is an object -u in V, called a negative of u, such that u+(-u) = (-u)+u = 0
    6. If k is any scalar and u is any object in V, then ku is in V
    7. k(u+v) = ku+kv
    8. (k+m)u = ku+mu
    9. k(mu) = (km)(u)
    10. 1u = u

    Definition of subspace:
    A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication defined on V.

    I may be wrong by this but to prove something is a subspace, we must only show it passes axiom 1 and axiom 6, because all the other axioms are inherited.

    I totally understand all the other review questions except these two, any help or advice will be very much appreciated, thanks.
  2. jcsd
  3. Mar 12, 2012 #2


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    Hey mosawi and welcome to the forums.

    Just a note: for the subspace rule, you have to show that the zero vector is also in the subspace: if it isn't then you don't have a real subspace.

    For these kinds of questions we always ask the poster to provide any work they have done as well as the thinking behind their work and any additional thinking that might be relevant.

    If you don't know where to start, I suggest you just use the definition of what matrix multiplication and scalar multiplication is.

    Chances are if you get the first one or two for the first question, you'll probably know what do to do for the rest.

    So just to start you off, what is u + v is u is a matrix and v is a matrix of a given type? (Hint: U = [a1 0; 0 b1] V = [a2 0; 0 b2] u + v = UV)
  4. Mar 12, 2012 #3
    thanks for the warm welcome and prompt reply chiro. I had to run to the market to get batteries for my calculator, but I have verified axiom 1 at least I think. I uploaded a picture of my work for axiom one, please let me know if I am correct or if I am wrong.

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  5. Mar 12, 2012 #4


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    Yep! Looks good! :)
  6. Mar 12, 2012 #5
    Hmmm, I'm kind of stuck at axiom 4 (0+u=u+0=u). I believe it fails since adding a zero vector to u will result in a zero vector (as addition is defined to multiply). Please take a look at my work and tell me if I am correct. I am almost certain I am right but you guys are smarter than me :) Attachment is below

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  7. Mar 12, 2012 #6


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    Before I give an absolute answer, is there any restriction on the a's and b's in your matrices? Does the a's and b's have to be non-zero or can they be zero as well? This is an important issue in determining the answer to your question.
  8. Mar 12, 2012 #7
    Oh my gosh! I thought of the SAME EXACT THING!!! because a1 or b1 could be zeros (if it were within its domain). I do not know, the question does not state that, it only says:

    Let V be the set of all diagonal 2x2 matrices i.e. V = {[a 0; 0 b] | a, b are real numbers} with addition defined as A ⊕ B = AB, normal scalar multiplication. Prove all 10 axioms. Is V a vector space? If it is not a vector space, which axiom(s) fail?

    "a, b are real numbers" so is zero a real number? This is getting a little tricky...
  9. Mar 12, 2012 #8


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    Ok then this will clearly fail.

    If it had to be non-zero then your 0 vector would be the identity and your -u would be [1/a 0; 0 1/b] which would give [a 0; 0 b] x [1/a 0; 0 1/b] = [1 0; 0 1] but if you have zero you can't do this.
  10. Mar 12, 2012 #9
    That makes perfect sense. I finished the problem, I am sure I am correct but again I need your counsel to reinforce my confidence for tomorrow's exam. I'm glad I did this on my own, otherwise I wouldn't have really understood this. Below are pictures of all my work, please let me know if I am correct, much thanks to you chiro

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  11. Mar 12, 2012 #10


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    I've just read your work and I have a few comments. You got the correct answer, but not for the right reasons.

    When we deal with vector spaces, we are not dealing with standard arithmetic as a general rule.

    When you do your vector space operations, the addition and scalar multiplication must be defined in terms of the actual objects and you can't assume that they will follow the intuition of arithmetic that we have been made so used to all throughout our primary and highschool years.

    I'll give a specific example of where you went wrong in your thinking.

    Let's say we want to prove that a zero vector exists. It turns out that indeed a zero vector does exist, but if we have zero's then we don't have inverses (the -u term) which means that axiom will fail. Let's prove that a zero vector exists:

    Let u = [a1 0; 0 b1], 0 = [a2 0; 0 b2]. We want to find an element of our space that satisfies u + 0 = 0 + u = u. Using our definition of addition which is just matrix multiplication of a special kind of 2x2 matrix we get:

    [a1 0; 0 b1] x [a2 0; 0 b2] = [a2 0; 0 b2] [a1 0; 0 b1] = [a1 0; 0 b1]

    [a1a2 0; 0 b1b2] = [a1a2 0; 0 b1b2] = [a1 0; 0 b1]

    This implies a1a2 = a1 and b1b2 = b1 which implies a2 = 1 and b2 = 1. Since this holds true for any a1 and b2 the zero vector exists and it equals [1 0; 0 1] or the 2x2 identity matrix.

    So the above shows that the zero vector exists with the definition. We can't just use our intuition and think that there will be a 'minus' term, we have to plug in the definition and make sure.

    It seems counter-intuitive, but this is what usually happens in higher mathematics: what you end up doing is getting rid of the idea of treating everything like 'arithmetic' and you end up 'trusting' the proof rather than your intuition. At first it seems kinda pointless at times, but the thing is that although our intuition is often right in some ways and a good place to start, when you get to the higher mathematical stuff like infinite-dimensional spaces, you can get lost because we don't really have intuition for things like infinity.

    But just in case you are wondering, one of the ideas of vector spaces is to basically classify things that 'work like arrows'. Just like a vector, if you can use your addition and scalar multiplication in a way that they 'behave like arrows', then that gives you an idea of what a vector space is.

    The other reason is that it also relates to linear algebra. It turns out that for things that 'act like arrows' we have very well defined theories of how to deal with these kinds of problems so if you end up showing that a particular thing 'acts like a vector in a vector space', then you actually have other tools to help you analyze those systems just like you have the chain rule and the substitution rule to help you analyze calculus problems for functions that can be integrated and differentiated.

    Now let's see why this is not a vector space. Consider the axiom to find an element -u where u + (-u) = 0 for all u in the vector space.

    Let -u = v = [a2 0; 0 b2], u = [a1 0; 0 b1]

    u+v = [a2 0; 0 b2] [a1 0; 0 b1] = [a1a2 0; 0 b1b2] = [1 0; 0 1] since we figured out our zero vector.

    Now a2a1 = 1 and b2b1 = 1 which means a2 = 1/a1 and b2 = 1/b1. But if a1 or b1 is zero then we get an undefined number that is not a real number. This is a contradiction which means the axiom fails.
  12. Mar 12, 2012 #11
    Okay, I see. You're right, I was using a lot of my intuition to prove the axioms (kinda funny how you knew my exact thought process). I guess I need to revisit the vector space chapter in the text. I'm using the anton 10th edition linear algebra book which I find not to be very concise. I need to see problems worked out and explained, perhaps I should look into finding a solutions manual.

    I started problem 2 but I am not sure how to proceed. I tried writing everything down into words so it'll make more sense for me but I am still kind of confused as to how to prove W is not a subspace of ℝ^2. Perhaps you can? Btw, thanks for taking all this time in helping me.

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  13. Mar 12, 2012 #12


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    With this problem, you have to use the definition again. You can't use an argument like that either in undergraduate or when you are working doing mathematical work: it has to be precise enough so that other people see how it is true.

    So we have x^2 + y^2 <= 1 for (x,y) which is in R^2. Now again you have use the definition of addition and scalar multiplication in your space, which unlike the last problem is intuitive: but don't fall into the trap of thinking like this for this kind of thing!

    So lets take U,V as part of the subspace being tested. This means U = (x1,y1) where (x1^2 + y1^2 <= 1) and V = (x2,y2) where (x2^2 + y2^2 <= 1). So if W = U + V we have to test that W = (x3,y3) where x3^2 + y3^2 <= 1. If this holds, then this axiom holds but it's gotta hold for all possible U and V. So let's check:

    W = U+V = (x1+x2,y1+y2). Now we have to have (x1+x2)^2 + (y1+y2)^2 <= 1 for the relationships x1^2 + y1^2 <=1 and x2^2 + y2^2 <= 1. So we have three inequalities and we have to show if we get a contradiction or not. Let's expand the first one out and compare it with the others:

    (x1+x2)^2 + (y1+y2)^2 = x1^2 + x2^2 + 2x1x2 + y1^2 + y2^2 + 2y1y2 <= 1.

    Now you have a few options: you can show one example that fails or you can do a general proof to show that there is a contradiction in terms of the general variables (and if you want show what values satisfy it being a subspace).

    So I'm going to go the easy way: lets consider U = (1,0) and V = (1,0). x1^2 + y1^2 = 1 which is ok and x2^2 + y2^2 = 1 which is ok. But if we plug those in then x3^2 + y3^2 = 2^2 = 4 > 1 which is a contradiction. Since we have shown one contradiction that is enough for that axiom to fail.
  14. Mar 12, 2012 #13
    wow, okay. Thank you soo much chiro, you should be a college professor, if your not one already. You explain it very well. I'm glad I ran into you here, this whole physicsforum place is a very useful tool I didn't know existed. I'm definitely going to come back here more often, thanks again! :smile:

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