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Homework Help: Proving vector space

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that if V is a vector space, a is any scalar and u is a member of V then
    1) (-1)x = -x
    2) a(-u) = -au
    3) -(-u) = u

    2. Relevant equations
    The ten axioms of vector space.

    3. The attempt at a solution
    I have solved a0 = 0, but I couldn't figure out how to start answering these. What is the line of thought necessary to be able to prove something so fundamental? I'm not asking you to solve these, I'm just asking how should I 'go about' solving this. Thanks in advance.
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2


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    Well, the first problem tells you to prove that (-1)x is the additive inverse of x.

    To prove this, you should notice a small property, which can easily be proven:

    0u = 0

    And, to prove that (-1)x is the additive inverse of x, we just need to prove that:

    (-1)x + x = 0, right? :)


    The second problem tells you to prove that a(-u) is the additive inverse of au, which is similar to proving:

    a(-u) + au = 0


    The third problem is just saying that u is the additive inverse of -u. Which is like, so obvious. Let's see if you can show this. :)
    Last edited: Sep 22, 2009
  4. Sep 22, 2009 #3
    Let's see...

    1) (-1)x = -x
    (-1)x + x = 0
    (-1)x + x + (-x) = 0 + (-x)
    (-1)x + 0 = -x
    (-1)x = -x

    2) a(-u) = -au
    a(-u) + au = 0
    a(-u) + au + (-au) = 0 + (-au)
    a(-u) + 0 = -au
    a(-u) = -au

    3) -(-u) = u
    -(-u) + (-u) = 0
    -(-u) + (-u) + u = 0 + u
    -(-u) + 0 = u
    -(-u) = u

    Am I doing it right? I notice the workings are pretty similar, but man, this stuff is really abstract :grumpy:
  5. Sep 22, 2009 #4


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    This is all wrong... :bugeye:

    Did you read my post carefully? What you are doing is not a proof!!!

    You go from what you need to prove, and then come back to it?


    You go from (-1)x = -x, and then arrive to (-1)x = -x. Well, I'm very sorry to inform you that this is no where to be called a proof. :(


    Well, but don't lose hope, everything starts hard. It should get better as you are more familiar with it. So what you need to do is to look back at your text-books to get some ideas of how a proof might be.

    As I tell you, to solve the first problem, you need to prove a small property:

    0u = 0.

    And to prove the second problem you need to prove, a0 = 0

    Now, start from these 2 easy ones.

    Notice that: 0 = 0 + 0, and 0 = 0 + 0.

    The bolded ones are vectors.
  6. Sep 22, 2009 #5
    What? Aren't they the same thing? :confused:

    Proving 0u = 0:
    0u = (0 + 0)u = 0u + 0u = 0
    -0u + 0u = (-0u + 0u) + 0u
    0 = 0 + 0u = 0u

    Proving a0 = 0:
    a0 = a(0 + 0) = a0 + a0 = 0
    -a0 + a0 = (-a0 + a0) + a0
    0 = 0 + a0 = a0

    Is this correct?
  7. Sep 22, 2009 #6


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    No, they are not, one is a vector, and one is a scalar.

    Well, you don't need the part "= 0" here. Since you are proving that 0u = 0, you haven't known if it's 0, or not.

    Well, ok. :)

    When writing down, you should write it formally. In your proof, you seem to be missing some [itex]\Rightarrow[/itex] signs.

    Same comment goes here. You don't need the "= 0" part.


    Ok, so, back to your problem. The first one, you need to prove that:

    (-1)x = -x, i.e (-1)x is the additive inverse of x. (Note that: -x is interpreted to be the additive inverse of x)

    Now, try to prove that: (-1)x + x = 0. If you can prove this, it means that (-1)x is the additive inverse of x, and everything is done, right?

    Hint, one of the axiom is: x = 1x.
  8. Sep 22, 2009 #7
    (-1)x + x = (0 - 1)x + x
    = 0x - 1x + x
    = 0x - x + x
    = 0x - 0
    = 0x [tex]\Rightarrow[/tex] 0

    How's this?
  9. Sep 22, 2009 #8


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    Well, no. It's still wrong though. :(

    How can you go from this line:

    to, this line:


    You haven't known that (-1)x = -x. This is what you need to prove!!!

    Note that: When writing (-1)x = (0 - 1)x means that (-1)x = (0 + (-1))x = 0x + (-1)x. Subtraction is just the inverse of addition.

    And, btw why is there an imply sign at the end of your proof?


    As I said, there's an axiom telling you that x = 1x.

    I'll give you a push then:

    (-1)x + x = (-1)x + (1)x = ...

    I think you should be able to go from here.
  10. Sep 22, 2009 #9
    I'm not sure how to use the imply sign formally as you said.

    (-1)x + x = (-1)x + (1)x
    = (-1 + 1)x
    = 0x
    = 0

    This question is pretty frustrating, since it doesn't look hard at all. I hope it's right this time, took me long enough. :bugeye:
  11. Sep 22, 2009 #10


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    Yup, looks fine this time. :)


    Well, if you haven't learned how to use [itex]\Rightarrow[/itex] sign, then, just don't worry about it. :)


    Let's see if you can solve b, and c. :)
  12. Sep 22, 2009 #11
    Thanks a lot for your help and patience, I'm beginning to see the light. :smile:
    Do you think it's necessary to show 0x = 0 when asked in an exam?

    2) a(-u) = -au
    a(-u) + au = 0
    = a(-u) + a(u)
    = a(-u + u)
    = a(0)
    = 0

    3) -(-u) = u
    -(-u) - u = 0
    = -(-u + u)
    = -0
    = 0

    Not sure about (3), it's slightly different.
  13. Sep 22, 2009 #12


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    Well, I think yes, you can put it as a lemma, or some kind of small note (with proof, of course) :)

    Well, again, don't put the "= 0" part there, you haven't proven it yet.

    You can say something like:

    Problem 2: a(-u) = -au

    We'll now prove that a(-u) is the additive inverse of au. It's true, since:

    a(-u) + au = a(-u + u) = a0 = 0.

    Same comment here. Just drop out the "= 0" part.

    Btw, when going from this line:

    to this line:

    you can make it "clearer" by using Problem 1.

    -(-u)) - u = (-1)(-u) + (-1)u = (-1)(-u + u) = (-1)0 = 0


    Or, you can do like this:

    Since -u is the additive inverse of u, we have: -u + u = 0, which also means that u is the additive inverse of -u, and hence -(-u)) = u.


    Since, I still think that you're messing up with some concept, I'm repeating this again, when writing: -u, it means that the writer is referring to the additive inverse of u, i.e, some element that when sum it with u will result in 0.

    -(-u) simply means that it's the additive inverse of -u, or similarly, the additive inverse of the additive inverse of u.

    -(-u)) = u is supposed to mean: u is the additive inverse of -u, or u is the additive inverse of the additive inverse of u.
    Last edited: Sep 22, 2009
  14. Sep 22, 2009 #13
    Why isn't there an open bracket for -(-u))?

    Anyway, this is awesome. Not sure I get everything you said, but I'll learn in due time.
    Thanks a million! (cám ơn) :smile:
  15. Sep 22, 2009 #14


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    Yah, my bad, there's an extra closing bracket.. =.="

    Trust me, this would be much easier when you get familiar with it.

    And, good luck. :)

    Geez. You make me surprised. It's no problem. Glad that I can help you. :smile:
  16. Sep 23, 2009 #15
    Sorry to revive this thread, but another relevant question stumps me.

    Let V be the set of all functions from f:R -> R such that f(0) = 1. Show that V is not a vector space with vector addition and scalar multiplication. Determine all the axioms which are not satisfied.

    I'm guessing I'd have to start like post 5 above, then after determining that f(0) = 0, V is shown to not be a vector space. I have no idea about the axiom thing, maybe it's the one where all vector spaces have a zero vector?
    Last edited: Sep 23, 2009
  17. Sep 23, 2009 #16


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    Well, you can start with the Definition of a Vector Space. What's a Vector Space? And what are its axioms?

    To prove that V is not a Vector Space, just point out that it does not satisfy requirements from the definition.

    And about the 'which axioms are not satisfied', you can scan through the set of axioms, and test each one if it's satisfied.


    I'll give you an example, about the Associativity of addition Axiom:
    [tex]\forall f, g, h \in V[/tex], we have:
    [tex]\forall x \in \mathbb{R}; ((f + g) + h)(x) = (f + g)(x) + h(x) = (f(x) + g(x)) + h(x) = f(x) + (g(x) + h(x)) = f(x) + (g + h)(x) = (f + (g + h)(x)[/tex]

    So, it means that: [tex]\forall f, g, h \in V; (f + g) + h = f + (g + h)[/tex]

    Hence, the first axiom is satisfied.


    You should show your work first, so that I can know where you get stuck, and provide more accurate help. That's how I would start off the problem, let's see how far can you get. :)
    Last edited: Sep 23, 2009
  18. Sep 23, 2009 #17
    what is meant by the notations f : R -> R?

    I think of vector space as a set of vectors that satisfy the 10 axioms:
    1. closed under addition
    2. addition is commutative
    3. addition is associative
    4. there is a unique zero vector
    5. for every component in the vector space there is an additive inverse
    6. closed under scalar multiplication
    7 and 8. distributive law
    9. scalar multiplication is associative
    10. 1u = u

    I believe your equation also proves the distributive law, but is multiplying by x the only way to prove associative law under addition?

    Basically, all that I've done is to do what I did on post 5, proving f(0) to be 0:

    f(0) = f(0 + 0) = f(0) + f(0)
    -f(0) + f(0) = -f(0) + f(0) + f(0)
    0 + f(0) = 0
    f(0) = 0

    It's not 1, so the definition f(0) = 1 is wrong. Yet I don't see how it's related to any of the axiom. Or am I doing it the wrong way? Should I prove it under each axiom?

    Anyway, give me some time to reread my notes. It still isn't making much sense. :(
    Last edited: Sep 23, 2009
  19. Sep 23, 2009 #18


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    Oh no, you seem to have lack so many basic concepts. [tex]f(0 + 0) \neq f(0) + f(0)[/tex]

    The notation: f: R -> R, means that f is a real function, of which domain is R.

    Some examples are:

    [tex]\begin{align*}f : \mathbb{R} & \rightarrow \mathbb{R} \\
    x &\mapsto x ^ 2\end{align*}[/tex]

    This means that f(x) = x2.
    f(0) = 02 = 0, f(2) = 22 = 4, blah blah blah..

    The following function, h is almost the same as f, except that its domain is restricted to positive real numbers.

    [tex]\begin{align*}h : \mathbb{R ^ {+}} & \rightarrow \mathbb{R} \\
    x &\mapsto x ^ 2\end{align*}[/tex]

    Another example is:

    [tex]\begin{align*}g: \mathbb{R} \backslash \{ 0 \} & \rightarrow \mathbb{R} \\
    x &\mapsto \frac{1}{x} \end{align*}[/tex]

    In this case: g(x) = 1/x
    g(1) = 1/1 = 1, and g(5) = 1/5, g(1/2) = 1/(1/2) = 2.

    Some trig functions are:

    [tex]\begin{align*}\sin : \mathbb{R} & \rightarrow [0; 1] \\
    x &\mapsto \sin(x) \end{align*}[/tex]

    [tex]\begin{align*}\cos : \mathbb{R} & \rightarrow [0; 1] \\
    x &\mapsto \cos(x) \end{align*}[/tex]

    Well, you should try to scan through some of your old books, to find if there's a Chapter on Functions.

    You can also have a look here: http://en.wikipedia.org/wiki/Function_(mathematics)" [Broken].

    (1) The sum of 2 function can be defined by:

    (f + g)(x) = f(x) + g(x)

    (2) And scalar multiplication can be defined as:

    [tex](\alpha f)(x) = \alpha [f(x)][/tex]

    (3) The 2 functions f1, and f2 are said to be equal (i.e f1 = f2) if and only if:
    • They have the same domain.
    • For every x in their domain, f1(x) = f2(x).

    Lacking some fundamental concepts may cause a lot of difficulties in the future. I really think that you should slow down a bit. Are you self-studying, or something?


    After having read the concepts about function, let's try to answer the 2 following questions: (pay attention to (1), (2), and (3))

    1. Is V closed under addition? (Hint: use f(0) = 1)
    2. Is V closed under scalar multiplication?
    Last edited by a moderator: May 4, 2017
  20. Sep 23, 2009 #19
    I'm just beginning my Further Mathematics module. First chapter was on Matrices, and Vector Space is the second chapter. The prerequisites for this module were topics like Calculus, Matrices, Linear Algebra, etc. I have no problem with Matrices, but as you can see, I'm having real trouble with this chapter. The professor was just finishing on Subspace, and I'm doing the exercises on this chapter.

    I have no problem in Maths in general, but I find this hard as it requires really deep understanding of the relevant concepts. Well, at least I understand what you wrote about functions, but the notations take some time getting used to. In fact, f(0+0) was a stupid thing to write. I was blinded by my previous workings on u(0+0). :shy:

    Please give me some time to read up a bit. I shall give it another go tomorrow.
    Last edited: Sep 23, 2009
  21. Sep 23, 2009 #20
    I don't get your notations under g, but anyway:

    For all f, x under R,

    1. {f(x)} + {f(x)} = {f(x) + f(x)}
    let x = 0, {f(0) + f(0)} = {1 + 1} = {2} (closed under addition)

    2. (af)(x) = a[f(x)]
    let x = 0,
    (af)(0) = 0
    a[f(0)] = a(1) = a (not closed under scalar multiplication?)
    Last edited: Sep 23, 2009
  22. Sep 24, 2009 #21


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    Err, why are you writing everything in brackets {...}?

    No, it's not closed under addition. Note that V is the set of all function from R to R, such that f(0) = 1. That means, any function, such that f(0) = 1 will be in V. And if some other function g, such that we have [tex]g(0) \neq 1[/tex], then [tex]g \notin V[/tex].

    So, for any [tex]f, g \in V[/tex], (f + g)(0) = f(0) + g(0) = 1 + 1 = 2.

    So, is (f + g)(0) = 1? Is [tex]f + g \in V[/tex]?

    Consider a function [tex]f \in V[/tex]. If V is closed under scalar multiplication, then for all real a, we must have [tex]af \in V[/tex] too, right?

    So, unless a = 1, we have [tex](af)(0) = a \neq 1[/tex], hence [tex]af \notin V[/tex] (because [tex](af)(0) \neq 1[/tex], for [tex]a \neq 1[/tex]). So, it's not closed under scalar multiplication.

    Yes, your second part of the proof is correct. However, you should re-do the first part. It's wrong.


    Well, if you find out any of my explanations are hard to understand, or sound strange, just shout it out. :)

    Which part makes you confused? Is it the notation R \ {0}?

    R \ {0} is simply the real number set, with 0 excluded.


    Let A, and B be 2 sets, the notation: A \ B is a set defined as follow:

    [tex]A \backslash B = \{ x \in A \wedge x \notin B \}[/tex]

    ([itex]\wedge[/itex] stands for and, whereas [itex]\vee[/itex] stands for or).

    So, basically, the set A \ B is the set of all elements in A, such that it's not in B.

    Example: [tex]\mathbb{R} \backslash \{ 0 \} = (-\infty; 0) \cup (0; + \infty)[/tex]
    If A = {1, 3, 5, 7, 9}, B = {3, 7, 8}, then A\B = {1, 5, 9}.
  23. Sep 24, 2009 #22
    I meant for the brackets to show that it's closed under R.

    Oh wow, it's finally making sense. For all f, g under R: (do we have to declare vectors such as x as well?)
    (f+g)(x) = f(x) + g(x)
    Let x=0, (f+g)(0) = 0
    f(0) + g(0) = 1 + 1 = 2 (not closed under addition)

    What is the form of the vector space in this case? I only know it touches (0,1), and as f(0) + g(0) = 2, which intersects (0,2), the line is outside the vector space V. Assuming f and g are simple linear functions, how is (f+g) shaped? Is it possible to visualize vector spaces?

    I'm not sure of the proper usage of '{}' and ';'. And since 0 is 'excluded', why is it a 'union'? Shouldn't it be an 'intercept' and the sets compliments?
  24. Sep 24, 2009 #23


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    No, you don't need any brackets there.

    But btw, what do you mean by "closed under R"? I've heard of something being "closed under addition", or scalar multiplication, or blah blah blah. But to tell the truth, I haven't heard of anything that's "closed under R".

    Note that: Vector is an element of a Vector Space. So x is not a vector, instead, x is a variable of a function.

    Say sin(x), sin is a function, and x is just a variable.

    Well, let f be a function defined as follow:

    [tex]\begin{align*}f : X & \rightarrow Y \\
    x &\mapsto y = f(x) \end{align*}[/tex]

    A variable x can be any value of the domain X. And for every value x = x0 of the domain, there's one and only one (unique) value y0 in the co-domain Y, such that: y0 = f(x0). And we say that f maps x0 to y0.

    Example: The function g(x) = x2 maps 2 to 22 = 4; maps 3 to 9, maps 10 to 100, and so on...


    And as I told you before, let's say you have 2 functions f, and g, defined as:

    [tex]\begin{align*}f : \mathbb{R} & \rightarrow \mathbb{R} \\
    x &\mapsto f(x) \end{align*}[/tex]


    [tex]\begin{align*}g : \mathbb{R} & \rightarrow \mathbb{R} \\
    x &\mapsto g(x) \end{align*}[/tex]

    The function f + g, also call the sum of f and g is defined as:

    [tex]\begin{align*}f + g : \mathbb{R} & \rightarrow \mathbb{R} \\
    x &\mapsto f(x) + g(x) \end{align*}[/tex]

    V in this case is not a vector space.

    Well, (f + g)(x) = f(x) + g(x).

    Say f(x) = 2x + 1; and g(x) = 5x + 3.

    So, the sum of 2 functions is: (f + g)(x) = f(x) + g(x) = (2x + 1) + (5x + 3) = 7x + 4.


    You know http://en.wikipedia.org/wiki/Euclidean_vector" [Broken] right? Every Euclidean Vector is determined by a fix initial point, a direction, and a magnitude. We can add 2 Euclidean Vectors, and can also multiply them by a real constant (scalar).

    The Vectors (elements) of Vector Space is just the generalization of Euclidean Vectors. We grab some of its (most basic) properties (the 10 axioms), and defined every set with 2 binary operations (the sum of 2 elements in it, and scalar multiplication). And say if that set, along with 2 binary operations previously defined satisfy 10 axioms, then the set is a Vector Space, and its element is called Vector.

    Vector Space is hard to visualize, but that's how I understand it.

    {..} are often used to defined sets, and sometimes to group (like (...), and [...]).

    Eg: {[(2 + 3) * 5] + 9} * 7

    You should do some more readings though: http://en.wikipedia.org/wiki/Interval_(mathematics)" [Broken]

    [tex](-\infty; 0) \cup (0; + \infty)[/tex]

    is the union of x < 0, and x > 0, hence 0 excluded.


    You seem to have lack some basic maths concepts. I really think you should have a talk to your professor, and ask him to recommend you some good books to cover your missing knowledge. Since English is not my Mother Tough, recommending books (that is readable for you) is well beyond my ability. :(

    Wikipedia is good, but it's not best. It's just a place to have some reference, not to learn. And, honestly, I prefer reading books than keeping my eyes stick to the computer screen to read wiki. :)


    If you find everything a little bit clearer. Then let's try the second part. Which axioms are not satisfied?

    (I'm very sorry, if there's any misspells in my post. Gotta get some sleep now.. =.=")
    Last edited by a moderator: May 4, 2017
  25. Sep 24, 2009 #24
    Like I said, the concept of vectors, R, and many things you've been teaching me is very new to me. I'm bound to get mixed up from time to time until I get familiar with the subject.

    This I understand, but I've never been able to explain it in mathematical terms.

    Ouch. I forgot what I've been trying to prove all along.

    Since it's not a vector space, I couldn't map the graph, but I didn't realize it until you told me. :blushing:

    Nope, I saw that page when trying to read up on vectors, but so far I haven't reached there yet.

    The functions weren't sets?

    I haven't encountered these notations, so I misunderstood.

    English is not my mother tongue either. I've been able to get good grades in Maths because I know how to answer questions, despite not understanding much of their fundamentals. I've never read more than I needed in order to pass exams, so that probably explains my lack of knowledge when discussing concepts in this manner. I dislike Wikipedia when referencing advanced concepts as they are often too complicated for my understanding.

    1. closed under addition
    4. there is a unique zero vector (not completely sure with this one)
    6. closed under scalar multiplication

    It's alright, there's no rush. Thank you for your patience so far. :biggrin:

    By the way, since x is a variable, if you substitute it with a number such as 0, is it a vector then?
    Last edited by a moderator: May 4, 2017
  26. Sep 25, 2009 #25


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    No, they weren't. Function is, roughly speaking, the relation between 2 sets (domain, and co-domain).

    Believe me, don't write {f(x)}, f(x) is enough.

    Can you show why 4 is not satisfied?

    The 5-th axiom is not satisfied either. Let's see if you see why.
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