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Proofs of dimensions and subspaces check

  1. Jan 13, 2013 #1
    Hi, I'd be grateful if someone could tell me whether these proofs I've done are correct or not. Thanks in advanced.

    Let [itex]V[/itex] be an [itex]n[/itex]-dimensional vector space over [itex]\mathbb{R}[/itex]

    Prove that [itex]V[/itex] contains a subspace of dimension [itex]r[/itex] for each [itex]r[/itex] such that [itex]0 \leq r \leq n[/itex]



    Since [itex]V[/itex] is n-dimensional, there exists [itex]v_1, v_2,...,v_n \in V[/itex] such that [itex]\left\{v_1,v_2,...,v_n\right\}[/itex] is a basis for [itex]V[/itex].

    For an element [itex] x \in V[/itex], we can write [itex]x[/itex] as

    [itex]x = \lambda_1 v_1 + \lambda_2 v_2 + ... + \lambda_n v_n[/itex] for [itex]\lambda_i \in \mathbb{R}[/itex]

    Since [itex]Sp\left\{v_1,v_2,...,v_n\right\} = V[/itex]

    Therefore we can also say that

    [itex] V = \left\{\lambda_1 v_1 + \lambda_2 v_2 + ... + \lambda_n v_n\right\}[/itex]

    Since [itex]V[/itex] is a vector space [itex]\left\{(0)\right\} \in V[/itex] and [itex]\left\{(0)\right\}[/itex] is clearly a 0-dimension subspace of [itex]V[/itex] as it is closed under both addition and multiplication.

    Since any non-empty subset of a linearly independent set of vectors is also linearly independent we have that for [itex]q[/itex], where [itex] 1 \leq q \leq n[/itex]

    [itex]\left\{v_1, v_2,..., v_q\right\} = \left\{\mu_1 v_1 + \mu_2 v_2 + ... + \mu_q v_q : \mu_i \in \mathbb{R}\right\}

    \subseteq \left\{\lambda_1 v_1 + \lambda_2 v_2 + ... + \lambda_n v_n\right\} = V[/itex]




    ---
    Second question
    ---

    Let [itex]U,W[/itex] be subspaces of [itex]V[/itex] with [itex]U \subseteq W[/itex]. Show that there is some subspace [itex]W'[/itex] of [itex]V[/itex] such that [itex]W \cap W' = U[/itex] and [itex]W +W' = V[/itex]


    Firstly, since [itex]W \cap W' = U[/itex],[itex]\ \ \ U \subseteq W'[/itex] because if [itex]x \in U[/itex] then [itex]x \in W'[/itex]

    Therefore we can write that

    [itex]W' = U \cup \left\{v_1, v_2, ... , v_m\right\}[/itex]

    For some vectors [itex](v_1, v_2, ... , v_m) \in V \backslash U[/itex] but [itex](v_1, v_2, ... , v_m) \notin W[/itex] because if any [itex]v_i \in W[/itex] then [itex]v_i \in W \cap W'[/itex] and we would have that [itex]W \cap W' \neq U[/itex]

    Since [itex]dim(V) = n[/itex], if we set [itex]dim(W) = x[/itex] we get that

    [itex]dim(\left\{v_1, v_2, ... , v_m\right\}) = m = n-x[/itex]

    Hence we can find [itex]m[/itex] linearly independent [itex]\left\{v_1, v_2, ... , v_m\right\} \notin Sp(W)[/itex] to give

    [itex]W \cap W' = W \cap (U \cup \left\{v_1, v_2, ... , v_m\right\}) = U[/itex]

    and

    [itex] W + W' = \left\{w + w' : w \in W, w' \in W'\right\} = Sp(W \cup W') = V[/itex]

    As [itex]dim(W \cup W') = x + n-x = n[/itex]


    Again, thanks to anyone who helps!
     
  2. jcsd
  3. Jan 14, 2013 #2

    CompuChip

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    Your first answer sounds a bit .... confused, if I may call it that. You correctly noticed that V is spanned by n basis vectors, and that taking the first r of these should give you a subspace of dimension r. So what I would do, is write that down as a claim, rather than a conclusion. Proving the claim should then be straightforward.

    For the second exercise, you seem to be using the conclusion as the first step of the proof. You need to show that a W' satisfying those conditions exists. Until you've done that, you cannot assume it. However, just like I said before: you can make a claim ("define W' as ....") and then prove that it satisfies the conditions. That proof will be very similar to what you've already got (just make sure you don't use the conclusion, i.e. don't use that [itex]W cap W' = U[/itex]).
     
  4. Jan 14, 2013 #3
    Yeah, I've redone the second proof - currently working on the first.

    [itex]U = Sp\left\{u_1, u_2, ... , u_k\right\}[/itex]

    Where [itex]\left\{u_1, u_2, ... , u_k\right\}[/itex] is linearly independent

    We can extend this to a basis of [itex]W[/itex] as [itex]U \subseteq W[/itex]

    So for [itex]w_{k+1}, w_{k+2}, ... , w_l \in W \backslash U[/itex]

    We have that

    [itex]W = Sp\left\{u_1, u_2, ... , u_k, w_{k+1}, w_{k+2}, ... , w_l\right\}[/itex]

    Which we can, again, extend to a basis of [itex]V[/itex] as [itex]W \subseteq V[/itex].

    So for [itex]v_{l+1}, v_{l+2}, ... , v_n \notin W[/itex]

    We have that

    [itex] V = Sp\left\{u_1, u_2, ... , u_k, w_{k+1}, w_{k+2}, ... , w_l, v_{l+1}, v_{l+2}, ... , v_n\right\}[/itex]

    Since [itex]W' \subseteq V[/itex] is a subspace it can be written as the span of a subset of the vectors forming the basis of [itex]V[/itex]

    If we define

    [itex]W' = Sp\left\{u_1, u_2, ... , u_k, v_{l+1}, v_{l+2}, ... , v_n\right\}[/itex]

    [itex]W \cap W' = Sp[\left\{u_1, u_2, ... , u_k, w_{k+1}, w_{k+2}, ... , w_l\right\} \cap \left\{u_1, u_2, ... , u_k, v_{l+1}, v_{l+2}, ... , v_n\right\}] = Sp\left\{u_1, u_2, ... , u_k\right\} = U[/itex]

    and

    [itex]W + W' = Sp[\left\{u_1, u_2, ... , u_k, w_{k+1}, w_{k+2}, ... , w_l\right\} \cup \left\{u_1, u_2, ... , u_k, v_{l+1}, v_{l+2}, ... , v_n\right\}]

    \

    = Sp\left\{u_1, u_2, ... , u_k, w_{k+1}, w_{k+2}, ... , w_l, v_{l+1}, v_{l+2}, ... , v_n\right\} = V[/itex]
     
  5. Jan 15, 2013 #4

    CompuChip

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    I would switch those statements:

    This shows that W' is a subspace, which is one of the things that you are trying to prove.
     
  6. Jan 15, 2013 #5
    Ah yeah, that makes more sense - define W' and then state it clearly is a subspace. Also, for the first question regarding showing there is a subspace of dimension r for each r between 0 and n - can this also be best done by extending the basis n times from the 0-dimensional subspace S = Sp{0}?

    So say we can add some vector in V giving S1 = Sp{0, v1} which is 1-dimensional and linearly independent.

    Then add another vector v2 in V\S1 to give S2 = Sp{0, v1, v2} which is 2-dimensional and linearly independent

    We can continue this all the way until Sn = Sp{0, v1, v2, ..., vn} which is an n-dimensional linearly independent subspace?
     
  7. Jan 16, 2013 #6

    CompuChip

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    Yep. A common way of writing that down is starting with "Let [itex]0 \le r \le n[/itex]. Then define [itex]V_r := \cdots[/itex]" and proceed to show that it is the required subspace. Just like before, really.
     
  8. Jan 16, 2013 #7
    Thanks!
     
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