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Thread: Proofs where V is a finite-dimensional vector space

  1. #1
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    Proofs where V is a finite-dimensional vector space

    These are probably easier than I think they are.

    Suppose $\displaystyle V$ and $\displaystyle W$ are vector spaces over a field $\displaystyle F$.

    Now suppose $\displaystyle W_1,...,W_k$ are subspaces of $\displaystyle V$ and that $\displaystyle V$ is finite-dimensional.

    a) Prove that $\displaystyle \dim(W_1+...+W_k)\leq\dim(W_1)+...+\dim(W_k)$.

    The question recommends proving this via induction (although proving via bases is possible, albeit more difficult). I think what I have below works, but if it needs correction, I'd like to know.

    For $\displaystyle k=1, \dim(W_1)\leq\dim(W_1)$ is true. (base case)
    Assume the result is true for $\displaystyle k=n$,
    i.e. $\displaystyle \dim(W_1,...,W_n)\leq\dim(W_1)+...+\dim(W_n)$ (hypothesis)
    For $\displaystyle k=n+1$, (inductive case)
    $\displaystyle \dim(W_1,...,W_n,W_{n+1})\leq\dim(W_1)+...+\dim(W_ n)+\dim(W_{n+1})$
    $\displaystyle =\dim(W_1,...,W_{k-1},W_k)\leq\dim(W_1)+...+\dim(W_{k-1})+\dim(W_k)$
    Therefore, $\displaystyle \dim(W_1,...,W_k)\leq\dim(W_1)+...+\dim(W_k)$

    b) Prove that $\displaystyle V=W_1\oplus...\oplus W_k$ if and only if every vector $\displaystyle v\in V$ is equal to $\displaystyle w_1+...+w_k$ for a unique choice of $\displaystyle w_j\in W_j, 1\leq j\leq k$.

    I don't have an answer yet for this half, but I'm working on it.

    Any help on these would be appreciated.
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  2. #2
    Member HappyJoe's Avatar
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    Your inductive step looks weird.

    You have typed commas instead of plusses between the W_i. Anyway, I suggest that you prove your base case _also_ for k=2.

    Then the inductive case could go like: Suppose the result is true for k=n, and let k=n+1. Then

    $\displaystyle dim(W_1+...+W_n+W_{n+1})$
    $\displaystyle = dim((W_1+...+W_n)+W_{n+1})$
    $\displaystyle \leq dim(W_1+\cdots+W_n)+dim(W_{n+1})$
    $\displaystyle \leq dim(W_1)+\cdots+dim(W_n)+dim(W_{n+1}),$

    where the first equality is just putting in parentheses, the first inequality is using the base case k=2 (with W_1+...+W_n and W_{n+1} being the two vector spaces), and the next inequality is using the induction hypothesis that the result is true for k=n.

    As for b), what is your definition of direct sum?
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  3. #3
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    Quote Originally Posted by HappyJoe View Post
    Your inductive step looks weird.

    You have typed commas instead of plusses between the W_i. Anyway, I suggest that you prove your base case _also_ for k=2.

    Then the inductive case could go like: Suppose the result is true for k=n, and let k=n+1. Then

    $\displaystyle dim(W_1+...+W_n+W_{n+1})$
    $\displaystyle = dim((W_1+...+W_n)+W_{n+1})$
    $\displaystyle \leq dim(W_1+\cdots+W_n)+dim(W_{n+1})$
    $\displaystyle \leq dim(W_1)+\cdots+dim(W_n)+dim(W_{n+1}),$

    where the first equality is just putting in parentheses, the first inequality is using the base case k=2 (with W_1+...+W_n and W_{n+1} being the two vector spaces), and the next inequality is using the induction hypothesis that the result is true for k=n.

    As for b), what is your definition of direct sum?
    Thanks for clearing up my answer on the first half. I knew I'd done SOMETHING wrong.

    As for the second half, I have quite a bit done, but again it could use some proof-reading. I probably have a few things out of proper order, which I feel would be the main issue.

    We first have $\displaystyle v=w_1+...+w_k$, as provided by the question.
    Then, we suppose that $\displaystyle v=w'_1+...+w'_k$.
    This leads to the following: $\displaystyle w_1+...+w_k=w'_1+...+w'_k$
    This implies that $\displaystyle 0=(w_1-w'_1)+...+(w_k-w'_k)$,
    $\displaystyle \Rightarrow w_1-w'_1=0,...,w_k-w'_k=0$
    $\displaystyle \Rightarrow w_1=w'_1,...,w_k=w'_k$
    Assuming the above is true, since each $\displaystyle w_j\in W_j$ is a unique choice, there could only be one solution to $\displaystyle w_1+...+w_k=0$, which is that $\displaystyle w_1=0,...,w_k=0$ for all $\displaystyle k>0$.

    Therefore, $\displaystyle V=W_1\oplus...\oplus W_k$.

    -----

    Tell me if I've messed up anything. I'm suspecting right now that my proof just might be a bit out of order.
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  4. #4
    Member HappyJoe's Avatar
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    Here's a quickly written answer, I hope you can read my non-LaTeX'ified scribbles.

    What's your definition of direct sum? Is it that V is the direct sum of W_1, W_2, ..., W_k, iff W_1+...+W_k = V and the W_i have only the 0-vector in common?

    For the first part of your proof, where you show that (V is a direct sum of the W_i) implies (each v is a unique sum of elements w_i), how do you argue from 0 = (w_1-w_1') + ... + (w_k - w_k') to w_1-w_1'=0, ..., w_k-w_k'=0?

    I don't quite get the other part. It may be because I'm not sure which definition you are working from, or maybe because I'm busy.
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