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Math Help - Orthogonal vectors and the Pythagorean theorem

  1. #1
    Member Mollier's Avatar
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    Orthogonal vectors and the Pythagorean theorem

    Problem statement:
    The Pythagorean theorem asserts that for a set of n orthogonal vectors {x_i},
    ||\sum^{n}_{i=1}x_i||^2 = \sum^{n}_{i=1}||x_i||^2

    (a) Prove this in the case n=2 by an explicit computation of ||x_1 + x_2||^2.
    (b) Show that this computation also establishes the general case by induction.

    -------------------------------------------------------------------------
    Attempt at solution:
    (a)
    ||x_1+x_2||^2 = ||x_1||^2+||x_2||^2
    x^{T}_{1}x_1+2x^{T}_{1}x_2+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2

    Since the dot product of two orthogonal vectors is zero, we have:

    x^{T}_{1}x_1+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2

    (b)
    Basis step, n=1: ||x_1||^2 = ||x_1||^2
    I assume that this holds up to k for 1 \leq k < n.
    Now I show that it holds for k+1:

    ||x_1+x_2+\cdots+x_k+x_{k+1}||^2=||x_1||^2+||x_2||  ^2+\cdots+||x_k||^2+||x_{k+1}||^2

    By orthogonality x_ix_j=0 for i \neq j:

    x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1} = x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1}

    Since the statement is true for n=1, n=k and n=k+1, it is also true for n.
    ----------------------------------------------------------------------

    I do not have much experience with proofs, and would appreciate some help with this.
    Thank you.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Mollier View Post
    Problem statement:
    The Pythagorean theorem asserts that for a set of n orthogonal vectors {x_i},
    ||\sum^{n}_{i=1}x_i||^2 = \sum^{n}_{i=1}||x_i||^2

    (a) Prove this in the case n=2 by an explicit computation of ||x_1 + x_2||^2.
    (b) Show that this computation also establishes the general case by induction.

    -------------------------------------------------------------------------
    Attempt at solution:
    (a)
    ||x_1+x_2||^2 = ||x_1||^2+||x_2||^2
    x^{T}_{1}x_1+2x^{T}_{1}x_2+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2

    Since the dot product of two orthogonal vectors is zero, we have:

    x^{T}_{1}x_1+x^{T}_2x_2 = x^{T}_{1}x_1 + x^{T}_{2}x_2

    (b)
    Basis step, n=1: ||x_1||^2 = ||x_1||^2
    I assume that this holds up to k for 1 \leq k < n.
    Now I show that it holds for k+1:

    ||x_1+x_2+\cdots+x_k+x_{k+1}||^2=||x_1||^2+||x_2||  ^2+\cdots+||x_k||^2+||x_{k+1}||^2

    By orthogonality x_ix_j=0 for i \neq j:

    x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1} = x^{T}_{1}x_1 + x^{T}_{2}x_2 + \cdots + x^{T}_{k}x_k + x^{T}_{k+1}x_{k+1}

    Since the statement is true for n=1, n=k and n=k+1, it is also true for n.
    The proof of (a) is correct. But in (b) you have not used the inductive hypothesis. What you should say is something like this:

    Let y = x_1  + \ldots + x_k. The inductive hypothesis is that \|y\|^2 = \|x_1\|^2 + \ldots + \|x_k\|^2. We want to find \| x_1  + \ldots + x_k + x_{k+1}\|^2 = \|y + x_{k+1}\|^2. But by (a) this is equal to \|y\|^2 + \|x_{k+1}\|^2. Now use the inductive hypothesis to see that this is equal to   \|x_1\|^2 + \ldots + \|x_{k+1}\|^2. That completes the inductive step.
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  3. #3
    Member Mollier's Avatar
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    I see.
    My inductive hypothesis is that the results I get in my basis step (n=1),hold for n=k. Then I check that I get the same result when n = k+1.

    Thank you very much!
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