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Math Help - vector with euclidian norm of 1

  1. #1
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    vector with euclidian norm of 1

    Let's say I have a vector x of positive real values and

    ||x||_2=1

    Then I have a matrix X=x^Tx

    My question is: are all elements of X outside of diagonal X \leq 0.5?

    Some more information I have about matrix X: X\succeq 0 and rank(X)=1
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  2. #2
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    Re: vector with euclidian norm of 1

    Quote Originally Posted by robustor View Post
    Let's say I have a vector x of positive real values and

    ||x||_2=1

    Then I have a matrix X=x^Tx

    My question is: are all elements of X outside of diagonal X \leq 0.5?
    The answer is Yes.

    The (i,j)-element of X is x_ix_j. Use the inequality x_ix_j\leqslant\tfrac12(x_i^2+x_j^2) (which follows from the fact that (x_i-x_j)^2\geqslant0) to see that x_ix_j\leqslant \tfrac12\|x\|_2^2=\tfrac12.
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  3. #3
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    Re: vector with euclidian norm of 1

    Quote Originally Posted by Opalg View Post
    The answer is Yes.

    The (i,j)-element of X is x_ix_j. Use the inequality x_ix_j\leqslant\tfrac12(x_i^2+x_j^2) (which follows from the fact that (x_i-x_j)^2\geqslant0) to see that x_ix_j\leqslant \tfrac12\|x\|_2^2=\tfrac12.
    Thanks.

    So now I have further that

    \sum_{i,j} x_{ij} \leq r, where 0 \leq x_{ij} \leq 0.5 and r \geq 0

    and I also have

    \sum_{i,j} y_{ij} \leq d, where y_{ij} \geq 0 and d \geq 0 and y is zero-diagonal

    What do we know about?

    \sum_{i,j} x_{ij} y_{ij}  \leq ?

    So far the best I can prove is

    \sum_{i,j} x_{ij} y_{ij}  \leq \frac{d}{2}

    Is there a tighter bound?
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