Results 1 to 3 of 3

Thread: vector with euclidian norm of 1

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    12

    vector with euclidian norm of 1

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

    $\displaystyle ||x||_2=1$

    Then I have a matrix $\displaystyle X=x^Tx$

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

    Some more information I have about matrix $\displaystyle X$: $\displaystyle X\succeq 0$ and $\displaystyle rank(X)=1$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10

    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

    $\displaystyle ||x||_2=1$

    Then I have a matrix $\displaystyle X=x^Tx$

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

    The (i,j)-element of X is $\displaystyle x_ix_j.$ Use the inequality $\displaystyle x_ix_j\leqslant\tfrac12(x_i^2+x_j^2)$ (which follows from the fact that $\displaystyle (x_i-x_j)^2\geqslant0$) to see that $\displaystyle x_ix_j\leqslant \tfrac12\|x\|_2^2=\tfrac12.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2011
    Posts
    12

    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 $\displaystyle x_ix_j.$ Use the inequality $\displaystyle x_ix_j\leqslant\tfrac12(x_i^2+x_j^2)$ (which follows from the fact that $\displaystyle (x_i-x_j)^2\geqslant0$) to see that $\displaystyle x_ix_j\leqslant \tfrac12\|x\|_2^2=\tfrac12.$
    Thanks.

    So now I have further that

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

    and I also have

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

    What do we know about?

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

    So far the best I can prove is

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

    Is there a tighter bound?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expansion of Vector Two Norm
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Mar 2nd 2010, 04:42 PM
  2. Prove that function is a vector norm.
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: Dec 29th 2009, 01:26 AM
  3. What is a Norm? Norm of a vector? P-Norm?
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Dec 10th 2009, 06:06 AM
  4. Euclidian norm
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Sep 30th 2009, 04:12 AM
  5. Vector Norm and Matrix Norm
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 18th 2008, 10:49 AM

Search Tags


/mathhelpforum @mathhelpforum