# vector with euclidian norm of 1

• Dec 1st 2011, 08:36 PM
robustor
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 $\displaystyle rank(X)=1$
• Dec 2nd 2011, 04:39 AM
Opalg
Re: vector with euclidian norm of 1
Quote:

Originally Posted by robustor
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 $\displaystyle x_ix_j\leqslant \tfrac12\|x\|_2^2=\tfrac12.$
• Dec 4th 2011, 07:46 AM
robustor
Re: vector with euclidian norm of 1
Quote:

Originally Posted by Opalg
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?