vector with euclidian norm of 1

**robustor** · December 1st 2011, 08:36 PM

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 $rank(X)=1$

**Opalg** · December 2nd 2011, 04:39 AM

Originally Posted by robustor

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 $x_ix_j\leqslant \tfrac12\|x\|_2^2=\tfrac12.$

**robustor** · December 4th 2011, 07:46 AM

Originally Posted by Opalg

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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