# Maximum of a quantity for two normal orthogonal vectors in $\mathbb{R}^n$

Let's define for every pair of vectors $$u,v\in\mathbb{R}^n$$, a quantity as follows: $$f(u,v) = \sum_{1\leq i,j\leq n}|u_iu_j-v_iv_j|.$$

I want to find: $$M(n)= \max \{f(u,v): u,v\in \mathbb{R}^n, |u|=|v|=1, u\perp v\}.$$

An easy estimate using the triangle inequality gives $$M(n) <2n$$, but it seems there should be better upper bounds.

Remark. 1) Let $$a = \cos(\pi/8), b = \sin(\pi/8)$$, then if for even numbers $$n$$ we define the vectors $$u = \sqrt{\frac{2}{n}}(a,b,a,b,\dots)$$ and $$v = \sqrt{\frac{2}{n}}(b,-a,b,-a,\dots)$$, we have $$f(u,v)=\sqrt{2}n\leq M(n)$$. 2)A similar construction shows that for every positive integers $$n,k$$, $$\frac{M(n)}{n}\leq \frac{M(nk)}{nk}.$$

Just use Cauchy-Schwarz and the identity $$\sum_{i,j}(u_iu_j-v_iv_j)^2=\left[\sum_i u_i^2\right]^2+\left[\sum_i v_i^2\right]^2-2\left[\sum_i u_iv_i\right]^2=2$$ to get $$f(u,v)\le \sqrt2 n$$ for all $$n$$. As you have observed yourself, this is sharp for even $$n$$. For odd $$n$$ you can, probably, do a bit better but the estimates get messy and it is not clear (to me) what the sharp bound really is.

• How does one apply Cauchy-Schwarz to the problem? Nov 3, 2018 at 11:51
• @ClaudeChaunier The sum of absolute values of $n^2$ real numbers does not exceed $n$ times the square root of the sum of their squares. Nov 3, 2018 at 17:59

Numerical approximations suggest $$M(n)=\sqrt{2n^2-1}$$ when $$n$$ is odd, with the max reached for some $$u=(a,b,a,b,\dots,a)$$, $$v=(c,d,c,d,\dots,c)$$. Solving exactly for such $$u,v$$ is easy and only implies a quadratic equation in one variable, $$x=b^2$$ for instance. Assuming $$b>d>0$$ without loss of generality, one finds $$b^2 = \frac{1}{n-1}(1+\frac{n}{\sqrt{2n^2-1}})$$ and deduce $$a,c,d$$ from the constraints $$|u|=|v|=1$$ and $$u\cdot v=0$$, yielding $$M(n)\ge \sqrt{2n^2-1}$$ when $$n$$ is odd. Now we still must show that more $$a$$ in $$u$$ or more than two distinct components in $$u$$ cannot improve the bound.

EDIT: Here are elementary details on an analysis for $$u$$ and $$v$$ restricted to
$$u=(a,a,\dots,a,b,b,\dots,b)\in\{a\}^{k}\times\{b\}^{\ell}$$ $$v=(c,c,\dots,c,d,d,\dots,d)\in\{c\}^{k}\times\{d\}^{\ell}$$ with $$k+\ell=n$$ odd or even and $$0\lt d\le b$$ and $$0\lt a\le -c$$.

$$\langle u|u\rangle = \langle v|v\rangle = 1$$ and $$\langle u|v\rangle = 0$$ translate into $$ka^2+\ell b^2 = 1$$ $$kc^2+\ell d^2 = 1$$ $$kac+\ell bd = 0$$ hence $$(kac)^2 = (\ell bd)^2 = (\ell b^2)(\ell d^2) = (1-ka^2)(1-kc^2)$$ simplifies into $$ka^2+kc^2 = 1$$ and likewise $$\ell b^2+\ell d^2 = 1$$ Furthermore $$\begin{matrix} k\ell(ab)^2 &=& (ka^2)(\ell b^2) &=& (1-\ell b^2)(\ell b^2) & &\\ k\ell(cd)^2 &=& (kc^2)(\ell d^2) &=& (1-\ell d^2)(\ell d^2) &=& (\ell b^2)(1-\ell b^2)\\ \end{matrix}$$ hence $$0\lt 1-\ell b^2$$ and $$ab = -cd = \sqrt{\frac{b^2-\ell b^4}{k}}$$

This all enables to write $$f(u,v) = k^2(c^2-a^2)+\ell^2(b^2-d^2)+2k\ell(ab-cd)$$ $$= k(1-2ka^2)+\ell(2\ell b^2-1)+4k\ell ab$$ $$= k(2\ell b^2-1)+\ell(2\ell b^2-1)+4k\ell ab$$ and finally $$f(u,v) = g(x) := 2n\ell x -n + 4\ell\sqrt{k}\sqrt{x-\ell x^2}$$ where $$0\lt x=b^2\lt \frac{1}{\ell}$$. Actually $$b^2$$ cannot be arbitrarily close to $$0$$ because from our assumptions $$1 = \ell b^2+\ell d^2 \le 2\ell b^2$$ thus $$\frac{1}{2\ell}\le x=b^2\lt \frac{1}{\ell}$$

Now $$g'(x)=2n\ell + 2\ell\sqrt{k}\frac{1-2\ell x}{\sqrt{x-\ell x^2}}$$ starts positive and goes toward $$-\infty$$ on the mentioned interval for $$x$$, therefore $$g(x)$$ and $$f(u,v)$$ will be maximal when $$g'(x)=0$$, which happens when $$2\ell x-1 = \frac{2n\ell}{2\ell\sqrt{k}}\sqrt{x-\ell x^2}$$ or equivalently on the prescribed interval $$(2\ell x-1)^2 = \frac{n^2}{k}(x-\ell x^2)$$ which is the quadratic equation $$x^2-\frac{1}{\ell}x+\frac{k}{\ell(4k\ell+n^2)} = 0$$ whose solutions are $$x_\pm = \frac{1}{2\ell} \pm \frac{1}{2}\sqrt{\frac{1}{\ell^2}-\frac{4k}{\ell(4k\ell+n^2)}}$$ $$= \frac{1}{2\ell} \pm \frac{1}{2}\sqrt{\frac{4k\ell+n^2-4k\ell}{\ell^2(4k\ell+n^2)}}$$ $$= \frac{1}{2\ell} \left(1 \pm \frac{n}{\sqrt{4k\ell+n^2}}\right)$$ where only $$x_+$$ is in $$[\frac{1}{2\ell},\frac{1}{\ell})$$. The maximum is then $$g(x_+) = \frac{n^2}{\sqrt{4k\ell+n^2}} + 4\ell\sqrt{k}\sqrt{x_+-\ell x_+^2}$$ The constant coefficient in the quadratic equation is also $$x_+x_-$$ thus $$x_+-\ell x_+^2=x_+(1-\ell x_+)=x_+(\ell x_-)=\ell x_+x_- = \frac{k}{4k\ell+n^2}$$ yields $$g(x_+) = \frac{n^2+4\ell k}{\sqrt{4k\ell+n^2}} = \sqrt{4k\ell+n^2}$$

In the end, when $$u$$ and $$v$$ are restricted to the shapes above, $$f(u,v)$$ is maximal for $$a = +\sqrt{\frac{1}{2k} \left(1 - \frac{n}{\sqrt{4k\ell+n^2}}\right)}\;, \quad b = \sqrt{\frac{1}{2\ell} \left(1 + \frac{n}{\sqrt{4k\ell+n^2}}\right)}\\ c = -\sqrt{\frac{1}{2k} \left(1 + \frac{n}{\sqrt{4k\ell+n^2}}\right)}\;, \quad d = \sqrt{\frac{1}{2\ell} \left(1 - \frac{n}{\sqrt{4k\ell+n^2}}\right)}\\$$ and the maximum is $$f(u,v)=\sqrt{4k\ell+n^2}$$ Now allowing $$k$$ and $$\ell=n-k$$ to vary for $$n$$ fixed, one gets that $$f(u,v)=\sqrt{4k\ell+n^2}$$ is maximum over $$k$$ for $$k=n/2$$ when $$n$$ is even, in which case $$f(u,v)=\sqrt{2n^2}$$, and $$k=(n\pm1)/2$$ when $$n$$ is odd, in which case $$f(u,v)=\sqrt{2n^2-1}$$. We still have to show that other shapes for $$u$$ and $$v$$ cannot improve $$f(u,v)$$.