I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows:

Let $(E, \langle, \rangle)$ be a euclidean space of dimension $n \geq 2$.

1) For any $x \in E$, show that $||x|| = \sqrt{n}$ if and only if there exists $(e_1, \dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ \ldots + e_n$.

2) For any $(x,y) \in E$, show that $||x|| = \sqrt{n}$, $||y|| = \sqrt{\frac{n(n+1)(2n+1)}{6}}$ and $\langle x,y \rangle = \frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, \ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + \ldots + e_n$ and $y = e_1 + 2e_2 + \ldots + ne_n$

The first question can be solved relatively simply : one proceeds inductively. The induction step goes as follows : find a vector $e_{t_0}$ such that $\langle x, e_{t_0} \rangle = 1$ and $||e_{t_0}|| = 1$. Then, consider $x' = x-e_{t_0}$. It is easily checked that $x'$ is orthogonal to $e_{t_0}$ and that $x'$ satifies the induction hypothesis. To find $e_{t_0}$ use the intermediate value theorem with the hypotheses $\langle x, x/||x|| \rangle = \sqrt{n} >1$ and $\langle x, z \rangle = 0$ for any $z$ orthogonal to $x$.

For the second question, one could proceed also inductively. It would be sufficient to find $e_{t_0} \in E$ such that $\langle x,e_{t_0} \rangle = 1$, $\langle y, e_{t_0} \rangle = n$ and $||e_{t_0}|| = 1$. I have a vague idea on how to find such a $e_{t_0}$ but it seems horribly complicated and certainly not acessible to second year undergraduate student majoring in economics. I am wondering if there was a (realtively) simple way to find the solution for this second question?


1 Answer 1


The answer to the first question is of course implied by the transitivity of rotations on the sphere, which allows to think of the following, hopefully accessible, proof (I can't judge but feel free to say it is not). Fix one orthonormal basis $(e_1,\ldots,e_n)$ of $E$ and let $x_0=e_1+\cdots +e_n$. Consider the plane $P$ generated by $x$ and $x_0$. Complete $p_1:=\frac{x_0}{\sqrt{n}}$ to an orthonormal basis $(p_1,p_2)$ of $P$. Since $x$ has norm $\sqrt{n}$ we may write $$ x=\sqrt{n}(\cos(\theta)p_1+\sin(\theta)p_2) $$ i.e. it is the image of $x_0$ by the rotation of angle $\theta$ in $P$, leaving the orthogonal complement fixed. Then it remains to check (hopefully it is known) that the latter application sends an orthonormal basis to another orthonormal basis and write $x$ as the sum of the components of the image of the initial basis $(e_1,\ldots,e_n)$.

If this is considered OK, then the second question amounts essentially to the same proof working in the orthogonal complement to $x$. We first fix a basis $(e_1,\ldots,e_n)$ such that $x=e_1+\cdots +e_n$ provided by the previous answer. Let $y_0=e_1+2e_2+\cdots +ne_n$. Consider $$z=y-\frac{n(n+1)}{2\|x\|^2}x \quad \text{ and }\quad z_0=y_0-\frac{n(n+1)}{2\|x\|^2}x$$ which are both in the orthogonal to $x$ and have same norm (by the assumptions on $y$). Similarly to the first answer, working in $\mathbb{R}z\oplus \mathbb{R}z_0$ you can find an explicit rotation which sends $z_0$ to $z$ while leaving $x$ fixed. It thus sends $y_0$ to $y$ and the orthonormal basis $(e_1,\ldots,e_n)$ to another orthonormal basis with the expected property.

  • $\begingroup$ Great, thank you! $\endgroup$
    – Libli
    Feb 22, 2019 at 19:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .