TITLE: A specific type of orthogonal bases
QUESTION [7 upvotes]: In $\mathbb{R}^n$, can we find an orthogonal basis $v_1,v_2,\ldots,v_n$ with $\|v_i\|=1$ such that
$$\|\bar{v}_1\|=\|\bar{v}_2\|= \cdots =\|\bar{v}_n\|$$
where $\bar{v}_i\in\mathbb{R}^m$ is the vector formed by the first $m$ coordinates of $v_i$ with respect to the standard basis and $m$ is a given integer $m<n$.

REPLY [4 votes]: I think there's a trick that can make joriki's idea work in any case (I hope I'm not mistaken). Let's denote $A_{m,n}$ the assertion that there exists a solution to the problem for the values $m \le n$ (we take the convention that $A_{0,n}$ is always true). Obviously, $A_{m,m}$ is true for all $m$. By flipping rows, we see that
$A_{m, n} \Longleftrightarrow A_{n-m, n}$
And joriki showed that $A_{m,n}$ is equivalent to the fact that there exist a (non zero) $m$ by $n$ matrix whose rows are orthonormal, and columns have same norm. So by concatenation of such matrices (with suitable scaling factor so that the rows remain of norm 1, and column of both matrices have same norm), we get
$(A_{m, n_1} \textrm{ and } A_{m, n_2} ) \Longrightarrow A_{m, n_1 + n_2}$
Let $m < n$ be two integers, and $r$ the remainder of $m$ modulo $n$. Using the properties above, we see that 
$A_{r, m} \Longrightarrow A_{r, m+r} \Longrightarrow A_{m, m+r} \Longrightarrow A_{m, n}$
So checking if $A_{m,n}$ is true can be reduced to checking if $A_{r, m}$ is. Applying Euclide's algorithm, we can reduce this to a point where the remainder is $0$, in which case there is a solution.