Orthogonal vectors translation using standard vectors

Question

When $n=2m$, let us consider the following vectors $\mathbf{v}_1,\ldots, \mathbf{v}_n$ in $\mathbb{R}^n$
$$\mathbf{v}_q=(v_{1q},\ldots,v_{n,q})$$ $$v_{p,q}=\sin\Big(\frac{pq}{n+1}\pi\Big)$$ It is known that $\mathbf{v}_1,\ldots, \mathbf{v}_n$ are mutually orthogonal.

Suppose that $\sigma$ is a permutation on the set of $\{1,\dots,n\}$. Let us define $$\mathbf{w}_j=\mathbf{v}_{\sigma(j)}+\mathbf{e}_{\sigma(j)}$$ where $\mathbf{e}_j$ is the $j$th-standard vector in $\mathbb{R}^n$, with all entries equal to zero except for the $j$th component, which is $1.$

Q. Can we conclude that $\mathbf{w}_1,\ldots, \mathbf{w}_m$ form a linearly independent set in $\mathbb{R}^n$?

p.s. Nevertheless, this question deals with the given set of orthogonal real vectors $\mathbf{v}_1,\ldots,\mathbf{v}_n$, it is likely to be generalized for a broader class of orthogonal vectors.

When examining $\mathbf{w_j}$,it should be noted that we are concentrating on $m$ arbitrary vectors $\mathbf{v_j+e_j}$, which may not necessarily correspond to the first $m$ vectors. T o clarify this point, a permutation is applied. — ABB
– ABB, Commented Oct 22, 2023 at 17:11

Iosif Pinelis · Accepted Answer · 2023-10-22 21:21:33Z

$\newcommand\v{\mathbf v}\newcommand\w{\mathbf w}$Let us show that the vectors $\w_1,\dots,\w_n$ are linearly independent. Of course, then the first $m$ of these $n$ vectors are linearly independent.

Without loss of generality, the permutation $\sigma$ is the identity permutation, because changing the order of vectors does not affect their linear independence (or the lack thereof).

The Gram matrix of $\v_1,\dots,\v_n$ is $\frac n2\,I=mI$, where $I$ is the identity matrix. So, the matrix $V$ with columns $\v_1,\dots,\v_n$ is of the form $cQ$, where $c:=\sqrt{m}$ and $Q$ is an orthogonal matrix.

So, the Gram matrix of $\w_1,\dots,\w_n$ is $G:=(I+cQ)^\top(I+cQ)=(1+c^2)I+c(Q^\top+Q)$. So, for any $n=2m\ge4$ and any $n\times 1$ matrix $x$ with $x^\top x=1$, in view of the Cauchy--Schwarz inequality we have $$x^\top Gx=1+c^2+2cx^\top Qx\ge1+c^2-2c=(\sqrt{m}-1)^2>0.$$ So, $G$ is nonsingular and hence $\w_1,\dots,\w_n$ are linearly independent.

If now $n=2$, then the determinant of the matrix with columns $\w_1,\w_2$ is $-1/2\ne0$, so that $\w_1,\w_2$ are linearly independent.

Thus, $\w_1,\dots,\w_n$ are linearly independent for any even $n$. $\quad\Box$

Let us consider vectors $\mathbf{v_j}$s are all orthogonal (it can be assumed by scaling). Then $c=1$ and so $1+c^2-2c=0$. — ABB
– ABB, Commented Oct 23, 2023 at 8:12
Indeed, despite the fact that the vectors $\mathbf{w}_1, \ldots, \mathbf{w}_n$ are never linearly independent, it seems that $\mathbf{w}_1, \ldots, \mathbf{w}_m$ are linearly independent. — ABB
– ABB, Commented Oct 23, 2023 at 8:15
@ABB : Your $\mathbb v_j$'s are orthogonal, but $c=\sqrt m>1$ if $m>1$. Perhaps, you now want to rescale the $\mathbb v_j$'s to make them orthonormal, but that would be another problem, right? — Iosif Pinelis
– Iosif Pinelis, Commented Oct 23, 2023 at 13:25
The matrix $S$, whose columns are $\mathbf{v}_1, \ldots, \mathbf{v}_n$, is a symmetric matrix satisfying the equation $S^2 = c^2\mathbf{I}$ for some positive scalar $c$. This implies that it has only two eigenvalues, namely $\pm\sqrt{c}$. On the other hand, one may observe that all $n$-vectors $\mathbf{v}_j + \sqrt{c}~\mathbf{e}_j$ are eigenvectors corresponding to the eigenvalue $\sqrt{c}$. Therefore, they can not be linearly independent. — ABB
– ABB, Commented Oct 23, 2023 at 14:29
Hence, You are right, it could be considered as another problem and your argument works in this case. — ABB
– ABB, Commented Oct 23, 2023 at 14:33

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