First, the case where $n$ is odd, with $n = 2k + 1$: Let $\theta = 2 \pi /n$. Taking the real and imaginary parts of the columns of the DFT matrix gives us the following nice orthogonal basis for $u_1^\perp$: $$ c_1 = [1\ \ \cos \theta \ \ \cdots \ \ \cos ((n-1)\theta)]\\ c_2 = [1\ \ \cos (2\theta) \ \ \cdots \ \ \cos (2(n-1)\theta)]\\ \vdots \\ c_{k} = [1\ \ \cos (k\theta) \ \ \cdots \ \ \cos (k(n-1)\theta)]\\ s_1 = [1\ \ \sin \theta \ \ \cdots \ \ \sin ((n-1)\theta)]\\ s_2 = [1\ \ \sin (2\theta) \ \ \cdots \ \ \sin (2(n-1)\theta)]\\ \vdots \\ s_{k} = [1\ \ \sin (k\theta) \ \ \cdots \ \ \sin (k(n-1)\theta)].\\ $$\begin{align*} c_1 &= [1\ \ \cos \theta \ \ \cdots \ \ \cos ((n-1)\theta)]\\ c_2 &= [1\ \ \cos (2\theta) \ \ \cdots \ \ \cos (2(n-1)\theta)]\\ &\ \ \vdots \\ c_{k} &= [1\ \ \cos (k\theta) \ \ \cdots \ \ \cos (k(n-1)\theta)]\\ s_1 &= [1\ \ \sin \theta \ \ \cdots \ \ \sin ((n-1)\theta)]\\ s_2 &= [1\ \ \sin (2\theta) \ \ \cdots \ \ \sin (2(n-1)\theta)]\\ &\ \ \vdots \\ s_{k} &= [1\ \ \sin (k\theta) \ \ \cdots \ \ \sin (k(n-1)\theta)].\\ \end{align*} In the case that $n$ is even, we do essentially the same thing, but also include the vector $[-1,1,-1,\dots,1].$
First, the case where $n$ is odd, with $n = 2k + 1$: Let $\theta = 2 \pi /n$. Taking the real and imaginary parts of the columns of the DFT matrix gives us the following nice orthogonal basis for $u_1^\perp$: $$ c_1 = [1\ \ \cos \theta \ \ \cdots \ \ \cos ((n-1)\theta)]\\ c_2 = [1\ \ \cos (2\theta) \ \ \cdots \ \ \cos (2(n-1)\theta)]\\ \vdots \\ c_{k} = [1\ \ \cos (k\theta) \ \ \cdots \ \ \cos (k(n-1)\theta)]\\ s_1 = [1\ \ \sin \theta \ \ \cdots \ \ \sin ((n-1)\theta)]\\ s_2 = [1\ \ \sin (2\theta) \ \ \cdots \ \ \sin (2(n-1)\theta)]\\ \vdots \\ s_{k} = [1\ \ \sin (k\theta) \ \ \cdots \ \ \sin (k(n-1)\theta)].\\ $$ In the case that $n$ is even, we do essentially the same thing, but also include the vector $[-1,1,-1,\dots,1].$
First, the case where $n$ is odd, with $n = 2k + 1$: Let $\theta = 2 \pi /n$. Taking the real and imaginary parts of the columns of the DFT matrix gives us the following nice orthogonal basis for $u_1^\perp$: \begin{align*} c_1 &= [1\ \ \cos \theta \ \ \cdots \ \ \cos ((n-1)\theta)]\\ c_2 &= [1\ \ \cos (2\theta) \ \ \cdots \ \ \cos (2(n-1)\theta)]\\ &\ \ \vdots \\ c_{k} &= [1\ \ \cos (k\theta) \ \ \cdots \ \ \cos (k(n-1)\theta)]\\ s_1 &= [1\ \ \sin \theta \ \ \cdots \ \ \sin ((n-1)\theta)]\\ s_2 &= [1\ \ \sin (2\theta) \ \ \cdots \ \ \sin (2(n-1)\theta)]\\ &\ \ \vdots \\ s_{k} &= [1\ \ \sin (k\theta) \ \ \cdots \ \ \sin (k(n-1)\theta)].\\ \end{align*} In the case that $n$ is even, we do essentially the same thing, but also include the vector $[-1,1,-1,\dots,1].$
First, the case where $n$ is odd, with $n = 2k + 1$: Let $\theta = 2 \pi /n$. Taking the real and imaginary parts of the columns of the DFT matrix gives us the following nice orthogonal basis for $u_1^\perp$: $$ c_1 = [1\ \ \cos \theta \ \ \cdots \ \ \cos ((n-1)\theta)]\\ c_2 = [1\ \ \cos (2\theta) \ \ \cdots \ \ \cos (2(n-1)\theta)]\\ \vdots \\ c_{k} = [1\ \ \cos (k\theta) \ \ \cdots \ \ \cos (k(n-1)\theta)]\\ s_1 = [1\ \ \sin \theta \ \ \cdots \ \ \sin ((n-1)\theta)]\\ s_2 = [1\ \ \sin (2\theta) \ \ \cdots \ \ \sin (2(n-1)\theta)]\\ \vdots \\ s_{k} = [1\ \ \sin (k\theta) \ \ \cdots \ \ \sin (k(n-1)\theta)].\\ $$ In the case that $n$ is even, we do essentially the same thing, but also include the vector $[-1,1,-1,\dots,1].$