# Find Matrix $S$ such that $S'CS=\left(\begin{array}{cc} 1_{N-1} & 0 \\ 0 & 0 \end{array} \right)$ where $C:=1_N-\iota \iota'$

The Centering Matrix $C:=1_N-\iota \iota'$ has eigenvalue $1$ of multiplicity $n − 1$ and eigenvalue $0$ of multiplicity $1$. Therefore a matrix $S$ with columns consisting of eigenvectors of $C$ can be chosen such that $$CS=S\left(\begin{array}{cc} 1_{N-1} & 0 \\ 0 & 0 \end{array} \right).$$

I understand that for the first $N-1$ eigenvectors $s$ it must hold that $$Cs=s-\frac{1}{N}\iota\iota's=s \Leftrightarrow \iota's=0$$ and for the last eigenvector $s_N$ it must hold that $$Cs_N=s_N-\frac{1}{N}\iota\iota's_N=0 \Leftrightarrow s_N=\alpha\iota, \alpha\in\mathbb{R}$$

Therefore to obtain $S'S=1_N$ it must hold that $s_N=\frac{1}{\sqrt{N}}\iota$. However, how can I choose $S=(s_1,\ldots,s_{N-1})$ such that $S'S=1_N$?

Found one possible characterization of $S$. As already mentioned $s_N=\frac{1}{\sqrt{N}}\iota$ to fullfill $S'S$ (Otherwise the last entry of $S'S$ is not equal to $1$.) If have chosen the following vectors $s_i$: $$s_1=\left(\begin{array}{c} \frac{1}{k_i}\\-\frac{1}{k_i} \\ 0\end{array} \right),s_2=\left(\begin{array}{c} \frac{1}{k_2}\\\frac{1}{k_2} \\ -\frac{2}{k_2}\\0\end{array} \right),\ldots,s_i=\left(\begin{array}{c} \frac{1}{k_i}\\\vdots \\ -\frac{i}{k_i}\end{array} \right).$$ These $s_i$ fullfill $\iota's_i$ and by construction $s_j's_k=0$. Therefore the $s_i$ form a basis of the subspace of Dimension $N-1$ spanned by range$(C)$. For $S'S=1_N$ to hold we need $s_i's_i=1$ therefore we have to choose $$k_i=\sqrt{i+i^2}$$.