Suppose $x_i\in\mathbb R^n$, $i=1,...,r$, $n>r$ and $x_i^Tx_j=0$ $\forall i\ne j$. I need to show that the matrix:

$$ \sum_{k=1}^rx_kx_k^T-\textbf{11}^T\succeq 0. $$

It's easy to see that this is the case by rewriting the above as $\sum_{k=1}^rx_kx_k^T\succeq \textbf{11}^T$. The left hand side has spectrum of $r$-times ones and the rest zeros, and the right hand side has spectrum $\{1,0,...,0\}$ thus the inequality holds. However, I'd like to know if one can show this via the Schur complement since:

$$ \sum_{k=1}^rx_kx_k^T-\textbf{11}^T\succeq 0\quad\Leftrightarrow\quad \begin{bmatrix} 1 & \mathbf{1}^T \\ \mathbf{1} & \sum_{k=1}^rx_kx_k^T \end{bmatrix}\succeq 0. $$

This is a "1-bordered on top and left matrix and we have $1>0$ and $\sum_{k=1}^rx_kx_k^T\succeq 0$. However, I am not able to think of a trik to show that the matrix is positive semidefinite. Any help much appreciated!


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