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Let $e_i$ the $i$-th column of the identity matrix. Is there an easy way to prove that the matrix

$$\left[\matrix{\mathbb{I}_n & e_1e_2^T & \cdots & e_1e_n^T\\ e_2e_1^T & \mathbb{I}_n & \cdots & e_2e_n^T \\ \vdots & \vdots & \ddots & \vdots\\ e_ne_1^T & e_ne_2^T &\cdots & \mathbb{I}_n}\right]$$

is positive semidefinite?

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The matrix is equal to

$$

\tilde{I}+\pmatrix{e_1\\e_2\\\vdots\\e_n}\pmatrix{e_1\\e_2\\\vdots\\e_n}^T,

$$

where $\tilde{I}$ is a block diagonal matrix where the $k$th diagonal block is the $n\times n$ identity matrix with the $(k,k)$-entry set to zero (compensated in the second term with $e_ke_k^T$). This “crippled” identity is positive semidefinite and a rank-one semidefinite matrix is added to it, so the result is positive semidefinite.

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