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Say we have a column vector $x=[x_1\ x_2\ x_3]^T$. Then is $ xx^T $ positive semi definite.

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closed as off-topic by Claude Leibovici, Shailesh, астон вілла олоф мэллбэрг, TheGeekGreek, Namaste Feb 9 '17 at 12:13

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  • $\begingroup$ It is not difficult to write down the eigenvalues of $x x^T$ explicitly (hint: think of the rank-nullity theorem), and symmetry is obvious. $\endgroup$ – Ian Feb 9 '17 at 6:54
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Let $y \in \mathbb{R}^3$,

$$y^Txx^Ty=(y^Tx)^2 \geq 0$$

Hence $xx^T$ is positive semidefinite.

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