Say we have a column vector $x=[x_1\ x_2\ x_3]^T$. Then is $ xx^T $ positive semi definite.
Is the outer product of a column vector with itself positive semi definite?
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0It is not difficult to write down the eigenvalues of $x x^T$ explicitly (hint: think of the rank-nullity theorem), and symmetry is obvious. – 2017-02-09
1 Answers
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Let $y \in \mathbb{R}^3$,
$$y^Txx^Ty=(y^Tx)^2 \geq 0$$
Hence $xx^T$ is positive semidefinite.