Outer products & positive (semi-) definiteness

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The discussion focuses on the positive (semi-) definiteness of the symmetric matrix defined by the outer products of two vectors, ##M = uv' + vu'##. An equivalent condition for ##M## to be positive (semi-) definite is that the product of the projections of any vector ##x## onto ##u## and ##v## must be non-negative, expressed as ##(x \cdot u)(x \cdot v) \geq 0##. The analysis further simplifies the problem by expressing the arbitrary vector ##z## in terms of ##u## and ##v##, leading to the formulation of the quadratic form ##z^\dagger M z = A^\dagger Q A##, where ##Q## is a Hermitian 2x2 matrix derived from the inner products of ##u## and ##v##. The positivity properties of ##M## can thus be determined by computing the eigenvalues of ##Q##.

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Let ##u,v## be vectors in the same Euclidean space, and define the symmetric matrix ##M = uv'+vu'##, the sum of their two outer products.

I'm interested in whether or not ##M## is positive (semi)definite.

Does anybody know of any equivalent conditions that I might phrase "directly" in terms of the vectors ##u,v##?
 
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An equivalent condition is ##(x\cdot u)(x\cdot v)\geq 0## for all vectors ##x##.
 
Let ##z## be an arbitrary vector, then we are interested in the properties of
$$z^\dagger M z = \langle z, u \rangle \langle v, z \rangle + \langle z, v \rangle \langle u, z \rangle .$$
Assume that ##v\neq u##, then we can write
$$ z = a u + b v + z_\perp,~~~\langle z_\perp, u \rangle = \langle z_\perp, v \rangle =0.$$
Then, with ## A = (a~~b)^T##, we have
$$z^\dagger M z = A^\dagger Q A,$$
where ##Q## is a Hermitian 2x2 form determined in terms of the inner products of ##u## and ##v##. Positivity properties of ##M## are reduced to those of ##Q## for which it is simple to compute the eigenvalues.
 

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