Inverse of the outer product of some vectors with their transpose Assume i have $n$ 3D unit vectors $v_s$, with different values.
Then i define a matrix $T$ as:
$$
T = \frac{1}{n} \sum_{s=1}^{n} v_s \times v_s' 
$$
where $v_s$ are $3\times 1$ vectors and therefore $T$ is a $3\times 3$ matrix.
How can i find a possible set of vectors $v_s$ if a positive definite diagonal matrix $T$ (with $trace=1$) is given? (any numerical or non-exact method is acceptable)
 A: Consider a unit vector $v=(\alpha,\beta,0)^\prime.$  Setting $\bar v = (\alpha,-\beta,0)^\prime$ (obtained by negating the middle value), the condition $\alpha^2 + \beta^2 =1$ implies
$$vv^\prime + \bar v {\bar v}^\prime = \pmatrix{2\alpha^2 & 0 & 0 \\ 0 & 2(1-\alpha^2)& 0 \\0 & 0 & 0}.$$
When we similarly take a unit vector $w = (0, \gamma,\delta)^\prime$ we find
$$ww^\prime + \bar w {\bar w}^\prime = \pmatrix{0 & 0 & 0 \\ 0 & 2(1-\delta^2)& 0 \\0 & 0 & 2\delta^2 }.$$
Averaging all four terms, we obtain a diagonal, unit-trace matrix
$$\left(vv^\prime + \bar v {\bar v}^\prime + ww^\prime + \bar w {\bar w}^\prime\right)/4 = \pmatrix{\alpha^2/2 & 0 & 0 \\ 0 & (2-\alpha^2-\delta^2)/2& 0 \\0 & 0 & \delta^2/2 }.$$
Thus, out of many possible solutions with $n=4,$

given a positive-definite unit trace diagonal matrix $T$ with values $\lambda_1,\lambda_2,\lambda_3$ on the diagonal, take $\alpha=\sqrt{2\lambda_1}$ and $\delta =\sqrt{2\lambda_3}$ to produce the solution $(v_1,v_2,v_3,v_4) = (v, \bar v, w, \bar w).$

This will make the middle values equal by virtue of the unit trace conditions.  Positive-definiteness is equivalent to the positivity of all the $\lambda_i,$ showing these solutions always exist.
