Let $\mathbf{x_i}=[ x_{i1}, x_{i2}, x_{i3} ]'$. Why can we say that $\sum_i^n \mathbf{x_i}\mathbf{x_i}'$ is positive definite?
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Take $n$ to be finite. – econ86 Oct 23 '19 at 12:43
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@RodrigodeAzevedo $n$ is not the size of the space; each vector has 3 entries. – Ben Grossmann Oct 23 '19 at 12:44
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@Omnomnomnom Sure. What if $n = 1$? In that case, how could the matrix be positive definite? – Rodrigo de Azevedo Oct 23 '19 at 13:12
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Related: https://math.stackexchange.com/q/1840801/339790 – Rodrigo de Azevedo Oct 23 '19 at 13:23
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A direct proof that your matrix is positive semidefinite is as follows: for any $v$, $$ v^T \left( \sum_{i=1}^n x_ix_i^T \right)v = \sum_{i=1}^n (v^Tx_i)(x_i^Tv) = \sum_{i=1}^n (x_i^Tv)^2 \geq 0. $$ Note also that this matrix is symmetric. Your matrix will be positive definite if and only if the vectors $x_i$ span $\Bbb R^3$, which is to say that the matrix $$ X = \pmatrix{x_1 & x_2 &\cdots & x_n} $$ has full row-rank. To see this, note that we can only have $x_i^Tv = 0$ for all $x_i$ if $v$ lies in the orthogonal complement of the span of the $x_i$.
Ben Grossmann
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