Positive definiteness of $\sum_{i=1}^n \boldsymbol{x}_i \boldsymbol{x}_i^\top$ Let $\{\boldsymbol{x}_i\}_{i=1}^n$ be a set of $n$ linearly independent vectors in $\mathbb{R}^n$. 


  My question: Is $\displaystyle\sum_{i=1}^n \boldsymbol{x}_i\boldsymbol{x}_i^\top$ always positive definite, i.e. $\displaystyle\sum_{i=1}^n \boldsymbol{x}_i\boldsymbol{x}_i^\top&gt;0$?


Clearly, if the set $\{\boldsymbol{x}_i\}_{i=1}^n$ is composed by orthogonal vectors the answer is in the affirmative. What about the general case?
The answer is yes, and the $x_i$ need not be linearly independent so long as they span $\Bbb R^n$.  For any vector $y$, we have
$$
y^T\left(\sum_{i} x_ix_i^T \right) y = \sum_{i}(x_i^Ty)^2 \geq 0
$$
so clearly the matrix is positive semidefinite.  As for positive definiteness, note that the matrix can be written as $AA^T$, where
$$
A = \pmatrix{x_1 &amp; x_2 &amp; \cdots &amp; x_N} \qquad (N \geq n)
$$
since $A$ has full row-rank, $AA^T$ must be invertible.  Also, for the first part, it suffices to note that
$$
y^T(AA^T)y = (A^Ty)^T(A^Ty) \geq 0
$$
You don't use your set anywhere in the question. The sum you write is just $\Vert x_i \Vert^2$ and yes, this is always greater than $0$ as long as $x_i \neq 0$.