Positive definite matrix implies a generating set

Question

If the matrix $$\sum_{i=1}^my^i(y^i)^T$$ is positive definite, then the set $\{y^1,\dots,y^m\}\subseteq \mathbb R^n$ is a a generating set for $\mathbb R^n$

Does anyone know how I can prove that?

Answer 1

Let $U$ denote the span of $\{y^1,\dots,y^m\}$. Note that the orthogonal complement $U^\perp$ is the set $$ U^\perp = \{x \in \Bbb R^n : x^Ty^1 = x^Ty^2 = \cdots = x^Ty^m = 0\}. $$ So, suppose that $M = \sum_{i=1}^my^i(y^i)^T$ is positive definite. It follows that for any non-zero $x \in \Bbb R^n$, we have $$ 0 < x^TMx = x^T\left(\sum_{i=1}^my^i(y^i)^T\right)x = \sum_{i=1}^m x^Ty^i(y^i)^Tx = \sum_{i=1}^m (x^Ty^i)^2. $$ In other words: for every $x \in \Bbb R^n$, we have $x^Ty^i \neq 0$ for some $i$ from $1$ to $m$. In other words, every non-zero $x \in \Bbb R^n$ is not an element of $U^\perp$. So, we have $U^\perp = \{0\}$, which means that $U = U^{\perp \perp} = \Bbb R^n$. So, the span of the vectors $y^1,\dots,y^m$ is $\Bbb R^n$, which is to say that these vectors form a generating set for $\Bbb R^n$ as desired.

Answer 2

The matrix is symmetric and thus has a basis consisting of eigenvectors. It is also positive semi-definite since $v^\top \left(\sum_{i=1}^m y^i (y^i)^\top \right) v = \sum_{i=1}^m (v^\top y^i)^2 \ge 0$ for all $v$.

Thus it is positive definite if and only if its nullspace is empty. One can check that the nullspace of this matrix consists of vectors that are orthogonal to each of the $y^i$. So if the matrix is positive definite, then the $y^i$ must span $\mathbb{R}^n$.