If the inner product of two vectors results in a positive definite matrix, does their commutative inner product result in a positive scalar? We are given two row vectors $x, y \in \mathbb R^n $ such that the product $ x^T \cdot y $ yields a positive-definite $n\times n$ matrix. Does the inner product $ x \cdot y^T $ yield a positive scalar?
 A: The general statement we are looking at here is the following.
Proposition. Let $A\in\mathbb R^{n\times n}$ be a positive-definite matrix, then $\operatorname{Tr}(A)>0$.
Proof. Let $e_i\in\mathbb R^n$ the $i$-th standard basis vector, then
$$
(e^i)^T A e^i = a_{ii} > 0,
$$
thus
$$
\operatorname{Tr}(A) = \sum_{i=1}^n a_{ii} > 0.
$$
This concludes the proof. $\square$
In your case $A=x^T y$ for $x,y\in\mathbb R^{1\times n}$, so $a_{ij} = x_i y_j$ and
$$
\operatorname{Tr}(A) = \sum_{i=1}^n a_{ii} = \sum_{i=1}^n x_i y_i = xy^T.
$$
A: The answer is yes. 
Consider the case where $n=2$. Let $x= (x_1,x_2)$ and $y=(y_1,y_2)$. 
Then $$x^Ty = \left ( \begin{array}{cc} x_1y_1 & x_1 y_2 \\ x_2 y_1 & x_2 y_2 \end{array} \right )$$
and
$$ xy^T = x_1y_1 + x_2 y_2$$
Use the standard basis vectors $e_1$ and $e_2$ to obtain $e_1^T x^Ty e_1 = x_1y_1 > 0$ and $e_2^T x^Ty e_2 = x_2y_2>0$. Now generalise to $n>2$.
A: If $\pmb{x} = \begin{pmatrix} x_1 \\ \vdots \\ x_n \end{pmatrix},\pmb{y} = \begin{pmatrix} y_1 \\ \vdots \\ y_n \end{pmatrix}$ are vectors in $\mathbb{R}^n$, their inner product $\pmb{x} \cdot \pmb{y}$ is defined by
$$\pmb{x} \cdot \pmb{y} = \pmb{x}^T\pmb{y}$$
where the product at the right denotes the usual product of matrices. Since $\pmb{x}^T$ is a $1 \times n$ matrix and $\pmb{y}$ is a $n \times 1$ matrix, their inner product is a $1 \times 1$ matrix, i.e. a scalar, not a bigger, positive definite matrix.
On the other hand, the product $\pmb{x} \pmb{y}^T$ is an $n \times n$ matrix. But this product is not an inner product, by definition.
EDIT: If $\pmb{x}$ and $\pmb{y}$ are row vectors then $\pmb{x}^T \pmb{y}$ is an $n \times n$ matrix while $\pmb{x} \pmb{y}^T$ is a scalar. In this case, the inner product would be the scalar $\pmb{x} \pmb{y}^T$. The product $\pmb{x}^T\pmb{y}$ is usually not called an inner product in this case. Also, the scalar $\pmb{x}\pmb{y}^T$ need not be positive. Take for instance $\pmb{x} = (1, 0)$ and $\pmb{y} = (-1, 0)$.
