Is a positive definite matrix times a positive semidefinite matrix positive semidefinite? I have a couple questions actually, couldn't get both of them in the title.
First suppose I have a positive definite matrix H and a positive semidefinite matrix D.  I believe that H*D would be a positive semidefinite matrix.  Is this correct?
Second suppose I have a vector $g$.  If I were to multiply $g$ by its transpose so that I get a matrix and not a single value, would the resultant matrix be positive semidefinite?  I believe that it will be.
I'm not looking for a proof (though it wouldn't be out of place for others looking for answers to similar questions) just yes or no answers and a brief explanation of why I am wrong, if am wrong.  I am working on a proof and want to make sure that my intuition is correct. 

The matrix $uu^T$ is a special case of $\mathsf{PSD}$, (Positive semidefinite), matrices called a symmetric dyad.  
Indeed, if $A = uu^T$ for some vector $u \in \mathbf{R}^n$, then for every $x$:
$$q_A (x)  = x^T uu^T x = (u^Tx)^2 \ge 0.   $$

Another way to look at this is thus:
$$\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \cdot \begin{bmatrix} x_1 & x_2 & \ldots & x_n \end{bmatrix} = \begin{bmatrix}x_1^2 & x_1x_2 & \ldots & x_1x_n \\ x_2x_1 & x_2^2 & \ldots & x_2x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_nx_1 & x_nx_2 &\ldots &x_n^2 \end{bmatrix}$$
What can you say about the above matrix? In particular, it's symmetric, and what property do all of its leading principal minors have in common? 
To answer your first question, I would direct you to an answer of Plainview's in 
the following question: Is the product of symmetric positive semidefinite matrices positive definite? 

The product of two symmetric $\mathsf{PSD}$ matrices is $\mathsf{PSD}$, iff the product is also symmetric. More generally, if $A$ and $B$ are $\mathsf{PSD}$, $AB$ is $\mathsf{PSD}$ iff AB is normal, ie, $(AB)^T AB = AB(AB)^T$.


Since your matrix $H$ is positive definite, it is also positive semidefinite. Therefore, the product $HD$ is positive semidefinite iff 
$$
(HD)^T HD = HD(HD)^T
$$
