Sum of two density matrices: $\rho=p_1\rho_1+p_2\rho_2$ Suppose we have
$$\rho=p_1\rho_1+p_2\rho_2$$
Where $\rho_1$ and $\rho_2$ are density matrices with $p_1+p_2=1$.
I'm trying to show this is also a density matrix.
If we let
$$\rho_1=\sum_i^n p_{\psi_i} |\psi_i \rangle \langle\psi_i|$$
and
$$\rho_2=\sum_i^n p_{\phi_i} |\phi_i \rangle \langle\phi_i|.$$
I'm assumsing these two density matrices are of size $n$, otherwise adding them wouldn't make any sense. I'm having trouble seeing how this produces a density matrix, if it were too then it would want to be describing the probabilities of the combinations of combined quantum states, which would be a $n^2\times n^2$ matrix? That's all I can see as being an physical interpretation of this as.
Approaching it more mathematically, each $n\times n$ matrix in the sum over both $p_1\rho_1$ and $p_2\rho_2$ has a factor of $p_1p_{\psi_i}+p_2p_{\phi_i}$ and
$$\sum_i^n p_1p_{\psi_i}+p_2p_{\phi_i}=\sum_i^n p_1(p_{\psi_i}-p_{\phi_i})+p_{\phi_i}=1,$$
Which is promising (and the only way i've found so far to use the condition on $p_1,p_2$), but as far as I can see this factor doesn't really mean anything. Any help would be greatly appreciated, I'm a little lost!
 A: Hints: 
A density operator is by definition a (semi-)positive operators with trace equal to one.
OP is essentially asking 

Is a convex linear combination of density operators again a density operator?

Answer: Yes, because:


*

*Semi-positive operators form a cone, and

*trace is linear. 
To see pt. 1, note that operators $A$ in complex$^1$ Hilbert spaces enjoy the characterizations
$$A ~\text{semi-positive}\qquad \Leftrightarrow \qquad  \forall v\in H~:~ \langle v| Av \rangle ~\geq~ 0, $$  
and
$$A ~\text{Hermitian}
\qquad \Leftrightarrow \qquad 
\forall v,w\in H~:~ \langle v| Aw \rangle ~=~\langle Av| w \rangle $$
$$\qquad \Leftrightarrow \qquad 
\forall v\in H~:~ \langle v| Av \rangle ~=~\langle Av| v \rangle 
\qquad \Leftrightarrow \qquad
\forall v\in H~:~ \langle v| Av \rangle ~\in \mathbb{R}. $$  
--
$^1$These characterizations do not hold for real Hilbert spaces, so here we are using that quantum mechanics are formulated in complex Hilbert spaces.
A: That depends on what you understand "density matrix" to mean. You seem to think that it refers to operators of the form
$$\rho=\sum_{k=1}^n p_k|\psi_k\rangle\langle\psi_k|,\tag{1}$$
where $p_k\geq0$ for all $k$, $\sum_{k=1}^n p_k=1$, and the $|\psi_k\rangle$ are vectors in some $N$-dimensional Hilbert space $\mathcal{H}$. As a contrast, QMechanic's answer relies on a characterization of density matrices as positive semidefinite hermitian operators with trace 1. Proving the equivalence of these definitions is a very informing exercise and it will probably teach you more than your current problem.
To prove that $\rho=p_1\rho_1+p_2\rho_2$ is a density matrix by your definition, you will need to rely on an eigenvalue-eigenvector decomposition. Writing $\rho$ as in equation (1) is possible because $\rho$ is hermitian; the problem is then proving the two conditions on the $p_k$. The first one is equivalent to $\rho$ being positive semidefinite (why?), and this you can prove using the (real) abstract definition of that: $$\langle v|\rho v\rangle\geq0\,\forall v\in \mathcal{H}.$$
The sum condition you can prove by taking the trace of the different expressions you have for $\rho$.
