Why is $tr((a_{ij})=\sum\limits_{i=1}^n a_{ii}$ k-positive for all $k$? Let $A$ be a $C^*$-algebra and $$Tr:M_n(A)\to A,$$
$$(a_{ij})\mapsto \sum\limits_{i=1}^n a_{ii}.$$ The claim is that this map is k-positive for all $k\in\mathbb{N}$.
Let $k\in\mathbb{N}$ and consider the induced map $$Tr^{(k)}:M_k(M_n(A))\to M_k(A)$$ $$(A_{i,j})_{i,j=1,..,k}\mapsto Tr^{(k)}((A_{i,j})_{i,j=1,..,k})=(Tr((A_{i,j}))_{i,j=1,..,k},$$ where $A_{i,j}\in M_n(A)$ and $M_k(M_n(A))\cong M_{kn}(A)$. 
Consider $(A_{i,j})^*(A_{i,j})\ge 0$, where $(A_{i,j})^*=(A_{j,i}^*)$, I'm stuck to prove that $(Tr((A_{j,i}^*)(A_{i,j}))_{i,j=1,..,k}\ge 0$ (I only know how to prove it for $k=1$). 
In order to gain a better overview I tried to do it for $k=2$. Let $A,B,C,D \in M_2(A)$, consider $\begin{pmatrix} A & B \\ C & D \end{pmatrix}\in M_4(A)$, then $$Tr^{(2)}((\begin{pmatrix} A^* & C^* \\ B^* & D^* \end{pmatrix})(\begin{pmatrix} A & B \\ C & D \end{pmatrix})=Tr^{(2)}(\begin{pmatrix} A^*A+C^*C & A^*B+C^*D \\ B^*A+B^*C & B^*B+D^*D \end{pmatrix})=\begin{pmatrix} Tr(A^*A+C^*C) & Tr(A^*B+C^*D) \\ Tr(B^*A+B^*C) & Tr(B^*B+D^*D) \end{pmatrix}.$$ 
Let $(x,y)\in H\oplus H$ for an appropiate Hilbert space $H$, it is 
$$\langle \begin{pmatrix} Tr(A^*A+C^*C) & Tr(A^*B+C^*D) \\ Tr(B^*A+B^*C) & Tr(B^*B+D^*D) \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix},\begin{pmatrix} x\\ y\end{pmatrix}\rangle =\langle Tr(A^*A+C^*C)x,x \rangle + \langle Tr(B^*A+B^*C)x,y \rangle + \langle Tr(A^*B+C^*D)y,x \rangle + \langle Tr(B^*B+D^*D)y,y \rangle.$$ It is $Tr(A^*A+C^*C), Tr(B^*B+D^*D)\ge 0$ and therefore $\langle Tr(A^*A+C^*C)x,x \rangle , \langle Tr(B^*B+D^*D)y,y \rangle \ge 0$ but I don't know what to do with the other summands. And I'm not sure if such a calculation is appropriate in the general case if $k$ is arbitrary. Can anyone help me? 
 A: The easy way is to notice that maps $a\longmapsto xax^*$ are always cp when they make sense (even if they are matrices and $x$ is rectangular). Then, with $x_j=\begin{bmatrix}0&\cdots0&&1&0\cdots&0\end{bmatrix}$ (the $1$ in the $j^{\rm th}$ position),
$$
\text{Tr}(a)=\sum_{j=1}^nx_jax_j^*
$$
is cp. 
The hard way is the following. Consider $\text{Tr}^{(k)}:M_k(M_n(A))\to M_k(A)$. Let $a\in M_r(M_n(A))$ be positive. The $a=b^*b$ for some $b\in M_r(M_n(A))$. Write $b$, with the canonical matrix units $e_{ij}$ for $M_r(\mathbb C)$, as a block matrix on $M_n(A)$:
$$
b=\sum_{i,j} b_{ij}\otimes e_{ij}
$$
Now
\begin{align}
\text{Tr}^{(k)}(b^*b)&=\text{Tr}^{(k)}\left(\sum_{i,j} b_{ij}^*\otimes e_{ji}\,\sum_{s,t} b_{st}\otimes e_{st} \right)
=\text{Tr}^{(k)}\left(\sum_{i,j,s,t} b_{ij}^*b_{st}\otimes e_{ji}e_{st}\right)
\\ \ \\
&=\text{Tr}^{(k)}\left(\sum_{j,s,t} b_{sj}^*b_{st}\otimes e_{jt}\right)
=\sum_{j,s,t} \text{Tr}^{(k)}\left(b_{sj}^*b_{st}\right)\otimes e_{jt}
\end{align}
Now, each $b_{st}$ is in $M_n(A)$, so using the canonical matrix units $\{f_{pq}\}$ for $M_n(A)$, we can write 
$$
b_{st}=\sum_{p,q}b_{s,t,p,q}\otimes f_{p,q},
$$
where $b_{s,t,p,q}\in A$. Then, following from above, 
\begin{align}
\text{Tr}^{(k)}(b^*b)&=\sum_{j,s,t} \text{Tr}^{(k)}\left(b_{sj}^*b_{st}\right)\otimes e_{jt}\\ \ \\
&=\sum_{j,s,t} \text{Tr}^{(k)}\left(\sum_{p,q}b_{sjpq}^*\otimes f_{qp}\sum_{v,w}b_{stvw}\otimes f_{vw}\right)\otimes e_{jt}\\ \ \\
&=\sum_{j,s,t} \sum_{p}\text{Tr}^{(k)}\left(\sum_{q,w}b_{sjpq}^*b_{stpw}\otimes f_{qw}\right)\otimes e_{jt}\\ \ \\
&=\sum_{j,s,t} \sum_{p}\left(\sum_{q}b_{sjpq}^*b_{stpq}\otimes f_{qq}\right)\otimes e_{jt}\\ \ \\
&=\sum_{p,s,q}\sum_{j,t} b_{sjpq}^*b_{stpq}\otimes f_{qq}\otimes e_{jt}\\ \ \\
&=\sum_{p,s,q}\left(\sum_{j} b_{sjpq}\otimes f_{qq}\otimes e_{1j}\right)^*\left(\sum_{j} b_{sjpq}\otimes f_{qq}\otimes e_{1j}\right)\geq0.
\end{align}
