Form of pure states on $M_n(\mathbb{C})$ Related question.
I have tried to show that pure states on $M_n(\Bbb{C})$ are of the form $\phi(A)=Tr(\rho A)$, where $\rho$ is one-rank projection. 
First, as in the related question, I've shown $\phi=\phi_\rho$, where $\phi_\rho(A)=Tr(\rho A)$ and $\rho$ is a positive matrix with $Tr(\rho)=1$. Nsxt I used the fact that if $\rho$ is positive then the representation as in the answer, $\rho=\sum_{k}\lambda_kP_k$, allows us to conclude that $\|\rho\| = \sup \lambda_k$, i.e. the maximum of the eigenvalues.
But, as $B(\Bbb{C}^n)=K(\Bbb{C}^n)=C_1(\Bbb{C}^n)$, from the isometry between the dual space of the trace class and bounded operators, we know that the norm is $1$, because $\phi$ is a state and $$1=\|\phi_\rho\|=\|\rho\|=\sup\{\lambda_k\}.$$
Meanwhile, from the isometry between the dual space of $K(H)$ and $C_1(H)$ we have  $$1=\|\phi_\rho\|=\|\rho\|_1=\sum_{k}\lambda_k.$$ 
As the eigenvales are non-negative, the only non-zero eigenvalue is $1$ with no multiplicity, so we got $\rho$ is a one-rank projection. This can't be true, because this is the case of pure states and not states, while by the above related answer, $\rho$ is also positive. Where am I wrong? 
 A: Your isometries are not such. Consider
$$
\rho_1=\begin{bmatrix}1/2&0\\0&1/2\end{bmatrix},\ \ \rho_2=\begin{bmatrix}1&0\\0&0\end{bmatrix}. 
$$
Then $$\|\rho_1\|=1/2,\ \ \|\rho_2\|=1,$$ while
$$
\|\rho_1\|_1=\|\rho_2\|_1=1.\ \ 
$$
For the proof, if $\rho=\sum_k\lambda_kP_k$ with $\sum_k\lambda_k=1$ and $0\leq\lambda_k$ for all $k$ and at least two of the lambdas distinct, then the state $\text{Tr}\,(\rho\cdot)$ is a proper convex combination of states:
$$
\text{Tr}\,(\rho A)=\sum_k\lambda_k\,\text{Tr}\,(P_kA).
$$
Conversely, if $\rho$ is a rank-one projection and 
$$
\text{Tr}\,(\rho A)=t\,\text{Tr}\,(\rho_1 A)+(1-t)\,\text{Tr}\,(\rho_2 A),
$$
we have
$$
\text{Tr}\,(\rho A)=\text{Tr}\,(\,(t\rho_1+(1-t)\rho_2) A)
$$
for all $A$. It follows that 
$$
\rho=t\rho_1+(1-t)\rho_2. 
$$
So
$$
0=(1-\rho)\rho(1-\rho)=t(1-\rho)\rho_1(1-\rho)+(1-t)(1-\rho)\rho_2(1-\rho).
$$
As the two summands are positive, we get $$(1-\rho)\rho_j(1-\rho)=0,\ \ j=1,2.$$ But $\rho_j$ is positive, so $\rho_j=x_j^*x_j$, so 
$$
0=(1-\rho)x_j^*x_j(1-\rho)=[x_j(1-\rho)]^*x_j(1-\rho).
$$
Then $x_j(1-\rho)=0$, and then $\rho_j(1-\rho)=(1-\rho)\rho_j=0$. Thus,
$$
\rho_j=\rho\rho_j\rho=\lambda_j\rho,
$$
since $\rho$ is rank-one. As $\text{Tr}\,(\rho_j)=1$, it follows that $\lambda_j=1$ and $\rho_j=\rho$. So $\text{Tr}\,(\rho\,\cdot)$ is pure. 
