Prove: the density operator of a pure state has exactly one non-zero eigenvalue equal to unity What is the proper way of proving : the density operator $\hat{\rho}$ of a pure state has exactly one non-zero eigenvalue and it is unity, i.e,

the density matrix takes the form (after diagonalizing):
\begin{equation}
\hat{\rho}=
{\begin{bmatrix}
1 &amp; 0 &amp; \cdots &amp; 0 \\
0 &amp; 0 &amp; \cdots &amp; 0 \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
0 &amp; 0 &amp; \cdots &amp; 0 \\
\end{bmatrix}}
\end{equation}

For mixed state: $\hat{\rho}=\sum \limits_{i}P_{i}|\psi_{i}\rangle\langle\psi_{i}|$ 

For any state: $Tr(\hat{\rho})=\sum\limits_{i}P_{i}=1$

For pure state: $\hat{\rho}=|\psi\rangle\langle\psi|$

$|\psi\rangle$ is the statevector of the system

$P_{i}$ is the probability to be in the state $|\psi_{i}\rangle$, which are the eigenvalues of the density operator.
Since $\langle \psi|\psi\rangle=1$, we have
$$
(|\psi\rangle\langle\psi|)^2=|\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi|
$$
Then $|\psi\rangle\langle\psi|$ is a projection, and its eigenvalues are $0$ and $1$. Since we know that the trace of $|\psi\rangle\langle\psi|$ is one we conclude that, counting multiplicities, the eigenvalues of $|\psi\rangle\langle\psi|$ are $1,0,\ldots,0$. So its diagonal form is 
\begin{equation}
\hat{\rho}=
{\begin{bmatrix}
1 &amp; 0 &amp; \cdots &amp; 0 \\
0 &amp; 0 &amp; \cdots &amp; 0 \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
0 &amp; 0 &amp; \cdots &amp; 0 \\
\end{bmatrix}}
\end{equation}
Do you not know how to find the eigenvalues of a matrix by solving the "characteristic equation" of the matrix?   The characteristic equation of any diagonal matrix is just the product of linear terms, each the number on the diagonal minus x so the eigenvalues [b]are[/b] the numbers on the diagonal.