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): $$\hat{\rho}= {\begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ \end{bmatrix}}$$

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.

• For a pure state, the density matrix is idempotent. Then its eigenvalues are either 0 or 1 – vnd Jan 4 '16 at 18:02
• @vnd but is there any way to prove it algebraically?. – ss1729 Jan 4 '16 at 18:06

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 $$\hat{\rho}= {\begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ \end{bmatrix}}$$