There are two questions I'm curious about, and both of them may be incredibly silly. I just haven't been able to convince myself otherwise.
We know that a density operator $\rho_E$ has the form $\rho_E = \sum_j \rho_j |x_i\rangle\langle x_i|$ where $\text{Tr}(\rho_E) = \sum_j \rho_j = 1$. Since $\rho_E$ is also normal, it must have a spectral decomposition $\rho_E = \sum_i \lambda_i\Pi_i$ where $\Pi_i$ are projections into the eigenspace associated with $\lambda_i$.
Now, is it the case that these two decompositions are same? Meaning $\lambda_i = \rho_i$ and $|x_i\rangle\langle x_i| = \Pi_i$?
If so, is there a good explanation why $|x_i\rangle\langle x_i|$ is actually a projection into eigenspace? Meaning that, assume we have some linear map $M: \mathbb{R}^2 \to \mathbb{R}^2$ with, say, eigenvector $v = (2,1)^T$. Will $(2,1)^T \cdot (2,1) = |v\rangle\langle v|$ actually be a projection to the eigensapce associated with $v$?
With the "ket-bra" notation $|u\rangle\!\langle v|$ we mean the linear operator $x\mapsto |u\rangle\langle v|x\rangle$, where $\langle v|x\rangle$ is the inner product between the vectors $|x\rangle$ and $|v\rangle$. In more standard mathematical notation, this might be denoted with $uv^*$, here $v^*$ is the dual of $v$.
Because of this, $|v\rangle\!\langle v|$, for any unit vector $|v\rangle$ (which is usually the case when one is using ket notation), is a trace-1 projector, which trivially projects onto the one-dimensional space spanned by $|v\rangle$.