Q:

Projection on cone of non-negative definite matrices

Ok, so if you have a real symmetric matrix $Q$ then the projection of that matrix on the cone of symmetric non-negative definite matrices $\mathcal{C}$ can be explicitly found if we do an eigendecomposition and set the negative eigenvalues to zero:
Mathematically, if $Q = V\Lambda V^T$, with $\Lambda = diagonal(\{\lambda_1,\dots,\lambda_n\})$,then the projection on $\mathcal{C}$ is $Q^+ = V\Lambda ^+V^T$ where $\Lambda^+ = diagonal(\{\lambda_1^+,\dots,\lambda_n^+\})$ and $\lambda_i^+=max\{0,\lambda_i\}$.
Now, my question is: why is that the projection on the symmetric non-negative definite cone ?  Can someone give me the intuition behind it (or refer me to a proof) ? 
Thanks

A:

I found a very nice article online, A Fresh Variational-Analysis Look at the Positive
Semidefinite Matrices World by
Jean-Baptiste Hiriart-Urruty and Jérôme Malick which I can recommend.  They refer to various notions from convex analysis, and they call your projection part of a Moreau Decomposition.  So, download that article, and buy or borrow the book   Hiriart-Urruty, J.-B., Lemaréchal, C.: Fundamentals of Convex Analysis (2001), which goes into more detail.  I don't have the book, but I can see roughly what they are doing; from Google books, your item is Theorem 3.2.5 on page 51, also exercise 15 on page 71. 
There is also this recent question Proof of the Moreau decomposition property of proximal operators?