# 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

• It is not entirely clear to me that this is well-defined (under different diagonalizations). What is your source for this? – Will Jagy Apr 22 '14 at 3:13
• Source is: staff.polito.it/roberto.tempo/pdf1/PT01.pdf (page 3 of the PDF) – DanielX2010 Apr 22 '14 at 3:16
• Well, it appears you want references 17 and 20. The language suggests that there are advantages to different methods for different investigations, and this method is best for this article. I'm looking through a standard book, Horn and Johnson, maybe they have something – Will Jagy Apr 22 '14 at 3:26
• I did take a look at 17, but it just seems to mention the same statement without proving it. 20 is a book and I don't have it. Please tell me if you find anything on your book. Really appreciate your help. – DanielX2010 Apr 22 '14 at 3:29