During my research I have come across matrices this type 


$$C=B\left(B^T B\right)^{-1}B^T\ ,$$


where $B$ is an $m\times n$ real matrix. If $B^TB$ is not invertible, then $\left(B^T B\right)^{-1}$ should be interpreted as the [Moore-Penrose pseudoinverse](http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse). the I am interested in the diagonal elements of $C$. 


Clearly, if $m=n$ and $B$ is invertible, then $C$ is the identity matrix $I\_{m\times m}$. It is pretty easy to show that this is still the case when $m<n$ and the rows of $B$ are linearly independent.


However, when the rows of $B$ are not independent, there is not much I can say. From the physical context where this came up, I have a very strong suspicion (supported by numerical experiments) that the diagonal elements of $C$ are bounded from above by $1$ (and clearly from below by $0$). My question is whether someone could give me a hint as to show this, or where to look for in the literature.



Upper bounds on elements of a matrix
$C$ is an idempotent matrix, so its eigenvalues are either 1 or zero. $C$ is also Hermitian so Schur's Theorem says the sum of the $k$ largest eigenvalues is greater then the sum of the $k$ largest diagonal elements of $C$ for $k=1,2,...,m$. No diagonal element of $C$ can therefore exceed unity.


