What is the effect of adding a constant to all diagonal members of a matrix on its Eigen decomposition? While reading a research paper [1], I came across the following result about Eigen decomposition:
Let $A=U \Lambda U^T$ where $U$ is the matrix containing the Eigen vectors and $\Lambda$ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues. Then $A+\gamma I = U(\Lambda +\gamma I) U^T + U^{\perp}(\gamma I) (U^{\perp})^{T}$ where $U^{\perp}$ is an orthonormal basis for the space orthogonal to the column space of $U$, and $I$ is identity matrix.
Is the above result correct? What is the intuition for this result if it is correct?
Thanks in advance!
Reference:
[1] Dolia, Alexander N., et al. "The minimum volume covering ellipsoid estimation in kernel-defined feature spaces." European Conference on Machine Learning. Springer Berlin Heidelberg, 2006.
In the above paper, the result in question is used to compute the Mahalanobis distance for a test point toward the end of Section 4.
 A: Yes, it's correct. Actually, I would phrase that a bit simpler:
Let $A=U\circ\Lambda\circ U^{\mathrm{T}}$ where $U$ is the matrix containing the† complete system of eigenvectors and $\Lambda$ is the diagonal matrix whose diagonal elements are the corresponding (possibly zero) eigenvalues. Then $A+\gamma\cdot I = U\circ (\Lambda +\gamma\cdot I)\circ U^{\mathrm{T}}$, where $I$ is identity matrix.
Now, well, this is fairly evident: since $U$ contains a complete basis, it's an orthogonal transformation. Thus
$$\begin{align}
  U\circ (\Lambda +\gamma\cdot I)\circ U^{\mathrm{T}}
  =& U\circ\Lambda\circ U^{\mathrm{T}} + \gamma \cdot (U\circ \underbrace{I\circ U^{\mathrm{T}}}_{\operatorname{id}\circ f \equiv f})
 \\ =& \underbrace{U\circ\Lambda\circ U^{\mathrm{T}}}_{\text{per def. $=A$}} + \gamma \cdot (U\circ \underbrace{U^{\mathrm{T}}}_{\text{orth.}})
 \\ =& A + \gamma \cdot (U\circ U^{-1})
+ \gamma \cdot (U\circ \underbrace{U^{\mathrm{T}}}_{\text{orth.}})
 \\ =& A + \gamma \cdot I.
\end{align}$$
Or, IMO preferrable, you might just note that matrix representations of linear mappings refer to an ultimately arbitrary basis. If you simply choose the eigenbasis of $A$, then $A$ is a diagonal matrix, and the above result is trivial.
All you need to change to adapt this to your phrasing with that orthogonal projection is to keep the eigenvectors with zero eigenvalue out of $U$.

†It would be more accurate to say “containing a complete system of eigenvectors” since, if multiple eigenvalues are zero, you actually have a degenerate situation and the choice of vectors with in this eigenspace is not unique. But that's an issue you can always have with eigensystems: they are generally only unique in the non-degenerate case, i.e. when all eigenvalues are different.
