Regularization of a Matrix Using a Diagonal Matrix For a $n\times n$ matrix $A$ it is well known that $A + \lambda I$ for suffiecientely large $\lambda >0$ makes $A$ positive definite. The proof is straightforward by looking at the characteristic polynomial. I wonder what happens if we look at a diagonal matrix $\Lambda$ with possibly distinct diagonal elements? Is the  conclusion still valid that $A + \Lambda$ becomes positive definite for large enough $\Lambda$?
Background: I have implemented a code that performs regularization using the former approach. By experimental studies I found that using different values for regularization improves my results. So I was wonderung whether this also backed by theory.
 A: By definition $A + \lambda I$ is positive definite if, for every vector
$\vec x \ne 0,  \tag 1$
we have
$\vec x^\bot(A + \lambda I) \vec x > 0; \tag{1.5}$
since
$\vec x^\bot(A + \lambda I) \vec x= \vec x^\bot A \vec x + \vec x^\bot (\lambda \vec x) = \vec x^\bot A \vec x + \lambda \vec x^T \vec x, \tag 2$
clearly (1.5) binds for $\lambda > 0$ sufficiently large, since 
$\exists M > 0, \; \vert \vec x^\bot A \vec x \vert < M \vec x^\bot \vec x; \tag 3$
we need only choose
$\lambda > M. \tag 4$
If now 
$\Lambda = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n), \tag 5$
then
$\vec x^\bot (A + \Lambda) \vec x = \vec x^\bot A \vec x + \vec x^\bot \Lambda \vec x; \tag 6$
with
$\vec x = (x_1, x_2, \ldots, x_n)^\bot, \tag 7$
we have
$\Lambda \vec x = (\lambda_1 x_1, \lambda_2 x_2, \ldots, \lambda_n x_n)^\bot, \tag 8$
whence
$\vec x^\bot \Lambda \vec x = \displaystyle \sum_1^n \lambda_i x_i^2; \tag 9$
choosing each $\lambda_i > \lambda$ yields
$\vec x^\bot \Lambda \vec x = \displaystyle \sum_1^n \lambda_i x_i^2 > \lambda \sum_1^n x_i^2 = \lambda \vec x^\bot \vec x, \tag{10}$
from which in light of (1.5) shows that $M + \Lambda$ is positive definite.
A: The conclusion is still correct, as 
\begin{align*}
\Lambda&=\text{diag}\left(\lambda_{1},\lambda_{2},\dots,\lambda_{n}\right)\\
&=\lambda_{\text{min}}I+\Xi\\
\Xi&=\text{diag}\left(\lambda_{1}-\lambda_{\text{min}},
\lambda_{2}-\lambda_{\text{min}},\dots,\lambda_{n}-\lambda_{\text{min}}\right)\\
\end{align*}
Here $\lambda_{\text{min}}=\min_{1\leq j\leq n}\lambda_{j}$. For large enough $\lambda_{\text{min}}$, we have that $A+\lambda_{\text{min}}I$ is positive definite, that is for any nonzero vector $u$, we have 
\begin{align*}
u^{\top}\left(A+\lambda_{\text{min}}I\right)u>0
\end{align*}
If $u$ is nonzero then 
\begin{align*}
u^{\top}\Xi u&=\sum_{j}u_{j}^{2}\left(\lambda_{j}-\lambda_{\text{min}}\right)\geq 0
\end{align*}
using this we have for every nonzero vector $u$
\begin{align*}
u^{\top}\left(A+\Lambda\right)u&=u^{\top}\left(A+\lambda_{\text{min}}I+\Xi\right)u>0
\end{align*}
thus $A+\Lambda$ is guaranteed to be positive definite, provided that the minimum entry of $\Lambda$ is sufficiently large.
