Sum of Symmetric Positive Definite Matrix and Scalar of Identity If $A$ is an $n\times n$ symmetric positive definite matrix with the smallest eigenvalue $\lambda$, then for any $\mu>-\lambda$, $A+\mu I$ is positive definite.
I am trying to show this, but I am stuck on one part. Here is what I have so far:
$$
\begin{align*}
\langle x,\left(A+\mu I\right)x\rangle&=\langle x,Ax+\mu x\rangle\\
&=\langle x,Ax\rangle+\langle x,\mu x\rangle\\
&>0+\mu\langle x,x\rangle\\
&>-\lambda\langle x,x\rangle.
\end{align*}
$$
I'm stuck on showing that $\mu\langle x,x\rangle$ is positive because I only know that $\mu>-\lambda$. Any help would be appreciated.
 A: If $A$ is a positive definite symmetric matrix with smallest eigenvalue $\lambda$, then for all vectors $x$ we have
$\langle x, Ax \rangle \ge \lambda \langle x, x \rangle; \tag 1$
the easiest way I know to see this is to diagonalize $A$ by a suitable orthogonal  matrix $O$, which will preserve the inner product:
$\langle Oy, Ox \rangle = \langle y, O^TOx \rangle = \langle y, Ix \rangle = \langle y, x \rangle, \tag 2$
where we have used the fact that
$O^TO = OO^T = I \tag 3$
in (2); then we have
$OAO^T = \Lambda = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n), \tag 3$
where $\lambda_1, \lambda_2, \ldots, \lambda_n$ are the eigenvalues of $A$.  It is well-known that $A$ is also possessed of an orthonormal eigenbasis of vectors $e_i$ such that
$A e_i = \lambda_i e_i, \; 1 \le i \le n; \tag 4$
we may then write
$x = \displaystyle \sum_1^n x_i e_i, \tag 5$
and
$\langle x, Ax \rangle = \left \langle \displaystyle \sum_1^n x_ie_i, \sum_1^n x_i Ae_i \right \rangle = \left \langle \displaystyle \sum_1^n x_ie_i, \sum_1^n x_i \lambda_i e_i \right \rangle = \displaystyle \sum_{i, j = 1}^n x_ix_j \langle e_i, \lambda_j e_j \rangle$
$= \displaystyle \sum_{i, j = 1}^n x_ix_j  \lambda_j \langle e_i,e_j \rangle = \sum_{i, j = 1}^n x_ix_j  \lambda_j \delta_{ij} = \sum_1^n \lambda_i x_i^2; \tag 6$
now if 
$\lambda = \min \{\lambda_1, \lambda_2, \ldots, \lambda_n \} > 0 \tag 7$
is the least eigenvalue, then (6) yields
$\langle x, Ax \rangle = \displaystyle \sum_1^n \lambda_i x_i^2 \ge \sum_1^n \lambda x_i^2 = \lambda \sum_1^n x_i^2 = \lambda \langle x, x \rangle; \tag 8$
therefore,
$\langle x, (A + \mu I)x \rangle = \langle x, Ax \rangle + \mu \langle x, x \rangle \ge \lambda \langle x, x \rangle + \mu \langle x, x \rangle = (\lambda + \mu) \langle x, x \rangle; \tag 9$
since
$\mu > - \lambda \Longleftrightarrow \mu + \lambda > 0, \tag{10}$
(9) becomes
$\langle x, (A + \mu I)x \rangle \ge (\mu + \lambda ) \langle x, x \rangle > 0, \tag{11}$
which shows that $A + \mu I$ is positive definite, the desired result.  $OE\Delta$.
