Numerical Linear Algebra Question

The objective function to be minimized is:
\[f(x) = \frac{1}{2} x^T H x + x^T c + \frac{1}{2\mu} \|Ax - b\|^2_2.\]
At the minimum points, the gradient should be zero:
\[ \nabla f(x) = Hx + c + \frac{1}{\mu} A^T (Ax - b) = 0.\]

The Hessian of $f$ is given by 
\[ \nabla^2 f(x) = H + \frac{1}{\mu} A^T A.\]
Next we show that for $\mu$ small enough, $\nabla^2 f(x)$ is positive definite. 

For any column vector $v$, we have $v^TA^TAv=(Av)^T(Av)=(Av)\cdot (Av)\geq 0$, therefore $A^TA$ is positive semi-definite.

Now note that $\mathbb{R}^n = \text{null}(A) \oplus \text{range}(A^T)$. We consider two cases: 

(i) If $z\in \text{null}(A)$ (i.e., $Az = 0$), the effect of the penalty term is zeros, i.e. 
$$\frac{1}{\mu} A^T A z=0,$$ 
and hence 
\[z^T \nabla^2 f(x) z = z^T H z>0\]
 for all nonzero $z$ in the null space of $A$, since $Z^T H Z$ is positive definite. 
(ii) Assume $x \notin \text{null}(A)$ and $||x||>0$. The matrix $A^TA: \text{range}(A^T) \rightarrow \text{range}(A^T)$ is invertable and positive definite, and hence 
\[ x^T\nabla^2 f(x)x = x^THx + \frac{1}{\mu} x^T(A^T A)x\geq \lambda_{min}^{H}||x||^2+\frac{\lambda_{min}^{+}}{\mu}||x||^2>0\]
provided $\mu$ is small enough so that 
\[\frac{\lambda_{min}^{+}}{\mu} >-\lambda_{min}^{H}.\]
Here $\lambda_{min}^{+}$ is the smallest positive eigencalue of $A^TA$ and $\lambda_{min}^{H}$ is the smallest eigenvalue of $H$, which could be a genative number. 

Hence for $\mu$ small enough the hessian $\nabla^2 f(x)$ is a positive definite matrix. Therefore $f$ is strictly convex, and will have a unique strict minimizer.