Show That Minimizer of Rayleigh Quotient Is the Smallest Eigenvalue - $ \min_{x : \left\| x \right\| = 1} {x}^{*} {A}^{*} A x = {\lambda}_{1}^{2} $ Show that (Some form of Rayleigh Quotient):
$$ \min_{x : \left\| x \right\| = 1} {x}^{*} {A}^{*} A x = {\lambda}_{1}^{2} $$
Where $ {\lambda}_{1} $ is the smallest singular value of $ A \in \mathbb{C}^{n \times n}$.

My attempt:
Let us define $ B = {A}^{*} A $, and say that $ \left( x, {\lambda}_{i}^{2} \right) $ is an eigen-pair of $ B $, i.e., $ B x = {\lambda}_{i}^{2} x $, such that the objective of the above optimization problem can be expressed as
\begin{align}
{x}^{*} \underbrace{B x}_{ ={\lambda}_{i}^{2} x } = {x}^{*} {\lambda}_{i}^{2} {x} = {\lambda}_{i}^{2} \hspace{-16mm}\underbrace{ {\left\| x \right\|}^{2} }_{\hspace{19mm}=1 \\ \text{due to the constraint that } \left\| x \right\|=1} = {\lambda}_{i}^{2}.
\end{align}
Also, $ B $ is Hermitian. So, the eigenvalues of $ B $ are non-negative and real-valued. Thus, the minimization problem boils down to the search over the minimum eigenvalue $ \min \{ {\lambda}_{i}^{2} \} $ which will render the smallest eigenvalue that can be said as $ {\lambda}_{1}^{2} $. 
Do you experts agree with this approach?

Can this be solved in some other ways, e.g., classical Lagrange multiplier based? 
Thank you so much in advance.
 A: I will sketch 2 approaches for the solution.
Option I


*

*Write the Lagrange of the problem and show that the dual variable must be an eigenvalue.

*Show that the Eigenvector of the Eigenvalue achieves the minimum value and hence the problem is solved.


Option II
Define $ B = {A}^{T} A $ which is PSD which means it has Eigen Decomposition $ B = {V}^{T} D V $ where $ V $ is Unitary matrix and $ D $ is diagonal matrix.
So the problem:
$$\begin{align*}
\min_{x} \quad & {x}^{T} B x \\
\text{subject to} \quad & \left\| x \right\| = 1
\end{align*}$$
Becomes:
$$\begin{align*}
\min_{x} \quad & {y}^{T} D y \\
\text{subject to} \quad & \left\| y \right\| = 1
\end{align*}$$
Where $ y = V x $.
Since $ V $ is Unitary which preserves $ {L}_{2} $ norm the constraint $ \left\| y \right\| = 1 $ is equivalent of $ \left\| x \right\| = 1 $.
Now, if you think about it, $ x $ like selecting and weighing columns of $ V $.
It is only logical it will select the column which matches the lowest value of $ D $ which is exactly the pair of Eigenvector and the smallest Eigenvalue.
A: I'll illustrate "Option I". Let $ x\in \mathbb{R}^n$ and $A$ be $n \times n$ (real) symmetric matrix. The problem is 
$$
\min_{x} x'Ax
$$
s.t. $ \| x \| = x'x = 1$. Constructing the Lagrange function you have to minimize the following expression 
$$
\min_x \left( x'A x+\lambda (x'x - 1) \right).
$$
Taking derivative w.r.t. $x$ vector you have 
$$
\frac{\partial }{\partial x}\left( x'A x+\lambda (x'x - 1) \right) = 2Ax + 2\lambda x = 0,
$$
or 
$$
Ax = \lambda x. 
$$
Namely, the vectors $x$ that minimize Rayleigh's quantity are those that satisfy this system of linear equations, i.e., the eigenvectors of $A$. Now, you have to choose among all the eigenvectors ($n$ at most). As such, note that if $x$ is an eigenvector then $Ax = \lambda x$, i.e., $x'Ax$ becomes $x'\lambda x = \lambda x'x = \lambda \| x\| = \lambda$. Namely, you just have to choose the eigenvector that corresponds to the minimal eigenvalue and then $ \min_{x} = x'Ax/\|x\| = \lambda_n$.    
