Doubt on a paragraph regarding Lagrange's multiplier. I've a topic in my notes "The method of Lagrange's multipliers" which is described as follows:  

Let $U$ be an open set in $\mathbb R^n$.Let $f\in C^1(U,\mathbb R)$ and let   $$g_i(x)=0,~~~~~~~~i=1,\ldots,m$$
  be a collection of equality constraints with $m\lt n$.Now consider the system of non-linear equations$$f(x)=a$$ $$g_i(x)=0,~~~~i=1,\ldots,m$$
  $x_0\in U$ is a local maximum if $f(x_0\geq f(x)$ for all $x\in U$ near $x_0$ which also satisfies the above constrints.A local minimum is defined similarly.  
Let $F:U\times \mathbb R\to \mathbb R^{m+1}$ be defined by   
$$F(x,a)=\left[ \begin{array}{ccc}
f(x)-a  \\ 
g_1(x)\\
\vdots \\ g_m(x)\end{array} \right] $$  
Now consider the $m+1\times n$ Jacobian matrix ,the matrix of linear transformation , $F'(x,a)$ w.rt. usual basis for $\mathbb R^n$ and $\mathbb R^{m+1}$ at $x_0$ :  
$$\left[ \begin{array}{ccc}
\partial_1 f(x_0) \ldots \partial_n f(x_0) \\ 
\partial_1g_1(x_0)\ldots \partial_n g_1(x_0)\\
\vdots ~~~~~~~~~~~~\vdots\\\partial_1 g_m(x_0) \ldots\partial_n g_1(x_0)\end{array} \right] $$    
If this matrix has rank $m+1$ then some $m+1\times m+1$ sub matrix has non-zero determinant.It follows from the implicit function theorem that there exist $m+1$ variables $x_{i_1},\ldots,x_{i_{m+1}}$ such that the system $$F(x,a)=0$$
  specifies these $m+1$ variables as a function of remaining $n-(m+1)$ variables and $a$ in an open set of $\mathbb R^{n-m}$ .  
Thus ,there is a solution $(x,a)$ to equation $F(x,a)=0$ for some $x$ close $x_0$ whenever $a$ is in some open interval .  
Therefore , $x_0$ cannot be local maximum or minimum.  If $x_0$ is either a local maximum or minimum ,then the above matrix must have rank less than $m+1$ which reqiures the rows to be linearly dependent. Thus
  ,$\exists m$ scalars $\lambda_1,\ldots,\lambda_m$ and a scalar $\mu$ not all zero s.t.: >$\mu\left[ \begin{array}{ccc}
\partial_1f(x_0)\\ 
\vdots \\ \partial_nf(x_0)\end{array} \right] $ =
  $\lambda_1\left[ \begin{array}{ccc}
\partial_1g_1(x_0)\\ 
\vdots \\ \partial_ng_1(x_0)\end{array} \right] $+$\ldots$ +$\lambda_m\left[ \begin{array}{ccc}
\partial_1g_m(x_0)\\ 
\vdots \\ \partial_ng_m(x_0)\end{array} \right] $

I have the following doubts from the above extract:  
$1.)$ for what purpose did we define  : $$F(x,a)=\left[ \begin{array}{ccc}
f(x)-a  \\ 
g_1(x)\\
\vdots \\ g_m(x)\end{array} \right] $$   
$2.)$ I can't understand how is it concluded that : " Therefore ,$x_0$ cannot be local maximum or minimum."  
Please if anyone can help me with these doubts...
 A: I like your question, thanks for taking the time to write this all out.  I'll try to explain as best I can:
1) The definition of $F(x,a)$ is designed so that when $F(x,a) = 0$ we are both satisfying the constraints $g_i$, and aware that $f(x)$ is taking on the value $a$.  Essentially, the solution set of $F(x,a) = 0$ is the intersection of the set satisfying the constraints $g_i$ with the level set of $f$.  For example, if we consider in $\mathbb{R}^3$
$$g_1 = x^2 + y^2 + z^2 - 4, ~~ f = x + y + z,$$ then $g_1 = 0$ describes a sphere centered at the origin of radius 2 while the level sets of $f$ are planes (specifically, $f = a$ is the plane which intersects each axis at a distance $a$ from the origin).  For example, the point $(2,0,0)$ is on the sphere, but $f(2,0,0) \neq 4$ so we should find that the constraint equation is satisfied (a zero in the second entry of $F$), but the first entry will only be satisfied if $a = 2$, since $(2,0,0)$ lies on the level set $f(x) = 2$.  And you compute 
$$F((2,0,0),4) = \left[\begin{matrix} x+y+z - a \\ x^2 + y^2 + z^2 - 4 \end{matrix} \right] = \left[\begin{matrix} -2 \\ 0 \end{matrix} \right]$$ while $$F((2,0,0),2) = \left[\begin{matrix} x+y+z - a \\ x^2 + y^2 + z^2 - 4 \end{matrix} \right] = \left[\begin{matrix} 0 \\ 0 \end{matrix} \right].$$
Our goal is to find an $x_0$ solving $F(x_0,a) = 0$ for the largest (or smallest) possible $a$.
2) The key to understanding this quote is that, for some neighborhood $U$ about $(x_0,a)$, we have a function $\phi:\mathbb{R}^{n-m} \rightarrow \mathbb{R}^{m+1}$ such that $\phi(u) = (x_u,a_u)$ which satisfies $$F(x_u,a_u) = 0 ~ \textrm{ for all } ~u \in U$$ and (because the Jacobian is non-singular) has an open image.  Since the image is open we can find some point $(x',a')$ in $\phi(U)$ with $a' > a$.
If this is true, then the level set $f = a'$ also intersects the solution of $g = 0$.  This shows $f = a$ is not the maximum value $f$ takes on.
To finish, look again at the example.  You already know that the largest value that $f = x + y + z $ can take on for points on the sphere $g_1 = 0$ is $a = 2\sqrt{3}$, and this happens at the point $({2\over\sqrt{3}},{2\over\sqrt{3}},{2\over\sqrt{3}})$ where the plane $f = 2\sqrt{3}$ is tangent to the sphere in the first quadrant.  At this point the Jacobian of $F$ breaks down:
$$ \left[ \begin{matrix} 1 & 1 & 1 \\ 2x & 2y & 2z \end{matrix} \right] = \left[ \begin{matrix} 1 & 1 & 1 \\ {4\over \sqrt{3}} & {4\over \sqrt{3}} & {4\over \sqrt{3}} \end{matrix} \right]$$
is definitely rank one.  For any $a \in (-2\sqrt{3}, 2\sqrt{3})$, the set of intersection is a circle, the Jacobian of $F(x,a)$ is nonsingular (try one if you like), and you can scoot up to another circle (level set) with a higher value of $f$.
