Minimization of Sum of Squares Error Function Given that $y(x,{\bf w}) = w_0 + w_1x + w_2x^2 + \ldots + w_mx^m =  \sum_{j=0}^{m} w_jx^j$ and there exists an error function defined as $E({\bf w})=\frac{1}{2} \sum_{n=1}^{N} \{y(x_n, w)-t_n\}^2$ (where $t_n$ represents the target value). I'm having trouble making sense of a passage in my textbook. (Note: ${\bf w}$ represents a vector of the polynomial's coefficients.) I've listed the passage below: 

We can solve the curve fitting problem by choosing the value of ${\bf w}$ for which $E({\bf w})$ is as small as possible. Because the error function is a quadratic function of the coefficients ${\bf w}$,   its derivatives with respect to the coefficients will be linear in the elements of ${\bf w}$, and so the minimization of the error function has a unique solution, denoted by ${\bf w^*}$, which can be found in closed form. 

How do we know that the minimal solution exists and is unique? What guarantees this? Any help understanding this would be appreciated.  
 A: Let's, as the cited passage suggest, look at the derivative of $E$. First we note that $y$ is linear in $w$, as 
$$ y(x,w+\mu w') = \sum_i (w_i +\mu w_i')x^i = \sum_i w_ix^i +\mu \sum_i w'_ix^i= y(x,w) + \mu y(x,w') $$
Now we have for $w, h \in \mathbb R^{m+1}$ that
\begin{align*}
  E(w+ h) &= \frac 12\sum_{n=1}^N \bigl(y(x_n, w) + y(x_n, h) - t_n\bigr)^2\\
          &= E(w) + \sum_{n=1}^N \bigl(y(x_n, w) - t_n)y(x_n, h) + \sum_{n=1}^N y(x_n, h)^2\\
          &= E(w) + \sum_{n=1}^N \bigl(y(x_n, w) - t_n)\bigr)y(x_n, h) + o(h)
\end{align*}
so $E'(w)h = \sum_{n=1}^N \bigl(y(x_n, w) - t_n)\bigr)y(x_n, h)$. The second derivative is 
$$ E''(w)[h,k] = \sum_{n=1}^N y(x_n, k)y(x_n, h) $$
Now for $h \in \mathbb R^{m} -\{0\}$ 
$$ E''(w)[h, h] = \sum_{n=1}^N y(x_n, h)^2 $$
and this is positive if $N \ge m+1$ and all $x_i$ are different (as a polynomial of degree $m$ cannot have $N \ge m+1$ zeros). So $E''(w)$ is positive definite for every $w$, as $E''$ is constant, hence every zeros of $E'$ is a minimum for $w$. Now lets look at $E'$, we have $E'(w) = 0$, if $E'(w)e_i = 0$ for each $i$ ($e_i$ denoting the $i$th standard basis vector), it holds 
$$ E'(w)e_i = \sum_{n=1}^N \bigl(y(x_n, w) - t_n\bigr) x_n^i $$
That is we want $w$ to be such that $y(x,w)- t$ is orthogonal to $(x_1^i, \ldots, x_N^i)$ for all $i$. Projection of $t$ onto the subspace generated by this vectors, gives us an unique point, as $w \mapsto y(x,w)$ is injective, $w$ is unique.
