# demonstrating convexity of set defined by quadratic inequality

So i'm trying to solve this problem and I don't have any idea how to proceed.

A is a Positive-definite matrix, $$\alpha \in \Bbb R$$,

Show that the set $$S(A,a; \alpha) = \{x \in \Bbb R^n \mid \frac 12 \langle x,Ax \rangle + \langle a,x \rangle \leq \alpha \}$$ is convex.

How?

P.S. I know what convex means , I know what a Positive-definite matrix is , but they don't give me the form of the matrix, no nothing.

What you need to show is that for $$x_1, x_2 \in S$$ and $$0 \leq t \leq 1$$, the element $$(1-t)x_1 + t x_2$$ is also an element of $$S$$. The only piece of information about $$A$$ required to show this is that $$A$$ is positive definite. In particular, if $$A$$ is positive definite, then the function $$f(x) = \sqrt{\langle x,Ax \rangle}$$ is a norm over $$\Bbb R^n$$.
It is also helpful to observe that $$\langle (x-x_0),A(x-x_0) \rangle = \langle x,Ax\rangle - \langle (A + A^T)x_0,x\rangle + \langle x_0,Ax_0\rangle$$ Now, if we take $$x_0 = (A + A^T)^{-1}\alpha$$, then we can rewrite the set $$S$$ as $$S = \{x \in \Bbb R^n \mid \langle x-x_0, A(x-x_0) \rangle - \langle x_0,Ax_0 \rangle \leq \alpha \} \implies\\ S = \{x \in \Bbb R^n \mid \langle x-x_0, A(x-x_0) \rangle \leq \alpha + \langle x_0,Ax_0 \rangle\} \implies\\ S = \{x \in \Bbb R^n \mid [f(x-x_0)]^2 \leq \beta \} \implies\\$$ where we have defined $$\beta = \alpha + \langle x_0,Ax_0 \rangle$$. If $$\beta < 0$$, $$S$$ must be empty. If $$\beta \geq 0$$, we have $$S = \{x \in \Bbb R^n \mid f(x-x_0) \leq \sqrt{\beta}\}$$ It is now straightforward to show that $$S$$ is convex using the fact that $$f$$ is a norm.
The set $$S$$ is convex as a lower-level set of a convex function $$f:x\mapsto \frac{1}{2}\langle x,Ax\rangle + \langle\alpha,x\rangle.$$ (If $$x,y$$ is such that $$f(x), f(y) \leq c$$, then $$f(tx+(1-t)y) \leq tf(x) +(1-t)f(y) \leq c$$ for all $$0.)
To see that $$f$$ is convex, it suffices to show $$g:t\in \mathbb{R}\mapsto f(x+ty)$$ is convex for all $$x$$ and $$y\neq 0$$. This follows from $$g''(t) = \langle y, Ay\rangle>0.$$