Coerciveness and Positive definiteness relation? Let $A ∈ \mathbb{R}^{n×n}$ be  a symmetric matrix. How can I demonstrate that A is positive definite iff the function $q(x) := x^TAx$ is coercive .
I know the eigenvalues of A have to be positive for a positive definite matrix. But I don't see how this relates to coerciveness.
Does't $x^TAx$ need to be greater than $0$ to be positive definite?
 A: The definition of positive definite is that $x^\top A x > 0$ for all $x \neq 0$.
The definition of coercive is that $x^\top A x > c\lVert x \rVert^2$ for some constant $c >0$.
Obviously, coercive implies positive definite.  For the reverse impliciation, I suggest thinking about $x^\top A x$ restricted to the unit sphere.  What can you say about its minimum value?
Another approach, if you know it, is to use the real spectral theorem.
A: There are several definitions that come into play. Let's take $V$ - our vector space with scalar product
A linear operator $A$ is called positive, if $$\forall x\in V \quad (Ax,x)>0.$$
A linear operator $A$ is called positive definite, if there exists a positive constant such that $$\forall x\in V \quad (Ax,x)\ge c \|x\|^2.$$
Let $B$ be a bilinear form over a normed space $W$, then we call it coercive if there exists a positive constant $b$ such that $$\forall x\in W \quad B(x,x)\ge b \|x\|^2.$$ Note that we don't need scalar product to define a coercive bilinear form.
Note also that to a linear operator $A$ in a space with inner product you can associate a bilinear form $$ x,y \to (Ax,y).$$ For $A$ to be Positive definite or for its associated bilinear form to be coercive is the same thing.
On a bright side, in finite dimensioned space all these notions are equivalent, because all norms are equivalent and we have a natural scalar product in $\Bbb C^n$.
