Minimum in $(0,0)?$ For which $a \in \mathbb{R}$, does  the polynomial $p (x, y) = x^2 + 2axy + a^3y^2$ have a minimum in $(0, 0)?$
So here is my problem. 
I used the hessian matrix, and showed that p is positive definit at $(0,0)$ for $a>1.$ But I also need to consider the fact that that $p$ is positive semidefinite. So with $det(H(x,y))=0$ I also get $a=0$ and $a=1$.
Now I need to check if there is a minimum for $a=0$ and $a=1$.
How do I do that?
My Idea:
$a=0 $
So we have $p(x,y)=x^2$ since $n$ is even $(2)$ we have $p(x,y)=x^2\ge0 $ and a minimum in $(0,0).$ 
For $a=1$ we have:
$p(x,y)=(x+y)^2$  n is also even $(2)$ here so we have $p(x,y)=(x+y)^2$ $\ge0 $ and a minimum in $(0,0) $
We can also check this with $p(x,0)=(x)^2..$ 
Is this correct? Thanks in advance! 
 A: Let $f(x,y) = x^2 + 2axy + a^3y^2$. Then
$$df = 2xdx + 2aydx + 2axdy + 2a^3ydy \\= (2x + 2ay)dx + (2ax+2a^3y) \\= 2(x + ay)dx + 2a(x+a^2y)dy$$
Then $f$ has a critical point in $(x,y)$ iff
$$x+ay = 0$$ and $$2a(x+a^2y) = 0$$
If $a\neq 0$ and $a\ne 1$ then this implies $x=y=0$. If $a = 0$ this implies $x=0$ but $y$ can be anything. It $a=1$ this implies $x=-y$. So, for any $a\in \mathbb{R}$,  $(0,0)$ is a critical point. This critical point will be the only one iff $a\ne 0$ and $a\ne 0$. So $(0,0)$ will be :
a global minimum
$\Leftrightarrow$ $(0,0)$ is a local minimum and $a\ne 0$ and $a\ne 1$
$\Leftrightarrow$ the determinant of the Hessian at $(0,0)$ is positive and $a\ne 0$ and $a\ne 1$
$\Leftrightarrow$ $4a^3 - 4a^2$ is positive and $a\ne 0$ and $a\ne 1$
$\Leftrightarrow$ $a\ne 0$ and $a-1$ is positive and $a\ne 1$
$\Leftrightarrow$ $a\ne 0$ and $a>1$ and $a\ne 1$
$\Leftrightarrow$ $a>1$.
Edit : if $a=0$ or $a=1$ the point $(0,0)$ will be a local minimum but not a global minimum.
A: You can rewrite the function as
$$
p(x,y)=(x+ay)^2+(a^3-a^2)y^2
$$
If $a^3-a^2\ge0$, that is, $a\ge1$ or $a=0$, then $p(x,y)\ge0$ for every $(x,y)$. Since $p(0,0)=0$, the function has a minimum at $(0,0)$.
Suppose, instead, that $a<1$ and $a\ne0$. Then
$$
p(-a,1)=a^3-a^2<0
$$
Since $p(0,0)=0$, the function has no minimum at $(0,0)$.

You're considering the quadratic form having matrix
$$
A=\begin{bmatrix}
1 & a \\
a & a^3
\end{bmatrix}
$$
The function has a minimum at $(0,0)$ if and only if the quadratic form is positive semidefinite, because otherwise it takes on negative values.
Since the first principal minor is $1>0$, the form is positive definite if and only if $\det A>0$, that is
$$
a^3-a^2>0
$$
by Sylvester’s criterion. If $\det A=0$, the form is positive semidefinite.

With partial derivatives and the Hessian. Find the critical points:
$$
\frac{\partial p}{\partial x}=2x+2ay=0
\qquad
\frac{\partial p}{\partial y}=2ax+2a^3y=0
$$
Let's look at $a\ne0$, for the moment. The determinant of the linear system is $4(a^3-a^2)$. If $a\ne0$ and $a\ne1$, the solution is unique, so $(0,0)$ is the only critical point.
Let's assume $a\ne0$ and $a\ne1$. The Hessian is
$$
\det\begin{bmatrix}
2 & 2a \\
2a & 2a^3
\end{bmatrix}=4(a^3-a^2)
$$
so the unique critical point is a minimum if and only if $a^3-a^2>0$, so when $a>1$ (taking into account $a\ne0$ and $a\ne1$).
If $a=0$ the function is $p(x,y)=x^2$ that certainly has a minimum at $(0,0)$. If $a=1$, the function is $p(x,y)=(x+y)^2$ that again has a minimum at $(0,0)$.

You can clearly see that the Hessian is, apart from a positive factor, the same as the matrix of the quadratic form. It's not by chance, of course.
