hessian matrix and conditions for semidefinite, definite In the lecture notes this was given as a theorem.
Let A be  a $n \times n$ matrix and let A$_k$ be the submatrix of A obtained by taking the upper left hand corner $k \times k$ matrix of A.Furthermore let $\det(A_k)$,be the kth principal minor of A.Then


*

*A is positive definite if and only if $\det(A_k)>0$ for $k=1,2,3,\ldots,n$

*A is negative  definite if and only if $(-1)^k\det(A_k)>0$ for
$k=1,2,3,\ldots,n$

*A is positive semi definite if and only if $\det(A_k)>0$ for
    $k=1,2,3,\ldots,n-1$ and $\det(A)=0$

*A is negative semi definite if and only if $(-1)^k\det(A_k)>0$ for
$k=1,2,3,\ldots,n-1$ and $\det(A)=0$

*Indefinite if $\det(A)<0$  
But I have a few problems with these definitions and some examples given in the note.
1)  $$
    \begin{bmatrix}
    1 & 1 & 1 \\
    1 & 1 & 1 \\
    1 & 1 & 1/2 \\
    \end{bmatrix}
$$
It is said that this this is indefinite.Why?According to the theorem this can't be because $\det(A)=0$ and not less than $0$?What category does this fall into?
2) $$
    \begin{bmatrix}
    0& 0  \\
     \\
    0 & 0 \\
    \end{bmatrix}
$$
This zero zero matrix is said to be positive semi definite.But $\det(A_1)\le 0$ therefore how is this positive definite according to the definition?
Is there a problem with the given theorem?What are the exact conditions I should check if looking for definite,semidefinite,indefiniteness?
Any help is appreciated.
 A: I don't know how you get these theorems. But apparently some of them are wrong.
In my opinion, the first, second and the last is true.
The 3-5 is hold only for "if" but not for "only if". 
So for example. $\det(A)<0 \Rightarrow$ $A$ is indefinite, but not vise versa.
A much better way to understand positive definite and positive semi definite is through the eigenvalues. 
If all eigenvalues are (positive/negative) $\Leftrightarrow$ matrix is (positive/negative definite)
If all eigenvalues are (non negative/non positive) $\Leftrightarrow$ matrix is (positive/negative) semi definite.
A: Matrix 
$$A=\begin{bmatrix}
        1 & 1 & 1 \\
        1 & 1 & 1 \\
        1 & 1 & 1/2 \\
        \end{bmatrix} $$
is indeterminate because the $2\times2$ minor is zero. 
In fact if we let
$$ v= 
\begin{bmatrix} 1\\1\\-2 \end{bmatrix}$$
then $v^T A v < 0$.
For 
$$ v= 
\begin{bmatrix} 1\\-1\\0 \end{bmatrix}$$
 $v^T A v =0$
Finally, for
$$ v= 
\begin{bmatrix} 1\\1\\1 \end{bmatrix}$$
 $v^T A v > 0$
The zero matrix, $B$, is clearly positive semi definite since $v^T B v \ge 0$ for any $v$.
It is also negative semi definite since  $v^T B v \le 0$ for any $v$. Moral of the story: Zero is both non-negative and non-positive. It is just zero!
