hessian matrix and conditions for semidefinite, definite

Question

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

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

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

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

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

5. 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.

Answer 1

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.

Answer 2

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!