Theorem. A $M_{n \times n}$ matrix is positive (negative) definite iff the determinate of all upper-left sub matrices are positive (negative).
However, consider this matrix:
\begin{pmatrix}
3 & 1 & 2 \\
2 & 4 & 3 \\
-1 & -2 & 1 \\
\end{pmatrix}
If I am not mistaken I take the upper-left submatrices and their respective determinant to be
$$\begin{pmatrix}
3 \\
\end{pmatrix}
\Longrightarrow 3$$
$$\begin{pmatrix}
3 & 1 \\
2 & 4
\end{pmatrix}
\Longrightarrow 10$$
$$\begin{pmatrix}
3 & 1 & 2 \\
2 & 4 & 3 \\
-1 & -2 & 1 \\
\end{pmatrix}
\Longrightarrow 25$$
But this matrix isn't strictly positive definite of negative definite. It you find the eigenvalues (I'd suggest using some type of software – can also try to plug in for the Cholesky Decomposition and should given an error) you get eigenvalues with real and complex parts.
The real parts are all $> 0$, but the complex parts are not. Does this mean the matrix is positive definite in the reals, but not strictly positive definite?
Best Answer
As was stated in comments, the Sylvester's criterion requires the matrix to be symmetric. A simpler example with a non-symmetric matrix would be $$ A = \begin{pmatrix} 1 & 4 \\ 0 & 1\end{pmatrix} $$ which makes it clear that none of the upper-left determinants are influenced by the entry $2$; thus they do not detect its effect on the signature of the matrix.
To determine positive-definiteness of a non-symmetric matrix one can apply Sylvester's criterion to $(A+A^T)/2$, which generates exactly the same quadratic form as $A$ itself.