This might be really dumb question but I've just started dealing with such matrices. I would like to know why $$\begin{bmatrix} 0 & 1 \\ 1 & x \end{bmatrix}$$ cannot be a positive semidefinite matrix.
Also, given the condition
$$
\begin{bmatrix}
0 & x & y\\
x & z & p\\
y & p & 1-x
\end{bmatrix}
\succeq 0$$
how do we get following matrix
$$\begin{bmatrix}
0 & 0 & 0\\
0 & z & p\\
0 & p & 1\\
\end{bmatrix}$$
Any help would be much appreciated.
Best Answer
Have you heard of Sylvester's Condition? So, in a nutshell, what it says is that you start looking at (top left) 1x1 2x2 3x3 ... nxn blocks of matrices all have positive determinants.
As a corollary, a matrix with negative determinant can never be positive definite. In your example, no matter what x is, the determinant is always -1 which by Sylvester's Condition (Note : It is IFF) is not positive definite.
EDIT :
Now that you clarified in the comments what you mean by "get". It makes more sense.
Think about it, if the entire top row wasn't zero: Apply the Sylvester's Criterion.
The first 1x1 block is just 0 which is OK (but the matrix can't be positive definite, at most semidefinite) The next block is 2x2 which has determinant $-x^2$. As long as you choose a real value of $x$, the criterion won't hold and you don't even need to go ahead if $x$ wasn't zero. However since we want a positive definite matrix, let's set $x=0$ and continue. If you carry out the same determinant criterion on the entire matrix you'll realize that $y$ needs to be zero as well (Assuming $z\geq0$. Else the determinant will be negative. (The determinant is $-zy^2$)