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.


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.


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 $-zy^2$)

  • $\begingroup$ That means that if I implement the condition on the matrix to be positive semidefinite then it would yield the matrix that follows with zeros in top row. $\endgroup$ – mat_123 May 8 '13 at 4:48
  • $\begingroup$ I think by your argument of Sylvester's Condition all the top row entries should be zeros. $\endgroup$ – mat_123 May 8 '13 at 4:51
  • $\begingroup$ @mat_123, edited. $\endgroup$ – Inquest May 8 '13 at 4:55

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