# For which $x \in \Bbb{R}$ matrix $A$ has all positive eigenvalues?

For which $$x \in \Bbb{R}$$ matrix $$A$$ has all positive eigenvalues, if matrix $$A$$ is given $$A= \begin{bmatrix} 1 & 2 & 3 \\ 2 & x & 4 \\ 3 & 4 & 5 \\ \end{bmatrix}$$ $$-$$ We know that when matrix is symmetric and positive definite then it has positive eigenvalues. Positive definite matrices have positive determinant, trace and positive diagonal elements..so, I can conclude that det$$(A)>0$$ then $$12-4x>0$$, and $$x<3$$, and we have $$x>0$$ if matrix is positive definite. At the end conclusion is $$x \in (0,3)$$. My problem is when I use 1 of 2 for x I don't get all positive eignevalues, and that is contradiction (maybe I'm wrong).  My question is, can I say that symmetric matrix can have all positive eigenvalues if it's not positive definite? And can someone help me to find $$x$$ in this problem..

Since $$A$$ is symmetric, if all eigenvalues of $$A$$ are positive, then $$A$$ is positive definite. We know that for $$A$$ to be positive definite, this submatrix of A $$\begin{bmatrix} 1 & 2 \\ 2 & x \end{bmatrix}$$ needs to have a positive determinant, i.e. $$x > 4$$. (This is known as Sylvester's criterion)
As you already mentinoned, we also need $$\det(A) > 0$$, and that implies $$x < 3$$. From this we can conclude, that no matter the value of $$x$$, $$A$$ will never be positive definite.
Alternatively, you can also note that for $$v = (2, 0, -1)^T$$, $$v^T A v = v^T \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = -3 < 0,$$ which implies that $$A$$ is not positive definite, and this calculation doesn't actually depend on $$x$$.