linear algebra - 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..

Answer

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 $x$.