TITLE: Find all real numbers $t$ such that the quadratic form $f$ is positive definite.
QUESTION [5 upvotes]: Where $$f(x_1,x_2,x_3)=2x_1^2+x_2^2+3x_3^2+2tx_1x_2+2x_1x_3$$.
This is a problem in my Matrix Analysis homework. Below is my effort.

Let $x=(x_1,x_2,x_3)^T$, then we have $$f=x^*Sx$$, in which $$S=\left(\begin{matrix}2&t&1\\t&1&0\\1&0&3\end{matrix}\right)$$. $f$ is positive definite is equivalent to $S$ is positive definite which is equivalent to all the eigenvalues of $S$ is positive.
The characteristic polynomial of $S$ is: $$\begin{align}|\lambda I-S|&=-\lambda^3+6\lambda^2+(3t^2-10)\lambda+(-3t^2+5)\\&=(-3+3\lambda)t^2+(-\lambda^3+6\lambda^2-10\lambda+5)\end{align}$$.
Now the only problem left is that how do I find all the possible real values of $t$ that makes this polynomial of $\lambda$ only has positive roots?

REPLY [4 votes]: An equivalent condition for a matrix to be positive definite is that the leading principal minors of the matrix are all positive. This condition is easier to check: You just need to check that $2>0$, $2-t^2>0$ and that $-3t^2+5>0$. For more info see characterizations of positive definite matrices