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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?

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up vote 2 down vote accepted

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

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Hint: If all roots of $p(X)=-X^3+a X^2+bX+c$ are positive then necessarily $p(0)>0$ because $p(x)\to+\infty$ as $x\to-\infty$. To rule out the possibility that two roots are complex, you may check that $p'=-3X^2+2aX+b$ has two real roots (i.e. $4a^2+12b>0$) and that $p(x)>0$ where $x$ is the greater of the roots of $p'(x)$.

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We can solve it writing $x^tSx$ as a sum of squares. We have \begin{align} x^tSx&=(x_2+tx_1)^2-t^2x_1^2+2x_1^2+3x_3^2+2x_1x_3\\ &=(x_2+tx_1)^2+3\left(x_3^2+\frac 23x_1x_3\right)+(2-t^2)x_1^2\\ &=(x_2+tx_1)^2+3\left(x_3+\frac{x_1}3\right)^2-3\frac{x_1^2}9+(2-t^2)x_1^2\\ &=(x_2+tx_1)^2+3\left(x_3+\frac{x_1}3\right)^2+\left(\frac 53-t^2\right)x_1^2. \end{align}

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