For what $t$ quadratic form $Q$ is positive definite supposing that: $$Q =(x_1, x_2)\left( \begin{array}{cc} 1 & 0\\ 1 & t \end{array} \right)\left( \begin{array}{cc} x_1\\ x_2 \end{array} \right)$$

I simplified it to the form: $x_1^2 +x_1x_2+tx_2^2$, but at this point I get confused and I am not sure how I am supposed to solve it. I would appreciate any help. Thanks.


The answers by @Mostafa Ayaz and @gimusi are incorrect. As @Patryk showed in a comment to the answer by @Mostafa Ayaz, x1 = -1, x2 = 2, t = 1/8 is a counterexample to the incorrect answers.

Because the matrix is not symmetric, its eigenvalues being positive is not the criterion for positive definiteness of $x^TMx$.

$x^TMx \ge 0$ implies $x^TM^Tx \ge 0$, and therefore also implies $x^T(M + M^T)x \ge 0$. Because $M + M^T$ is symmetric, we have reduced the problem to finding the conditions under which $M + M^T$ has positive eigenvalues. In this case, that comes out to $t > 1/4$ as the criterion for positive definiteness of $Q$.

  • $\begingroup$ Note: my first paragraph referred to the original answer by @gimusi , not that poster's current corrected answer. $\endgroup$ – Mark L. Stone Feb 23 '18 at 17:37

Note that the matrix associated to Q can be divided into two parts symmetric and skew-symmetric as follow

$$A=\frac12(A+A^T)+\frac12(A-A^T)=\left(\begin{array}{cc} 1 & \frac12\\ \frac12 & t \end{array}\right)+\left(\begin{array}{cc} 0 & -\frac12\\ \frac12 & 0 \end{array}\right)$$

Since for the skew-symmetric part $Q=0$ the positive definiteness depends upon the symmetric part and thus since

$$\det(1)=1>0 \quad \quad \det\left(\begin{array}{cc} 1 & \frac12\\ \frac12 & t \end{array}\right)=t-\frac14>0\iff t>\frac14$$

thus the $Q$ is positive definite $\iff t>\frac14$.

  • $\begingroup$ See my answer as to why your answer is incorrect. $\endgroup$ – Mark L. Stone Feb 23 '18 at 17:15
  • 1
    $\begingroup$ @MarkL.Stone oh yes of course you are right, thanks to have pointed it out, I fix! $\endgroup$ – user Feb 23 '18 at 17:21

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