0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $Q$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $\iff t>\frac14$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.