0
$\begingroup$

Please help me solving this question:

Let $m$ be the minimal value of the quadratic form $\mathbf{x}^t A \mathbf x$ subject to the contraint $\mathbf{x}^t \mathbf x = 1$.

Given a vector on the unit circle in which this minimal value is attained.

$$A=\left[\begin{array}{cc} -\frac{7}{4} & \frac{3}{4}\\ \frac{3}{4} & \frac{1}{4} \end{array}\right] $$

$$\left[\begin{array}{c} \frac{3}{\sqrt{10}}\\ -\frac{1}{\sqrt{10}} \end{array}\right] \text { or } \left[\begin{array}{c} -\frac{3}{\sqrt{10}}\\ -\frac{1}{\sqrt{10}} \end{array}\right]$$

I calculated the eigenvectors and eigenvalues and tried to fill them in to the formula: $$\mathbf x_k = c_1 \lambda_1^k \mathbf v_1 + c_2 \lambda_2^k \mathbf v_2$$ and this is not really solving the problem.

What do I need to do tho solve this question?

$\endgroup$
  • 1
    $\begingroup$ Look here $\endgroup$ – A.Γ. Aug 6 at 11:37
  • $\begingroup$ The answer is eigenvectors, but your eigenvectors are not correct. $\endgroup$ – mert Aug 6 at 12:30
  • $\begingroup$ @A.Γ. Make that an answer, since it solves the problem entirely $\endgroup$ – Ant Aug 6 at 13:14
0
$\begingroup$

The Lagrange function is $$ L\left(x,\lambda\right)=x^{\top}Ax+\lambda\left(1-x^{\top}x\right) $$ Derivative wrt $x$ yields $2Ax-2\lambda x=0\implies\left(A-\lambda I\right)x=0$ means $x$ is in the null space of $A-\lambda I$ and $\left|A-\lambda I\right|=0$. So the rule is to find the eigenvectors of $A$. Then, normalize them to $x^{\top}x=1$.

$\endgroup$

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.