# Minimal value of the quadratic form $\mathbf{x}^t A \mathbf x$ subject to the contraint $\mathbf{x}^t \mathbf x = 1$

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?

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

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 $$x^{\top}x=1$$.