$x^{2} - 4xy + y^{2} = 1$ ; $p\lambda$ MOTRISFORM

\begin{array}{r l} & [ x y ] \left[ \begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = 1; \quad \overline{x} = \left[ \begin{array}{c} x \\ 0 \end{array} \right] \\ & \therefore \quad \overline{x}^T A \overline{x} = 1 \end{array} \quad A = \left[ \begin{array}{cc} 1 & -2 \\ -2 & 1 \end{array} \right]

BESTÄM EGENVÄNDEN OCH EGENVEKTORER FÖR A.
KOR. EUV.: $\det(A - \lambda I) = 0 \iff

\begin{pmatrix} 1 - \lambda & -2 \\ -2 & 1 - \lambda \end{pmatrix}

= 0$

\begin{array}{l} (=) (1 - \lambda)^2 - 4 = 0 \quad (=) (1 - \lambda)^2 = 4 \quad (=) \lambda = 1 \pm 2 \\ (=) \lambda = 3; -1 \quad \text{KOU}: 3 + (-1) = 2, \quad \text{tr} A = 1 + 1 = 2 \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 3(-1) = -3, \quad \det A = 1 - 4 = -3 \\ \therefore \quad \text{DET BLIR EN HYPERBEL.} \end{array}

EGENVEKTORER:

\begin{array}{l} \lambda = 3: A \overline{\sigma} = 3 \overline{\sigma} \quad (=) (A - 3I) \overline{\sigma} = \overline{0} \\ \left[ \begin{array}{cc} -2 & -2 \\ -2 & -2 \end{array} \right] \sim \left[ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right] \quad \therefore \quad \overline{\sigma} = t \left[ \begin{array}{cc} -1 & 1 \\ 1 & 1 \end{array} \right] \quad \text{VÄLAND} \quad \left[ \begin{array}{c} -1 \\ 1 \end{array} \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{VÄLAND} \quad \left[ \begin{array}{c} -1 / \sqrt{2} \\ 1 / \sqrt{2} \end{array} \right] \end{array} \begin{array}{l} \lambda = -1: A \overline{\sigma} = -1 \overline{\sigma} \quad (=) (A + I) \overline{\sigma} = \overline{0} \\ \left[ \begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array} \right] \sim \left[ \begin{array}{cc} 1 & -1 \\ 0 & 0 \end{array} \right] \quad \therefore \quad \overline{\sigma} = t \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right] \quad \text{VÄLAND} \quad \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{VÄLAND} \quad \left[ \begin{array}{c} 1 / \sqrt{2} \\ 1 / \sqrt{2} \end{array} \right] \end{array} \because A = P D P^{-1} = P D P^T \text{MED } P = \begin{bmatrix} 1 / \sqrt{2} & -1 / \sqrt{2} \\ 1 / \sqrt{2} & 1 / \sqrt{2} \end{bmatrix}, D = \begin{bmatrix} -1 & 0 \\ 0 & 3 \end{bmatrix} \text{OCH } \overline{x}^T A \overline{x} = 1 \iff \overline{u}^T D \overline{u} = 1 \text{MED } \overline{u} = P \overline{x} \quad \text{då} \quad \overline{u} = \begin{bmatrix} u \\ v \end{bmatrix} \text{ BLIR NU} \overline{x}^T A \overline{x} = 1 \iff \overline{u}^T D \overline{u} = 1 \iff -u^2 + 3v^2 = 1