91影视

• $$\begin{gathered} \mathrm{q}\left(\mathrm{x}\right)=\mathrm{q}\left({\mathrm{x}}_1{\mathrm{x}}_2\right)=3{\mathrm{x}}_1^2-{\mathrm{x}}_1{\mathrm{x}}_2+3_2^2=\left[\begin{array}{cc}{\mathrm{x}}_1& {\mathrm{x}}_2\end{array}\right]\left[\begin{array}{cc}3& -5\\ -5& 3\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\end{array}\right]={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax} \\ \mathrm{A}=\left[\begin{array}{cc}3& -5\\ -5& 3\end{array}\right] \end{gathered}$$
• It is indefinite as$\left|{\mathrm{A}}^{\left(1\right)}\right|=3>0\mathrm{and}\left|{\mathrm{A}}^{\left(2\right)}\right|=-16<0$.

• Here, the eigenvalues are given as$\left|\mathrm{A}-\mathrm{位濒}\right|=0$ , that is

${\left(3-\mathrm{位}\right)}^2-25=0鈬�\left(\mathrm{位}-8\right)\left(\mathrm{位}+2\right)=0鈬�{\mathrm{位}}_1=8>0,{\mathrm{位}}_2=-2<0$

• For${位}_1=8$, we have$(\mathrm{A}-8\mathrm{l}){\stackrel{鈫�}{\mathrm{v}}}_1=0鈬�\left[\begin{array}{cc}-5& -5\\ -5& -5\end{array}\right]\left[\begin{array}{c}{\mathrm{v}}_{11}\\ {\mathrm{v}}_{12}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]鈬�{\mathrm{v}}_{11}+{\mathrm{v}}_{12}=0$
• one normalized eigenvector corresponding to ${位}_1=8$is
• For ${位}_2=-2$, we have $(\mathrm{A}+2\mathrm{l}){\stackrel{鈫�}{\mathrm{v}}}_2=0鈬�\left[\begin{array}{cc}-5& -5\\ -5& -5\end{array}\right]\left[\begin{array}{c}{\mathrm{v}}_{21}\\ {\mathrm{v}}_{22}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]鈬�{\mathrm{v}}_{21}+{\mathrm{v}}_{22}=0$
• Hence, one normalized eigenvector corresponding to ${位}_2=-2$is ${\stackrel{鈫�}{v}}_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right]$.
• The orthonormal eigenbasis of A corresponding to eigenvalues ${位}_1=8$and ${位}_2=-2$is formed by $\mathrm{B}=\left\{{\stackrel{鈫�}{\mathrm{v}}}_1,{\stackrel{鈫�}{\mathrm{v}}}_2\right\}$
• The vectors from Exercise 65 is ${\stackrel{鈫�}{\mathrm{w}}}_1=\frac{{\stackrel{鈫�}{\mathrm{v}}}_1}{\sqrt{\left|{\mathrm{位}}_1\right|}}=\frac{1}{4}\left[\begin{array}{c}1\\ -1\end{array}\right]\mathrm{and}{\stackrel{鈫�}{\mathrm{w}}}_2=\frac{{\stackrel{鈫�}{\mathrm{v}}}_2}{\sqrt{\left|{\mathrm{位}}_2\right|}}=\frac{1}{2}\left[\begin{array}{c}1\\ -1\end{array}\right]$ forms an orthogonal basis of A , which gives us $\mathrm{q}\left({\mathrm{c}}_1{\stackrel{鈫�}{\mathrm{w}}}_1+{\mathrm{c}}_2{\stackrel{鈫�}{\mathrm{w}}}_2\right)={\mathrm{c}}_1^2={\mathrm{c}}_2^2$
• Here comes the sketch of $\mathrm{q}\left(\mathrm{x}\right)=\mathrm{q}(\mathrm{x},\mathrm{y})=3{\mathrm{x}}^2-10\mathrm{xy}+3{\mathrm{y}}^2=1$, which is a rotated hyperbola.

Result:

The vectors ${\stackrel{鈫�}{\mathrm{w}}}_1=\frac{1}{4}\left[\begin{array}{c}1\\ -1\end{array}\right]\mathrm{and}{\stackrel{鈫�}{\mathrm{w}}}_2=\frac{1}{2}\left[\begin{array}{c}1\\ 1\end{array}\right]\mathrm{such}\mathrm{that}\mathrm{q}\left({\mathrm{c}}_1{\stackrel{鈫�}{\mathrm{w}}}_1+{\mathrm{c}}_2{\stackrel{鈫�}{\mathrm{w}}}_2\right)={\mathrm{c}}_1^2-{\mathrm{c}}_2^2$