Question

For the quadratic form $\mathbf{q}\mathbf{(}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{3}{\mathbf{x}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}\mathbf{10}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{x}}_{\mathbf{2}}^{\mathbf{2}}$, find an orthogonal basis ${\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{,}{\overrightarrow{\mathbf{w}}}_{\mathbf{2}}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{R}}^{\mathbf{2}}\mathbf{}\mathbf{such}\mathbf{}\mathbf{that}\mathbf{}\mathbf{q}\mathbf{(}{\mathbf{c}}_{\mathbf{1}}{\overrightarrow{\mathbf{w}}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\overrightarrow{\mathbf{w}}}_{\mathbf{2}}\mathbf{)}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}^{\mathbf{2}}$. Use your answer to sketch the level curve $\mathbf{q}\left(\overrightarrow{x}\right)\mathbf{=}\mathbf{1}$.Exercise 65 is helpful.

Short answer

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

Step-by-step solution

0f 2: Given information

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

of 2: Application

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

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

• For${\lambda }_1=8$, we have$(\mathrm{A}-8\mathrm{l}){\overrightarrow{\mathrm{v}}}_1=0\Rightarrow \left[\begin{array}{cc}-5& -5\\ -5& -5\end{array}\right]\left[\begin{array}{c}{\mathrm{v}}_{11}\\ {\mathrm{v}}_{12}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\Rightarrow {\mathrm{v}}_{11}+{\mathrm{v}}_{12}=0$
• one normalized eigenvector corresponding to ${\lambda }_1=8$is
• For ${\lambda }_2=-2$, we have $(\mathrm{A}+2\mathrm{l}){\overrightarrow{\mathrm{v}}}_2=0\Rightarrow \left[\begin{array}{cc}-5& -5\\ -5& -5\end{array}\right]\left[\begin{array}{c}{\mathrm{v}}_{21}\\ {\mathrm{v}}_{22}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\Rightarrow {\mathrm{v}}_{21}+{\mathrm{v}}_{22}=0$
• Hence, one normalized eigenvector corresponding to ${\lambda }_2=-2$is ${\overrightarrow{v}}_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right]$.
• The orthonormal eigenbasis of A corresponding to eigenvalues ${\lambda }_1=8$and ${\lambda }_2=-2$is formed by $\mathrm{B}=\left\{{\overrightarrow{\mathrm{v}}}_1,{\overrightarrow{\mathrm{v}}}_2\right\}$
• The vectors from Exercise 65 is ${\overrightarrow{\mathrm{w}}}_1=\frac{{\overrightarrow{\mathrm{v}}}_1}{\sqrt{\left|{\mathrm{\lambda }}_1\right|}}=\frac{1}{4}\left[\begin{array}{c}1\\ -1\end{array}\right]\mathrm{and}{\overrightarrow{\mathrm{w}}}_2=\frac{{\overrightarrow{\mathrm{v}}}_2}{\sqrt{\left|{\mathrm{\lambda }}_2\right|}}=\frac{1}{2}\left[\begin{array}{c}1\\ -1\end{array}\right]$ forms an orthogonal basis of A , which gives us $\mathrm{q}\left({\mathrm{c}}_1{\overrightarrow{\mathrm{w}}}_1+{\mathrm{c}}_2{\overrightarrow{\mathrm{w}}}_2\right)={\mathrm{c}}_1^2={\mathrm{c}}_2^2$
• 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 ${\overrightarrow{\mathrm{w}}}_1=\frac{1}{4}\left[\begin{array}{c}1\\ -1\end{array}\right]\mathrm{and}{\overrightarrow{\mathrm{w}}}_2=\frac{1}{2}\left[\begin{array}{c}1\\ 1\end{array}\right]\mathrm{such}\mathrm{that}\mathrm{q}\left({\mathrm{c}}_1{\overrightarrow{\mathrm{w}}}_1+{\mathrm{c}}_2{\overrightarrow{\mathrm{w}}}_2\right)={\mathrm{c}}_1^2-{\mathrm{c}}_2^2$