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Expert-verified Found in: Page 402 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Show that for every indefinite quadratic form q on ${{\mathbf{R}}}^{{\mathbf{2}}}$, there exists an orthogonal basis ${\stackrel{\mathbf{\to }}{\mathbf{w}}}_{{\mathbf{1}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{\stackrel{\mathbf{\to }}{\mathbf{w}}}_{{\mathbf{2}}}$ of such that ${\mathbf{q}}\mathbf{\left(}{\mathbf{c}}_{\mathbf{1}}{\stackrel{\mathbf{\to }}{\mathbf{w}}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\stackrel{\mathbf{\to }}{\mathbf{w}}}_{\mathbf{2}}\mathbf{\right)}{\mathbf{=}}{{\mathbf{c}}}_{{\mathbf{1}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbf{c}}}_{{\mathbf{2}}}^{{\mathbf{2}}}$, Hint: Modify the approach outlined in

The diagonalizability of quadratic form and indefiniteness of the quadratic form are used to prove it.

See the step by step solution

## Step 1 of 2: Given information

• Let $\mathrm{q}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2}\right)$ be an indefinite quadratic form which is defined by the symmetric matrix $A\in {\mathrm{ℝ}}^{2×2}$, where $\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{T}}\mathrm{Ax}$ and A is indefinite.
• It has one positive and one negative eigenvalue.
• As per the theorem , orthonormal eigenbasis is $\mathrm{B}=\left\{\stackrel{\to }{{\mathrm{v}}_{1}},\stackrel{\to }{{\mathrm{v}}_{2}}\right\}$ and its corresponding eigenvalues are role="math" localid="1659676117779" ${\mathrm{\lambda }}_{1}>0,{\mathrm{\lambda }}_{2}<0$ .
• The quadratic form is diagonalizable as $\mathrm{q}\left(\mathrm{x}\right)={\mathrm{\lambda }}_{1}{\mathrm{c}}_{1}^{2}+{\mathrm{\lambda }}_{2}{\mathrm{c}}_{2}^{2}.....\left(1\right)$
• The coordinates of X with respect to the eigenbasis B are ${\mathrm{c}}_{\mathrm{i}}^{\text{'}}\mathrm{s}$ .

## Step 2 0f 2: Application

• Let us define $\mathrm{C}=\left\{{\stackrel{\to }{\mathrm{w}}}_{1},{\stackrel{\to }{\mathrm{w}}}_{2}\right\}$ , where ${\mathrm{w}}_{\mathrm{i}}=\frac{{\stackrel{\to }{\mathrm{v}}}_{\mathrm{i}}}{\sqrt{\left|{\mathrm{\lambda }}_{\mathrm{i}}\right|}}$, such that is the orthogonal basis of A. Now,
• $\mathrm{q}\left({\mathrm{c}}_{1}{\stackrel{\to }{\mathrm{w}}}_{1}+{\mathrm{c}}_{2}\stackrel{\to }{\mathrm{w}}\right)=\mathrm{q}\left({\mathrm{c}}_{1}\frac{\stackrel{\to }{{\mathrm{V}}_{1}}}{\sqrt{\left|{\mathrm{\lambda }}_{1}\right|}}+{\mathrm{c}}_{2}\frac{\stackrel{\to }{{\mathrm{V}}_{2}}}{\sqrt{\left|{\mathrm{\lambda }}_{2}\right|}}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{q}\left(\frac{{c}_{1}}{\sqrt{\left|{\mathrm{\lambda }}_{1}\right|}}\stackrel{\to }{{\mathrm{v}}_{1}}+\frac{{c}_{2}}{\sqrt{\left|{\mathrm{\lambda }}_{2}\right|}}\stackrel{\to }{{\mathrm{v}}_{2}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{\lambda }}_{1}{\left(\frac{{\mathrm{c}}_{1}}{\sqrt{\left|{\mathrm{\lambda }}_{1}\right|}}\right)}^{2}+{\mathrm{\lambda }}_{2}{\left(\frac{{\mathrm{c}}_{2}}{\sqrt{\left|{\mathrm{\lambda }}_{2}\right|}}\right)}^{2}\phantom{\rule{0ex}{0ex}}={c}_{1}^{2}\frac{{\mathrm{\lambda }}_{1}}{\left|{\mathrm{\lambda }}_{1}\right|}+{c}_{2}^{2}\frac{{\mathrm{\lambda }}_{2}}{\left|{\mathrm{\lambda }}_{2}\right|}\phantom{\rule{0ex}{0ex}}={c}_{1}^{2}-{c}_{2}^{2}\left({\lambda }_{1}>0\mathrm{and}{\mathrm{\lambda }}_{2}<0\right)$

Result:

The problem is proved using diagonalizability of quadratic form and using indefiniteness of the quadratic form. ### Want to see more solutions like these? 