Short Answer

Show that for every indefinite quadratic form q on $R^{2}$ , there exists an orthogonal basis ${\vec{w}}_{1}, . . ., {\vec{w}}_{2}$ of such that $q (c_{1} {\vec{w}}_{1} + c_{2} {\vec{w}}_{2}) = c_{1}^{2} - c_{2}^{2}$ , Hint: Modify the approach outlined in

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

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

Let us define $C = \{{\vec{w}}_{1}, {\vec{w}}_{2}\}$ , where $w_{i} = \frac{{\vec{v}}_{i}}{\sqrt{|λ_{i}|}}$ , such that is the orthogonal basis of A. Now,
$q (c_{1} {\vec{w}}_{1} + c_{2} \vec{w}) = q (c_{1} \frac{\vec{V_{1}}}{\sqrt{|λ_{1}|}} + c_{2} \frac{\vec{V_{2}}}{\sqrt{|λ_{2}|}}) = q (\frac{c_{1}}{\sqrt{|λ_{1}|}} \vec{v_{1}} + \frac{c_{2}}{\sqrt{|λ_{2}|}} \vec{v_{2}}) = λ_{1} {(\frac{c_{1}}{\sqrt{|λ_{1}|}})}^{2} + λ_{2} {(\frac{c_{2}}{\sqrt{|λ_{2}|}})}^{2} = c_{1}^{2} \frac{λ_{1}}{|λ_{1}|} + c_{2}^{2} \frac{λ_{2}}{|λ_{2}|} = c_{1}^{2} - c_{2}^{2} (λ_{1} > 0 and λ_{2} < 0)$

Result:

The problem is proved using diagonalizability of quadratic form and using indefiniteness of the quadratic form.