Let g be a symmetric bilinear form given by the Gram matrix $
\begin{pmatrix}
3 & -2 & 0 \\
-2& 2& -2 \\
0 & -2 & 1 \\
\end{pmatrix}
$.
Find a basis A such that the Gram matrix in this basis is diagonal.
Find a basis B such that the Gram matrix is
$ \begin{pmatrix} 1 &
Sory, i didnt get you. Why you are looking for the conditions. Because for me it seems that i need to find Basis B such that $ABA^{T}=G$ where G is matrix in part b and A is set of orthogonal basis matrx from part (a). Please correct me if i am wrong
My point is you should see directly that the diagonal matrix with negative entry can't be a gramian matrix in any basis, because we will get negative distances along the dimension of the negative eigenvalue.
$v_{1}=(2, 1, -2)$, $v_{2}= (2,-2,1)$, $v_{3}=(1,2,2)$. These are orthogonal basis obtained from part a. then basis B is obtained by as follows:
$a^{2}v_{1}^{T}Av_{1}=1$ it will give a. like wise $b^{2}v_{2}^{T}Av_{2}=1$ and $c^{2}v_{3}^{T}Av_{3}=-1$ give $b$ and $c$. So the basis B is $(\frac{v_{1}}{a}, \frac{v_{2}}{b}, \frac{v_{3}}{c})$
Discussion between Lena and mathreadler
Discussion between Lena and mathreadler