For this bilinear form: $q(v)=q(x_1,x_2,x_3)=x_1^2+x_2^2+9x_3^{2}+2x_1x_2-6x_1x_3-5x_2x_3$ find a base $B$ so that $[q]^B_B=D$ diagonalizable matrix For this bilinear form: $q(v)=q(x_1,x_2,x_3)=x_1^2+x_2^2+9x_3^{2}+2x_1x_2-6x_1x_3-5x_2x_3$ I need to find a base $B$ so that $[q]^B_B=D$ will be diagonalizable matrix. So, I tried to look for eigenvalues after writing this bilinear form as a matrix $\begin{pmatrix}
1 & 1 &-3 \\  1&1  &-2.5 \\  -3& -2.5 & 9\end{pmatrix}$ and find eigenvalues but it's impossible mission, it's really messy. Is there any other method which with I can find it? maybe something with Jacobi method?
Thank you.

There is a general method to solve this kind of exercises, which works for $\mathbb R^n$, not only $\mathbb R^3$. Let $q\colon\mathbb R^n\rightarrow \mathbb R$ a quadratic form. We can write it as a sum of squares of linearly independent linear functionals, namely $\displaystyle q(x_1,\ldots,x_n)=\sum_{j=1}^r\alpha_jl_j(x_1,\ldots,x_n)$ (we may not have $n$ termes, for example look at $q(x_1,\ldots,x_n)=x_1^2$). Then $q(x_1,\ldots,x_n) = ^t(P(x_1,\dots,x_n))DP(x_1,\dots,x_n)$ where $D=\mathrm{Diag}(\alpha_1,\ldots,\alpha_r,0\ldots,0)$ and $P(x_1,\ldots,x_n) = (l_1(x),\ldots,l_r(x),l_{r+1}(x),\ldots,l_n(x))$ ($l_{r+1}\ldots,l_n$ are linear functionals chosen in order to make $P$ invertible). 
