Finding a basis $B$ such that $M_B(s)=$diag$(I_k,-I_l,O)$

Question

Let $V=\mathbb{R^3}$, $A=\begin{pmatrix} 0 & -2 & 1 \\ -2 & 1 & 2 \\ 1 & 2 & 0 \end{pmatrix} \in M_3(\mathbb{R})$ and $s$ the symmetric bilinear form on $V$ with respect to $A$, $s(v,w)=\langle v,Aw \rangle$.

How to determine a basis $B$ such that the transformation matrix is $M_B(s)=$diag$(I_k,-I_l,O)$?

Firstly, I computed the characteristic polynomial $p_A(\lambda)=-(\lambda-1)(\lambda+3)(\lambda-3)$ so I get the Eigenvalues $\lambda_1=1, \lambda_2=-3$ and $\lambda_3=3$ and the eigenvectors

$v_1=\begin{pmatrix} -1\\-1\\1 \end{pmatrix}, v_2=\begin{pmatrix} 1\\0\\1 \end{pmatrix} $ and $v_3=\begin{pmatrix} -1\\2\\1 \end{pmatrix}$.

Then I used the Gram-Schmidt process and got for $v_1=\begin{pmatrix} -1\\-1\\1 \end{pmatrix}$ the vector $w_1=\frac{v_1}{\sqrt{s(v_1,v_1)}}=\frac{v_1}{\sqrt{\langle v_1,Av_1 \rangle}}=\begin{pmatrix} \frac{-1}{\sqrt{-9}}\\\frac{-1}{\sqrt{-9}}\\\frac{1}{\sqrt{-9}} \end{pmatrix}$

Here I don't know how to continue.

I know the result is $B=\{v_2,\frac{1}{\sqrt{3}}v_3,\frac{1}{\sqrt{3}}v_1\} $ and

$M_B(s)=$diag$(1,1,-1)$

but I don't know how to get there.

Answer 1

You're almost there:

1. The indices of the eigenvalues ought to match those of the eigenvectors, so reorder them: $$\lambda_1=-3,\ \lambda_2=1,\ \lambda_3=3$$
2. We don't need the Gram-Schmidt process, as the eigenvectors are already orthogonal. We only need to norm these, according to $s$ and preserving the sign.
3. Now we have $s(v_1,v_1)=\langle v_1, Av_1\rangle=\lambda_1\langle v_1,v_1\rangle=-9$, so taking $w_1:=\frac13 v_1$ will give $s(w_1,w_1)=-1$.
   Similarly, $s(v_2,v_2)=2,\ s(v_3,v_3)=18$ so take $w_2=\frac1{\sqrt2}v_2$ and $w_3=\frac1{3\sqrt2} v_3$.