The task is the following:
Consider $\mathbb{R}^2$ equipped with the canonical dot-product $\langle \cdot , \cdot \rangle$, and also the symmetrical bilinear form $$\beta(u,v) := \left\langle u,\ \begin{pmatrix} 2 & -\sqrt{2} \\ -\sqrt{2} & 3 \end{pmatrix} v\right\rangle ~~~~~~~~~~~~~~ \forall u,v \in \mathbb{R}^2.$$ Calculate the normal form considering Sylvester's law of inertia. That means calculate the numbers $r_+$, $r_-$ and $r_0$ and an orthogonal basis for which the transformation matrix has this normal form.
I know that $r_+$, $r_-$ and $r_0$ are the positive, negative and $0$ Eigenvalues. In my case I think they have to be normed to $+1$ and $-1$ since a normal form is being searched for. Am I right?
Well what I did was calculating the eigenvectors that will form an orthogonal basis because we can consider the matrix a self-adjoint endomorphism on $\mathbb{R}^2$. And now consider the Matrix $S$ with those eigenvectors as columns. The matrix $S^t M S$ will be in diagonal form. In numbers:
I got the eigenvalues $\lambda_1 = 1$ and $\lambda_2 = 4$ with the corresponding eigenvectors $$v_{\lambda = 1} = \begin{pmatrix}\sqrt{2} \\ 1\end{pmatrix}~~~~~ \text{ and } ~~~~~ v_{\lambda = 4} = \begin{pmatrix} -\frac{1}{\sqrt{2}} \\ 1\end{pmatrix}$$
$$ \Rightarrow S = \begin{pmatrix} \sqrt{2} & -\frac{1}{\sqrt{2}} \\ 1 & 1\end{pmatrix} \quad\text{ and }\quad S^t = \begin{pmatrix} \sqrt{2} & 1 \\ -\frac{1}{\sqrt{2}} & 1\end{pmatrix}. $$
And calculating $S^t M S$ gives:
$$ \begin{pmatrix} \sqrt{2} & 1 \\ -\frac{1}{\sqrt{2}} & 1\end{pmatrix} \begin{pmatrix} 2 & -\sqrt{2} \\ -\sqrt{2} & 3 \end{pmatrix} \begin{pmatrix} \sqrt{2} & -\frac{1}{\sqrt{2}} \\ 1 & 1\end{pmatrix} = \begin{pmatrix}3 & 0 \\ 0 & 6\end{pmatrix}.$$
Well is it true then that $r_+ = 2$, $r_- = 0$ and $r_0 = 0$ since the eigenvalues are $1$ and $4$? But I struggle finding a basis for which that matrix has normal form. All I am capable to find is one with diagonal form - or is this already the normal form and the orthogonal basis is given by $B = \{v_{\lambda = 1},v_{\lambda = 4} \}$?
I hope someone can help me out a little.
Thank you very much for your help.
FunkyPeanut