Let $B$ be a $n\times n$ real symmetric positive definite form, and $A$ be a $n\times n$ real symmetric form. There exists an orthogonal matrix O such that $O^TBO=I$ and $O^TAO=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$, where $\lambda_i$ is a proper eigenvalue of $A$ with respect to $B$ (meaning that $A\mathbf v_i=\lambda_iB\mathbf v_i$). Also the columns of $O$ are made by proper eigenvectors ($\mathbf v_i$) of $A$ with respect to $B$.
So I just proved this other theorem that is very related to the first one :
Let $B$ be a $n\times n$ real symmetric positive definite form, and $A$ a $n\times n$ real symmetric form. Then there are $n$ generalised eigenvalues $\lambda_1,\ldots,\lambda_n$ of $A$ with respect to $B$. Let $\mathbf{w}_1,\ldots,\mathbf{w}_n$, each of which is standardised, i.e. $\mathbf{w}_i^TB\mathbf{w}_i=1$. Let $W$ be the $n\times n$ matrix whose columns are the generalised normalised (with respect to $B$) eigenvectors. Then \begin{align} W^TBW=I, && W^TAW=\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \end{align}
Proof:
First of all, we can get the normalised eigenvectors by considering some eigenvectors with different norms $\mathbf{v}_i,\ldots,\mathbf{v}_n$. Let \begin{align} \mathbf{w}_i = \frac{\mathbf{v}_i}{\sqrt{\langle\mathbf{v}_i,B\mathbf{v}_i\rangle}}. \end{align}
Then \begin{align} \mathbf{w}_i^TB\mathbf{w}_i = \Big\langle \frac{\mathbf{v}_i}{\sqrt{\langle\mathbf{v}_i,B\mathbf{v}_i\rangle}},B\frac{\mathbf{v}_i}{\sqrt{\langle\mathbf{v}_i,B\mathbf{v}_i\rangle}} \Big\rangle = \frac{\langle\mathbf{v}_i,B\mathbf{v}_i\rangle}{\langle\mathbf{v}_i,B\mathbf{v}_i\rangle}=1. \end{align}
Now, we know that there exists a matrix $P$ such that $PAP^T=\tilde \Lambda =: U_1$ and $PBP^T=I=:U_2$. Consider $U_1\mathbf{x}=\lambda U_2\mathbf{x}$. It's easy to see that the generalised eigenvalue of $U_1$ with respect to $U_2$ are the diagonal values of $\tilde \Lambda$. In particular, for some generalised eigenvalue $\tilde \Lambda_{ii}$: $\tilde{\Lambda}\mathbf{\tilde v}_i=\tilde \Lambda_{ii}\mathbf{\tilde v}_i$, only if $\mathbf{\tilde v}_i=\mathbf{e}_i$ is a generalised eigenvector. Then \begin{align} U_1I=U_2I\tilde \Lambda && \equiv&& PAP^T I = PBP^TI\tilde{\Lambda}. \end{align}
Let $W=P^T$, therefore last equation yields: \begin{align} TAW=TBW\tilde{\Lambda} \Longrightarrow AW=BW\tilde{\Lambda} \end{align}
and shows that $\tilde{\Lambda}=\Lambda$ is the matrix of generalised eigenvalues of $A$ with respect to $B$, and moreover $W$ is the matrix whose columns are the generalised eigenvectors of $A$ with respect to $B$. Moreover: \begin{align} PAP^T=\lambda= W^TAW && \text{and} && PBP^T=I=W^TBW. \end{align}
$$\text{________________________ QUESTION: ________________________}$$
My question is how can I prove that $W$ is indeed orthogonal? If it is not, how can I find this orthogonal matrix? Would a matrix $O$ with normal proper eigenvectors as columns satisfy $O^TBO=I$ and $O^TAO= \operatorname{diag}(\lambda_1,\ldots,\lambda_n)$?
