How many ways to diagonalize a symmetric real matrix? Given a symmetric matrix $A$ with real entries and non-vanishing determinant $det(A)\neq0$, one can use orthogonal transformations $R$ to diagonalize the matrix:
$$RAR^{-1}=B$$
where $B$ is diagonal. Now I wonder if this is the only kind of transformation that can be performed to obtain the diagonal matrix $B$ in this case? In particular, I wonder if something like the following is also possible:
$$UAV=B$$
where $U$ and $V$ have other properties (like for instance being unitary) and $UV\neq 1$? Is this a possibility, or are orthogonal transformations all that can be done in this case?
 A: If you want to use a transformation of the type $R^{-1}AR$ then the only possibility is to use an orthogonal matrix $R$ because, if you see this from the point of view of base changing, by multiplying by $R$ you are changing the base of domain and codomain. $R$ has columns made by the eigenvectors of $A$ and then it is necessarily orthogonal because the eigenvectors relative to different eigenvalues of a symmetrical matrix are orthogonal (check that if the columns of $R$ are made by orthogonal vectors then $R$ is orthogonal).
If you want to consider a generic transformation of the type $UAV$ with $UV\neq I$ and get the same diagonal matrix $B$: if the base for the domain is $v_1,...,v_n$ with the last $k$ vectors making a base for $ker(A)$ then you can choose the codomain base as $Av_1,...,Av_{n-k}$ and other $k$ vectors you can choose as you like. This way you obtain 
$$UA=\begin{pmatrix} 1 \\ & 1\\& & 1 \\ & & & ... \\ & & & & & 0 \\& & & & & & 0 \\ & & & & & & & ...\\ & & & & & & & & 0
\end{pmatrix}$$
being the number of ones $n-k$. You don't have the term $V$ bacause you haven't changed the domain base. Now multiply this by a diagonal matrix $D$ with the numbers arranged to obtain exactly the diagonal numbers of $B$ to get $UAD=B$. If $v_1,...,v_n$ wasn't a base of eigenvectors then you'll have $UD \neq I$.
