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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?

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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 $UD \neq I$.

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  • $\begingroup$ I expect $A$ to have $n$ non-zero eigenvalues, while the $UA$ you write seems to have rank $n-k$. It is a bit puzzling how all $n$ eigenvalues are then recovered just from a multiplication with a diagonal matrix $D$. Or maybe I missed something? $\endgroup$ – Kagaratsch Feb 21 '16 at 3:20
  • $\begingroup$ Sorry I forgot the fact that $det(A) \neq 0$: then it's much easier but my result still holds. The fact is that you can get the eigenvalues in a diagonal because you are actually not using the eigenvectors, but you are just adjusting the base choice in function of what you want to get. You can do it because you are not using similar matrixes, and this simplifies everything (though I've never seen an application of this) $\endgroup$ – Nicolò Feb 21 '16 at 15:39
  • $\begingroup$ So basically for matrices with $det(A)\neq0$ we have $k=0$ and the rank is fully present. I see. Thank you! $\endgroup$ – Kagaratsch Feb 22 '16 at 5:22

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