orthogonal matrix

Question

I have to show the following claim: Let $A\in Mat(n,\mathbb{R})$ be positive definite and symmetric. Show that there exists a Matrix $T\in Mat(n,\mathbb{R})$ such that $T^tAT$ is a diagonal matrix. My idea was to use the spectral theorem. Therefore I can find a set of orthogonal Eigenvectors$\{v_1,...,v_n\}$,which form a basis. Then I defined the "change of basis Matrix" T by $T:=(v_1,...,v_n)$. Clearly $TT^t=I$. Now I would like to triangulate T,i.e. $C=B^{-1} T B$, where C is an upper triangular matrix. Is it true that $CC^t=I$?In case this is true does it imply that C is the matrix I am looking for? I would be very grateful for any help! Thank you very much in advance!

Answer 1

The inverse of an upper (resp. lower) triangular matrix is upper (resp. lower) triangular. So $C C^T = I$ is only possible if $C$ is actually diagonal, and in this case its diagonal entries are all $\pm 1$.

So I think the short answer to your question is "no".