I have a question about the Gram-Schmidt process, and about the algorithm to find an orthogonal basis of eigenvectors (aka orthogonal diagonlization).
let $T:V \to V$ be a diagonlizable linear transformation, such that $T(x)=Ax$ for some given matrix $A$ (for the sake of argument let's say its symmetric, and so there is an orthonormal basis of eigenvectors). $dim(V)=n$. Assume the eigenvectors of $T$ are $v_1,v_2,v_3,...,v_n$. After running the Gram-Schmidt process, we get the vectors $u_1,u_2,u_3,...,u_n$ such that $span(u_1,u_2,u_3,...,u_n) = span(v_1,v_2,v_3,...,v_n)$ and for all $i\neq j$: $<u_i,u_j>=0$
Are these new vectors, $u_1,u_2,u_3,...u_n$ eigenvectors?
When you first think about it, there doesn't seem to be any apparent reason why this should be true. But, according to the orthogonal diagonlization process shown here: http://en.wikipedia.org/wiki/Orthogonal_diagonalization it is indeed true.Can someone explain this?