# Question about Gram-Schmidt algorithm. Orthogonal diagonalization. Does GS conserve eigen-ness property

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?

• No, I didn't mean they are not orthonormal. Yes, ofcourse the new vectors are orthonormal, what I'm asking is - are they eigenvectors? Will running Gram-Schmidt with an input thats a basis of eigenvectors, will it yield a basis of orthonormal eigenvectors, or is the eigen-ness property gone? – Oria Gruber Dec 29 '13 at 17:36
• oppsie, misread the question, sorry. – Gina Dec 29 '13 at 17:38
• In that case, then since we already assume that there exist an orthonormal eigenbasis, which means that all eigenspace are orthogonal. Hence any part you need to fix are the one inside the same eigenspace, and the result is still inside that space (ie. still an eigenvector). – Gina Dec 29 '13 at 17:44

Take $T=\begin{bmatrix}1 & 2 \\ 0 & 2 \end{bmatrix}$. Then the eigenvectors $v_1,v_2$ of $T$ are not orthogonal. If you use Gram Schmidt to orthogonalize $v_1,v_2$ one of them will no longer be an eigenvector.
If the resulting vectors were eigenvectors, then the matrix would have been orthogonally (or unitarily depending on the field) diagonalizable, which would imply that the matrix was normal. However, a quick check shows that $T T^* \neq T^* T$.