Eigenvectors of a symmetric matrix and orthogonality

Given a full rank symmetric matrix $A_{p\times p}$ we can build a matrix $U=[u_1,...,u_p]$ where $u_i$ is the eigenvector associated to the $i^{th}$ largest eigenvalue of $A$.

Assuming that eigenvectors have unit norm it is easy to prove that $U'U=I_p$ (eigenvectors are orthogonal). I am wondering if somebody knows under which conditions it is also true that $UU'=I_p$

• always............................ Commented Aug 28, 2018 at 22:43
• If $U'U=I_p$ the eigenvectors are even orthonormal, as you've additionally assumed that they have unit norm.
– blub
Commented Aug 28, 2018 at 22:43
• @WillJagy I don't get why.. Commented Aug 28, 2018 at 22:45
• @zzuussee Yes, they are orthonormal but $UU'$ is not the same as $U'U$.... Commented Aug 28, 2018 at 22:47
• $U'U = I$ implies that $U$ is injective. So, $U$ is invertible and $U^{-1} = U'$. Commented Aug 28, 2018 at 22:48

$$U' U = I$$ $$U (U'U) = UI = U$$ associativity $$(UU') U = U$$ cancel by multiplying on the right by the right inverse of $U$ $$UU' = I$$
• @amsmath I do not know what would satisfy this OP. One can, for example, multiply on the right so as to bring about elementary column operations, eventually taking $U$ to reduced column echelon form, which will be the identity matrix as $U$ is full rank Commented Aug 28, 2018 at 22:54
• @WillJagy Is it enough saying that since $U$ has a left inverse and $U$ has full rank then the left inverse and the right inverse have to be the same? Commented Aug 28, 2018 at 23:00
• @WillJagy The key here is that column rank = row rank. And since $U$ is quadratic, its column rank is $p$. Hence, it has a right inverse. Commented Aug 28, 2018 at 23:10