I've seen two very small distinctions in Schur's theorem, one is that for any $n\times n$ matrix $A, \exists$ a Unitary $U$ s.t. $U^*AU$ is upper triangular, the other is that $U^*AU$ is lower triangular. I've only learned the proof and can use the 'upper triangular' version and would like to show that Schur's theorem can also build a lower triangular matrix. I don't think I should have to go through the entire proof again, it seems I should be able to simply state something about "By Schure's theorem..."
Any ideas would be appreciated.
Thank you.
By Schur's theorem, there is a unitary $U$ such that $U^*A^*U$ is upper triangular. Hence $(U^*A^*U)^*=U^*AU^*$ is lower triangular.