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.

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  • $\begingroup$ That was my first thought but it seemed a bit odd, but thinking about it, we could take the approach that, before using Schur's theorem, we could state that $B = A^*$, use Schur's theorem, then we would get $U^*BU$ is upper triangular $\implies U^*AU$ is lower triangular. Does this sound right? $\endgroup$ – John Dec 5 '14 at 18:47
  • $\begingroup$ @Eric yes but it's correct also without introducing any additional notation. $\endgroup$ – Algebraic Pavel Dec 5 '14 at 20:12

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