# Why can we write $B=UAU^{-1}$?

This question is from the 1979 Berkeley Problems in Mathematics. It asked me to prove that every complex matrix $A$ can be written in the form $$B=UAU^{-1}$$ with $U$ unitary and $B$ upper triangular. The Jordan decomposition only gives me that $U$ invertible, which is not enough in this case. I do not know a simple trick to turn invertible matrices into unitary matrices (like the one that works for the real case $C=\sqrt{DD^{T}}$).

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fixed the typos. –  Bombyx mori Jul 31 '12 at 11:54
sorry........... –  Bombyx mori Jul 31 '12 at 12:10
Look up the Schur decomposition. –  Guess who it is. Jul 31 '12 at 12:14
Thanks for the guidance. –  Bombyx mori Jul 31 '12 at 12:17

## 1 Answer

Just so this question does not remain unanswered:

As I've already mentioned, the decomposition you are interested in is in fact the (somewhat) famous Schur decomposition; that is, any matrix is unitarily similar to a triangular matrix. (In the special case of normal matrices, the triangular matrix is diagonal, and your Schur decomposition is now the spectral decomposition). For proofs of the existence of this very useful decomposition, see this, this, or this.

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