It is true that if an upper triangular matrix $A$ with complex entries has distinct elements on the diagonal, then $A$ is diagonalizable. However, I don't think the converse is true. Is there a complete characterization of all diagonalizable upper triangular matrices?
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Yes, this is true. In case there are eigenvalues with multiplicity greater than $1$, both cases can occur: $A$ can be be diagonalizable or not diagonalizable. – Julien Feb 18 '13 at 17:01
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"Diagonalisable upper-triangular matrices" sounds like a pretty solid characterisation to me :P I don't know of any way to characterise them except by characterising diagonalisability separately (e.g. eigenvectors span, etc.) – Ben Millwood Feb 18 '13 at 17:19
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1A heuristic idea of why we shouldn't expect a good result: by Schur's decomposition, every complex square matrix is similar to an upper triangular matrix. If in addition to the eigenvalues we could read the geometric multiplicities from the entries of a triangular matrix, by these days we'd already have a well-known algorithm to look for general diagonalizability from Schur's decomposition. – Jose Brox Feb 01 '18 at 17:15
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Every upper triangular matrix with distinct elements on the diagonal is diagonalizable, because
$$\det(A-\lambda I)=\prod_{i=1}^n (a_{ii}-\lambda)$$
with $a_{ii}\neq a_{jj}$ for $i\neq j$, so every eigenvalue has multiplicity $1$.
The converse is not true. Take $A=I$. Then $A$ is diagonalized, but not with distinct values on the diagonal.
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