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For a triangular matrix, its diagonal entries are eigenvalues repeated with algebraic multiplicities.

I wonder if the reverse is true. In other words, a matrix whose diagonal entries are eigenvalues repeated with algebraic multiplicities must be triangular?

Thanks!

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    That was plain wrong, it is not equivalent but rather a special case, in which it is easy to find a counter-example since (for me :) it is easier to think of nilpotent matrices.2012-11-26

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No; consider $ a=\begin{bmatrix}2&1&0\\0&2&0\\0&1&3\end{bmatrix}. $ The eigenvalues are $2,2,3$ but it is not triangular.

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    I don't immediately see any general property. They most likely have lost o$f$ zero entries, but I wouldn't know how to quantify/prove that.2012-11-26
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As Martin already pointed out in his answer, this must not necessarily be true. However, if the diagonal entries of an $n\times n$ matrix $A$ coincide with its eigenvalues, the characteristic polynomial of $A$ can be completely factored into linear factors as there are exactly $n$ diagonal entries and at most $n$ eigenvalues (counted by multiplicity). $A$ is thus at least similar to a triangular matrix, i.e. there exists $S\in \operatorname{GL}(n,V)$, such that $ S^{-1}AS $ is triangular.

Edit: This holds for any (finite-dimensional) vector space $V$ over an arbitrary field $\mathbb{K}$. If $\mathbb{K}$ is algebraically closed, then $A$ is always similar to a triangular matrix (see comments).

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    There was $n$o mention in the initial question that only complex matrices were of interest. Sorry if my answer didn't turn out to be helpful.2012-11-26