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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!

Tim
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  • @DennisGulko: Thanks! Why is that? A linear mapping is the sum of a diagonalizable linear mapping and a nilpotent mapping. But I don't see why your comment? – Tim Nov 26 '12 at 13:24
  • 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. – Dennis Gulko Nov 26 '12 at 13:35

2 Answers2

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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.

Martin Argerami
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  • Thanks! What we can say about matrices that have such property, after we know they may not be triangular? – Tim Nov 26 '12 at 13:11
  • I don't immediately see any general property. They most likely have lost of zero entries, but I wouldn't know how to quantify/prove that. – Martin Argerami Nov 26 '12 at 14:02
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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).

Andy Brandi
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    But Andy, that's always true. In fact, for a general matrix $A$ one can even choose $S$ to be a unitary (that's the Schur Decomposition). – Martin Argerami Nov 26 '12 at 14:00
  • It may be false if you consider a matrix with entries from a non-algebraically closed field. For example, the matrix $\begin{bmatrix} 1&2\-3&4 \end{bmatrix}$ does not have any eigenvalues over $\mathbb{R}$ and thus cannot be similar to a triangular matrix. – Andy Brandi Nov 26 '12 at 14:16
  • There was no mention in the initial question that only complex matrices were of interest. Sorry if my answer didn't turn out to be helpful. – Andy Brandi Nov 26 '12 at 14:30