# Characterize triangular matrices by its eigenvalues?

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!

• @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

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.
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).
• 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