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?


  • $\begingroup$ @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? $\endgroup$ – Tim Nov 26 '12 at 13:24
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ Thanks! What we can say about matrices that have such property, after we know they may not be triangular? $\endgroup$ – Tim Nov 26 '12 at 13:11
  • $\begingroup$ 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. $\endgroup$ – Martin Argerami Nov 26 '12 at 14:02

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).

  • 1
    $\begingroup$ 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). $\endgroup$ – Martin Argerami Nov 26 '12 at 14:00
  • $\begingroup$ 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. $\endgroup$ – Andy Brandi Nov 26 '12 at 14:16
  • $\begingroup$ 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. $\endgroup$ – Andy Brandi Nov 26 '12 at 14:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.