I want to learn about matrices whose diagonal elements are the eigenvalues... but the matrix is neither diagonal nor triangular.

Is there a term for such matrices, and have they been researched?

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    $\begingroup$ Where did you encounter these? But here's something to think about: you can decompose any square matrix as $\mathbf A=\mathbf L+\mathbf U$, where $\mathbf L$ is strictly lower triangular, and $\mathbf U$ is upper triangular. If $\mathbf A\mathbf x=\lambda\mathbf x$, consider the result of substituting in the decomposition. $\endgroup$ Jun 28, 2016 at 1:47
  • $\begingroup$ Why are they interesting? Do they show up in any other contexts? $\endgroup$
    – Neal
    Jun 28, 2016 at 1:48
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    $\begingroup$ I was researching the determinantal conjecture: en.wikipedia.org/wiki/Determinantal_conjecture, and these types of situations came up. $\endgroup$ Jun 28, 2016 at 1:55

3 Answers 3


I am not aware of any relevant research. Yet, for any $n\ge3$, there always exists a matrix that is non-triangular but whose eigenvalues are any $n$ given scalars $\lambda_1,\lambda_2,\ldots,\lambda_n$. The construction is recursive. First, we begin with a triangular matrix $$A_2=\pmatrix{\lambda_1&1\\ 0&\lambda_2}.$$ Now, if $n\ge3$ is odd, we define $$A_n=\pmatrix{A_{n-1}&0\\ \mathbf1^T&\lambda_n},$$ where $\mathbf 1$ is a vector of ones of appropriate length. If $n\ge3$ is even, define $$A_n=\pmatrix{A_{n-1}&\mathbf1\\ 0&\lambda_n}.$$ To illustrate, we have $$ A_4=\pmatrix{\lambda_1&1&0&1\\ 0&\lambda_2&0&1\\ 1&1&\lambda_3&1\\ 0&0&0&\lambda_4}. $$ Clearly, $A_n$ is not triangular (although it is block triangular) when $n\ge3$, because it has both subdiagonal and superdiagonal nonzero elements. Furthermore, as $A_n$ is block triangular, its eigenvalues are $\lambda_n$ and those eigenvalues of $A_{n-1}$. In turn, $\lambda_1,\lambda_2,\ldots,\lambda_n$ are eigenvalues of $A_n$.

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    $\begingroup$ Huh. I think I've always seen "block triangular" to mean the pattern of which blocks are nonzero is triangular, rather than that the individual blocks are triangular. $\endgroup$
    – user14972
    Jun 28, 2016 at 7:18
  • $\begingroup$ @Hurkyl I don't understand your comment. Nowhere in my answer did I say that a matrix is called block triangular when every block is triangular. In fact, for every $n\ge4$, none of the nontrivial blocks of $A_n$ is triangular. $\endgroup$
    – user1551
    Jun 28, 2016 at 13:03
  • $\begingroup$ The depicted $A_4$ is an unfortunate example, because it partitions into four upper triangular $2 \times 2$ blocks, none zero. (and, IMO, that catches the eye sooner than the 3-1 split, though I see now that's what was intended) $\endgroup$
    – user14972
    Jun 28, 2016 at 14:27

matrix example of this type : Let $A$ and $B$ are two Jordan matrices form not diagonalizable, and $M$ the matrix obtained by concatenating $A$ and $^tB$, then $M$ is a matrix with two block in its diagonal and this block one are upper triangular matrix and the other are lower triangular matrix. Example $M =\left( \begin{array}{cccc} \alpha & 1 &0&0\\ 0 & \alpha&0&0\\ 0 & 0&\beta&0\\ 0&0&1&\beta \end{array} \right) $


There is a result due to Fillmore: Theorem 2 of On Similarity and the Diagonal of a Matrix that states:

The nonscalar matrix $A$ is similar to a matrix with main diagonal $\lambda_1,\lambda_2, \ldots, \lambda_n$ if and only if $\lambda_1+\lambda_2+ \cdots+ \lambda_n=\text{tr}(A)$.

This is valid over any field, and basically one can use this theorem to prove that any nonscalar matrix $A$ where the characteristic equation splits and of order $n \geq 3$ is similar to a non-triangular matrix with the eigenvalues of $A$ on its diagonal.

We start out by choosing our diagonal entries to be the eigenvalues of $A$ (I'm assuming here the trace of a matrix is equal to the sum of its eigenvalues - certainly this is true if the matrix has a Jordan Form - in cases where the characteristic polynomial does not split this might fail). Then, we use the construction employed in Fillmore's proof, which makes use of the fact that for any nonscalar matrix we can find a vector $x$ such that $Ax$ and $x$ are linearly independent. This means $x$ and $Ax-\lambda_1x$ are also linearly independent, and if we now complete this set of vectors to a basis, and represent $A$ with respect to this basis we get $$ \begin{bmatrix} \lambda_1 & C \\ E_1 & B \end{bmatrix}, $$ where $C$ is some row vector, $E_1$ is a column vector with 1 in the top entry and zeros elsewhere, and $B$ is a matrix such that $\text{tr}(B)=\text{tr}(A)-\lambda_1$ (similar matrices have the same trace). An inductive argument is then used on $B$, and if $B$ so happens to be scalar there is a simple similarity transform to fix that. I'm not going to complete the full induction argument here to prove Fillmore's theorem, but the point of showing the construction is that the resulting matrix when employing this construction contains the desired values on the diagonal and contains non-zero values below the diagonal which will be due to multiplication of the final change of basis matrix with $E_1$ (at every step). In particular the first column will contain non-zero entries below $\lambda_1$*. Let this matrix be $A'$.

Now suppose $A'$ is lower triangular. The lower right $2 \times 2$ submatrix will be of the form $$\begin{bmatrix} \lambda_{n-1} & 0 \\ 1 & \lambda_n \end{bmatrix}.$$ Let $$P = \begin{bmatrix} I_{n-2} & 0 & 0\\ 0 & 0&1\\0&1&0 \end{bmatrix}, $$ then $P^{-1}A'P$ still has the same diagonal, except that the last two diagonal entries are swapped and it will have an 1 in entry $(n-1,n)$, so it is no longer triangular, which then completes the proof.

It remains to prove this for scalar matrices (that is, of the form $A=kI_n$) if it is true in that case, and perhaps a proof or counter-example in the case $n=2$ - think one can get an example of a matrix which cannot be similar to a non-triangular matrix with eigenvalues on diagonal in this case maybe.

Just to clarify *: Suppose $Q$ is the change of basis matrix such that $Q^{-1}BQ$ has diagonal $\lambda_2,\lambda_3,\ldots,\lambda_n$, then $$\begin{bmatrix} 1 & 0 \\ 0 & Q^{-1}\end{bmatrix}\begin{bmatrix} \lambda_1 & C\\ E_1 & B\end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & Q\end{bmatrix} = \begin{bmatrix} \lambda_1 & CQ \\ \text{Col}_1(Q^{-1}) & Q^{-1}BQ\end{bmatrix}, $$ and $\text{Col}_1(Q^{-1})$ cannot have only zero entries since it is full rank.

  • $\begingroup$ Ok, I think I see... given a particular set of diagonal elements... if the sum of those diagonal elements adds to the trace of A, then there exists a matrix with those diagonal elements that is similar to A. Am I understanding right? $\endgroup$ Jun 28, 2016 at 19:30
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    $\begingroup$ yes that's right - the only "catch" being that $A$ must be nonscalar of course...it's a very useful result, used in proving a few other interesting results (eg sums of nilpotent matrices), and was obtained in 1969 - some time ago. $\endgroup$ Jun 28, 2016 at 19:36

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