Is it possible to construct a matrix with non-zero off-diagonal entries whose eigenvalues are nonetheless equal to its diagonal entries?

EDIT: @ajotataxe pointed out that this holds for triangular matrices. My follow up is - is the converse true? If the Eigen values are the diagonals, does it have to be triangular?

  • $\begingroup$ Do you mean "non-zero entries off its diagonals"? (Otherwise the question is kind of trivial.) $\endgroup$
    – Kenny Wong
    May 23, 2017 at 17:28
  • $\begingroup$ Yes, non-zero entries off its diagonals. $\endgroup$ May 23, 2017 at 17:29
  • 3
    $\begingroup$ A triangular matrix? $\endgroup$
    – ajotatxe
    May 23, 2017 at 17:42
  • $\begingroup$ Right, triangular is true.. Thanks! Is it only possible for triangular? Let me edit the question. $\endgroup$ May 23, 2017 at 17:49
  • 2
    $\begingroup$ What would be truly surprising is a symmetric real matrix that is not diagonal and whose eigenvalues are on the diagonal. $\endgroup$ May 23, 2017 at 17:55

1 Answer 1


Yes:$$\begin{pmatrix}0&\frac12&\frac12\\1&1&1\\1&-1&-1\end{pmatrix}.$$Its eigenvalues are $0$, $1$ and $-1$.

  • $\begingroup$ What Eigen vectors do you get for 1 and -1? $\endgroup$ May 23, 2017 at 17:53
  • $\begingroup$ Also, can you provide some intuition for how you constructed it? $\endgroup$ May 23, 2017 at 17:55
  • $\begingroup$ Ok, I got the Eigen vectors - [0.3015,0.3015,0] and [0.9,0.3015,-0.707]. How did you construct it though? $\endgroup$ May 23, 2017 at 17:59
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    $\begingroup$ I started with the matrix $\left(\begin{smallmatrix}0&a&b\\c&1&d\\e&f&-1\end{smallmatrix}\right)$. Its characteristic polynomial is $-x^3+a c x+b e x+d f x+x+a c-b e+a d e+b c f$. So, I searched for numbers $a$, $b$, $c$, $d$, $e$, and $f$ such that $ac+be+de=0$ and that $a c - b e + a d e + b c f=0$. In order to do that, I solved the system in order to $a$ and $b$ and then I chose more or less random values for $c$, $d$, $e$, and $f$. $\endgroup$ May 23, 2017 at 18:22

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