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If a matrix is upper-triangular, does its diagonal contain its eigenvalues? If yes, how can this be proven? My textbook and teacher just jumped over this statement (we are working over complex numbers, does the answer change if it's over reals?) and I was wondering if someone could provide a proof.

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  • $\begingroup$ Can you compute the characteristic polynomial? $\endgroup$ Oct 4, 2011 at 3:28
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    $\begingroup$ Well, in any case, you can try to write down some eigenvectors. Try some examples. $\endgroup$ Oct 4, 2011 at 3:32
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    $\begingroup$ One of the eigenvectors should be obvious. To construct the others I advise induction. Alternately, you can try to characterize when an upper-triangular matrix is invertible and use the fact that $\lambda$ is an eigenvalue iff $A - \lambda I$ is not invertible. $\endgroup$ Oct 4, 2011 at 3:37
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    $\begingroup$ An upper triangular $n \times n$ matrix with no $0$ on the main diagonal "obviously" has row rank equal to $n$. $\endgroup$ Oct 4, 2011 at 3:57
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    $\begingroup$ Enough with the comments. Someone, please turn them into answers. $\endgroup$ Oct 4, 2011 at 5:13

2 Answers 2

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The following steps lead to a solution:

1)If a matrix $A$ is upper triangular, prove that $A$ is invertible iff none of the elements on the diagonal equals zero.

Suppose you have a matrix $A$ that is upper triangular. Consider $A - \lambda I$. Then for $A$ to have a non-zero eigenvector, the kernel of $A - \lambda I$ must not be trivial, in other words $A - \lambda I$ must not be invertible.

2) Hence prove that the eigenvalues of a matrix that is upper triangular all lie on its diagonal.

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  • $\begingroup$ Do the eigenvalues show up on the diagonal with correct multiplicity? $\endgroup$
    – user7530
    Oct 4, 2011 at 7:21
  • $\begingroup$ @user7530: Yes, the dimension of he generalized eigenspace is equal to the number of times that $\lambda$ shows up in the diagonal. $\endgroup$ Oct 4, 2011 at 13:15
  • $\begingroup$ @user7530 I refer you to Axler's Linear Algebra Done Right, but as Arturo said the number of times that $\lambda$ appears on the diagonal is dim null $(T - \lambda I)^{dim V}$. You can show that this is equivalent to the multiplicity of the eigenvalue obtained from the characteristic polynomial. However I am not good in a position to talk of determinants as I am not so familiar with them. $\endgroup$
    – user38268
    Oct 4, 2011 at 13:17
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Hint 1: The determinant of a triangular matrix is the product of its diagonal elements.

Hint 2: To prove Hint 1 develop with respect to the first row resp. column and use induction.

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