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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.

upptrang
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  • Can you compute the characteristic polynomial? – Qiaochu Yuan Oct 04 '11 at 03:28
  • @QiaochuYuan: No we haven't gotten there yet. – upptrang Oct 04 '11 at 03:31
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    Well, in any case, you can try to write down some eigenvectors. Try some examples. – Qiaochu Yuan Oct 04 '11 at 03:32
  • @QiaochuYuan: Whenever I construct such a matrix over complex numbers, I can find the eigenvectors corresponding to the diagonal entries, I just can't prove that this will always work. – upptrang Oct 04 '11 at 03:35
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    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. – Qiaochu Yuan Oct 04 '11 at 03:37
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    An upper triangular $n \times n$ matrix with no $0$ on the main diagonal "obviously" has row rank equal to $n$. – André Nicolas Oct 04 '11 at 03:57
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    Enough with the comments. Someone, please turn them into answers. – Gerry Myerson Oct 04 '11 at 05:13
  • a "caveat"-observation to remind that this is for finite matrices only (but likely this exceeds the area for usual homework-problems): this need not be true for infinite matrices. Consider the infinite upper triangular Pascal-matrix $\small P$. Then there is an infinite (square) matrix $\small W$ and an infinite diagonal-matrix $\small E$ such that $\small P \cdot W = W \cdot E$ . The entries of $\small E $ $\small e_{r,r} = \exp(r) $ for $\small r=0\ldots \infty$ is one possible solution, and are all but one different from the diagonal entries of $\small P$. – Gottfried Helms Oct 04 '11 at 06:00

2 Answers2

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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.

  • Do the eigenvalues show up on the diagonal with correct multiplicity? – user7530 Oct 04 '11 at 07:21
  • @user7530: Yes, the dimension of he generalized eigenspace is equal to the number of times that $\lambda$ shows up in the diagonal. – Arturo Magidin Oct 04 '11 at 13:15
  • @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. –  Oct 04 '11 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.

Christian Blatter
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