# If a matrix is upper-triangular, does its diagonal contain all the eigenvalues? If so, why?

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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Can you compute the characteristic polynomial? –  Qiaochu Yuan Oct 4 '11 at 3:28
Well, in any case, you can try to write down some eigenvectors. Try some examples. –  Qiaochu Yuan Oct 4 '11 at 3:32
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 4 '11 at 3:37
An upper triangular $n \times n$ matrix with no $0$ on the main diagonal "obviously" has row rank equal to $n$. –  André Nicolas Oct 4 '11 at 3:57
Enough with the comments. Someone, please turn them into answers. –  Gerry Myerson Oct 4 '11 at 5:13

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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Do the eigenvalues show up on the diagonal with correct multiplicity? –  user7530 Oct 4 '11 at 7:21
@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 4 '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. –  fpqc Oct 4 '11 at 13:17