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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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    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 $\small P$.2011-10-04

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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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    @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.2011-10-04
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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.