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Let $V$ be an $F$-vector space with $\dim(V) = n$ (which is finite) and let $T$ be an element of the homomorphism from $V$ to $V$. Prove that the diagonals of the upper-triangular matrix of $T$ are the eigenvalues of $T$ without the use of determinants.

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Can you assume that an upper triangular matrix with a zero on the diagonal is not invertible? – wj32 Nov 15 '12 at 5:41
What happened to this question ? – beauby Nov 15 '12 at 6:11
@Philip Please do not vandalize questions. – Bill Dubuque Nov 15 '12 at 6:13

Recall that the eigenvalues are precisely the (non-trivial) vectors in the nullspace of $T-\lambda I$.

You know the nullspace is non-trivial if and only if $T- \lambda I$ is non-invertible. Furthermore, you know that $T-\lambda I$ is upper-triangular and that an upper-triangular matrix is non-invertible if and only if there is a $0$ on the main diagonal.

This should be enough to carry you through.

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