# Answer to Question #24838 in Linear Algebra for Jacob Milne

Question #24838

Prove that the eigenvalues of an upper triangular matrix T are the entries of its main diagonal, so that the trace of T is the sum of its eigenvalues.

Expert's answer

Eigenvalues are roots ofpolynomial det(T-kI)=0, where I is an identity matrix.

Since T is upper triangular matrix, then we deal with determinant of upper triangular

matrix again, and its determinant is product of diagonal entries.

So, k is one of the roots of (a_11-k)(a_22-k)...(a_nn-k) = 0.

Then all k varies over set {a_11,...,a_nn}, and sum of all k's is

a_11+....+a_nn=tr(T).

Since T is upper triangular matrix, then we deal with determinant of upper triangular

matrix again, and its determinant is product of diagonal entries.

So, k is one of the roots of (a_11-k)(a_22-k)...(a_nn-k) = 0.

Then all k varies over set {a_11,...,a_nn}, and sum of all k's is

a_11+....+a_nn=tr(T).

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