Introduction to Quantum Mechanics · Problem A.21

Afterword — David J. Griffiths

Physics — Quantum Mechanics · eigenvectors-eigenvalues

Problem

It's obvious that the trace of a diagonal matrix is the sum of its eigenvalues, and its determinant is their product (just look at Equation A.79). It follows (from Equations A.65 and A.68) that the same holds for any diagonalizable matrix. Prove that in fact

\det(\mathsf{T}) = \lambda_1 \lambda_2 \cdots \lambda_n, \quad \text{Tr}(\mathsf{T}) = \lambda_1 + \lambda_2 + \cdots + \lambda_n,

for any matrix. (The

\lambda

's are the

n

solutions to the characteristic equation---in the case of multiple roots, there may be fewer linearly-independent eigenvectors than there are solutions, but we still count each

\lambda

as many times as it occurs.) Hint: Write the characteristic equation in the form

(\lambda_1 - \lambda)(\lambda_2 - \lambda) \cdots (\lambda_n - \lambda) = 0,

and use the result of Problem A.20.

Related theory

Eigenvalues And Eigenvectors

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