The proof part I actually got through okay.
Start by proving that if λ is an eigenvalue of A with eigenvector v, then λ⁻¹ is an eigenvalue of A⁻¹ with the same eigenvector. Then use the fact that the sum of eigenvalues equals the trace, so the sum of reciprocals is the trace of A⁻¹. Finally, state the necessary assumptions: A must be invertible (no zero eigenvalues) and the eigenvalues must be considered over the complex numbers for completeness.
Pro tip: Connect the result to practical data science applications, such as using the trace of the inverse covariance matrix (precision matrix) in Gaussian graphical models or in regularization techniques, to show deeper understanding.
Let (λ, v) be an eigenpair of A. Since A is invertible, λ ≠ 0. Multiply Av = λv by A⁻¹ to get v = λ A⁻¹ v, hence A⁻¹ v = (1/λ) v. Thus 1/λ is an eigenvalue of A⁻¹.
The sum of eigenvalues of A⁻¹ is Σ (1/λ_i). By the trace property, this sum equals the trace of A⁻¹, i.e., tr(A⁻¹).
So the sum is tr(A⁻¹). This is a scalar obtained by summing the diagonal entries of the inverse matrix.
A must be invertible (det(A) ≠ 0), which implies all eigenvalues are nonzero. Also, for the trace property to hold, the eigenvalues must be counted with algebraic multiplicity, and the field should be algebraically closed (e.g., complex numbers) to ensure all eigenvalues exist.
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