Exercise 941Linear algebra › Eigenvalues and diagonalisationL1Added 3 Nov 2026

Sum of eigenvalues equals trace

#eigenvalues#trace#characteristic polynomial

Statement

Let $A$ be an $n\times n$ matrix with eigenvalues $\lambda_1,\ldots,\lambda_n$ (counted with multiplicity). Show that
$\sum_i\lambda_i = \textrm{tr}(A), \qquad \prod_i\lambda_i = \det(A).$

Hint

Expand $\chi_A(\lambda)=\prod(\lambda-\lambda_i)$ and compare coefficients with $\chi_A(\lambda)=\lambda^n-\textrm{tr}(A)\lambda^{n-1}+\cdots+(-1)^n\det A$ .

Show solutionHide solution

Solution

Expansion of $\prod(\lambda-\lambda_i)$ . The coefficient of $\lambda^{n-1}$ is $-\sum_i\lambda_i$ . The constant term is $(-1)^n\prod_i\lambda_i$ .

Identification. Comparing with
$\chi_A(\lambda)=\lambda^n-\textrm{tr}(A)\lambda^{n-1}+\cdots+(-1)^n\det A,$
we get
$\sum_i\lambda_i=\textrm{tr}(A), \qquad \prod_i\lambda_i=\det A.$

Conclusion. Trace and determinant are the elementary symmetric functions of degree $1$ and $n$ of the eigenvalues.

Sum of eigenvalues equals trace

Statement

Solution

More problems in Eigenvalues and diagonalisation