Exercise 1061Linear algebra › Matrices and linear systemsL1Added 3 Nov 2026

Similarity invariants

#similarity#characteristic polynomial#trace#determinant

Statement

Show that similar matrices have the same characteristic polynomial, trace, and determinant.

Hint

If $B = P^{-1}AP$ , compute $\det(B - \lambda I)$ by factoring $P$ and $P^{-1}$ .

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Solution

Characteristic polynomial.
$\det(B - \lambda I) = \det(P^{-1}AP - \lambda I) = \det(P^{-1}(A - \lambda I)P) = \det(P^{-1})\det(A - \lambda I)\det(P) = \det(A - \lambda I).$
So the characteristic polynomials coincide.

Trace and determinant. The trace is the coefficient of $\lambda^{n-1}$ in the characteristic polynomial (up to sign), and the determinant is the constant term (up to sign). Since the characteristic polynomials coincide, the trace and determinant coincide.

Conclusion. Similar matrices share the characteristic polynomial, the trace, and the determinant.

Similarity invariants

Statement

Solution

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