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Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.