Exercise 151Linear algebra › Linear mapsL1Added 3 Nov 2026

Invariants under similarity

#similarity#trace#determinant#rank#characteristic polynomial

Statement

Two $n\times n$ matrices $A$ and $B$ are similar if $B=P^{-1}AP$ for some invertible $P$ . Show that similar matrices have the same rank, trace, determinant, and characteristic polynomial.

Hint

For the rank, use the invariance under multiplication by invertible matrices. For the trace, use $\operatorname{tr}(XY)=\operatorname{tr}(YX)$ . For the determinant and characteristic polynomial, use multiplicativity of $\det$ .

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Solution

Rank. By the invariance of the rank under left and right multiplication by invertible matrices, $\operatorname{rank}(P^{-1}AP)=\operatorname{rank}(A)$ .

Trace. For square matrices, $\operatorname{tr}(XY)=\sum_{i,j}x_{ij}y_{ji}=\operatorname{tr}(YX)$ . With $X=P^{-1}$ and $Y=AP$ :
$\operatorname{tr}(B)=\operatorname{tr}\big(P^{-1}(AP)\big)=\operatorname{tr}\big((AP)P^{-1}\big)=\operatorname{tr}(A).$

Determinant. $\det B=\det(P^{-1})\det(A)\det(P)=\det(A)$ , since $\det(P^{-1})\det(P)=\det(I)=1$ .

Characteristic polynomial. Since $P^{-1}(\lambda I)P=\lambda I$ ,
$\det(B-\lambda I)=\det\big(P^{-1}(A-\lambda I)P\big)=\det(A-\lambda I).$

Conclusion. Rank, trace, determinant, and characteristic polynomial are invariants of the similarity class. In particular they are attributes of an endomorphism, independent of the chosen basis.

Invariants under similarity

Statement

Solution

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