UPSC 2017 Maths Optional Paper 1 Q1b — Step-by-Step Solution

10 marks · Section A

Congruence and similarity of matrices · Linear Algebra · asked 3× in 14 yrs · Read the full method →

Question

Show that similar matrices have the same characteristic polynomial.

Factor $B-\lambda I=P^{-1}(A-\lambda I)P$ and use multiplicativity of $\det$ ; the conjugating factors cancel.

Let $A,B$ be $n\times n$ matrices over a field, and suppose they are similar: there is a non-singular matrix $P$ with

B=P^{-1}AP.

The characteristic polynomial of a matrix $M$ is $\chi_M(\lambda)=\det(M-\lambda I)$ . We show $\chi_B(\lambda)=\chi_A(\lambda)$ .

Since $I=P^{-1}IP$ ,

B-\lambda I=P^{-1}AP-\lambda P^{-1}IP=P^{-1}(A-\lambda I)P.

Using $\det(XY)=\det X\det Y$ and $\det(P^{-1})=(\det P)^{-1}$ :

\chi_B(\lambda)=\det(B-\lambda I)=\det\!\big(P^{-1}(A-\lambda I)P\big)=\det(P^{-1})\,\det(A-\lambda I)\,\det(P).

The scalar factors cancel:

\det(P^{-1})\det(P)=\det(P^{-1}P)=\det I=1,

\chi_B(\lambda)=\det(A-\lambda I)=\chi_A(\lambda).

\boxed{\;B=P^{-1}AP\ \Longrightarrow\ \det(B-\lambda I)=\det(A-\lambda I)\ \text{for all }\lambda.\;}