UPSC 2018 Maths Optional Paper 1 Q2a — Step-by-Step Solution

12 marks · Section A

Congruence and similarity of matrices · Linear Algebra · asked 2× in 13 yrs · Read the full method →

Question

Show that if $A$ and $B$ are similar $n\times n$ matrices, then they have the same eigenvalues.

Technique

Show characteristic polynomials coincide using $B-\lambda I=P^{-1}(A-\lambda I)P$ and $\det(P^{-1})\det P=1$ .

Solution

Step 1 — Definition of similarity

$A$ and $B$ are similar if there is an invertible $n\times n$ matrix $P$ with

B=P^{-1}AP.

Step 2 — The characteristic polynomials are equal

Compute the characteristic polynomial of $B$ :

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

using $\lambda I=P^{-1}(\lambda I)P$ . By multiplicativity of the determinant,

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

Hence

\det(B-\lambda I)=\det(A-\lambda I)\quad\text{for all }\lambda.

Step 3 — Same eigenvalues

The eigenvalues are exactly the roots of the characteristic polynomial. Since $A$ and $B$ have identical characteristic polynomials, they have the same eigenvalues, with the same algebraic multiplicities.

Answer

\boxed{\;B=P^{-1}AP\ \Longrightarrow\ \det(B-\lambda I)=\det(A-\lambda I)\ \Longrightarrow\ A,B\text{ share all eigenvalues (with multiplicity).}\;}