Question

If $\lambda$ is a eigenvalue of A, then corresponding eigen value of $P^{-n}A{P}^n$ (P is a square matrix with same order as A) is

• A. ${\lambda }^n$
• B. 1
• C. ${\lambda }^{-n}$
• D. $\lambda$

Answer 1

Answer: (D)

$\lambda$

Let $P^{-1}AP=B$
$\therefore B-\lambda I=P^{-1}AP-\lambda I$
$=P^{-1}AP-P^{-1}\lambda IP$
$=P^{-1}(A-\lambda I)P$
$\Rightarrow |B-\lambda I|=|P^{-1}||A-\lambda I||P|$
$=|A-\lambda I||P^{-1}||P|$
$=|A-\lambda I||P^{-1}P|$
$=|A-\lambda I||I|=|A-\lambda I|$
Thus, the two matrices A and B have the same characteristic determinants and hence the same characteristic equations and the same characteristic
roots. The same thing may also be seen in another way. Now,
$AX=\lambda X$
$\Rightarrow P^{-1}AX=\lambda P^{-1}X$
$\Rightarrow \left(P^{-1}AP\right)\left(P^{-1}X\right)=\lambda \left(P^{-1}X\right)$