× Close
Join the Ques10 Community
Ques10 is a community of thousands of students, teachers, and academic experts, just like you.
Join them; it only takes a minute
Sign up
Question: If lambda is eigen value of A then prove that (lambda)^n is eigen value of A^n
0

Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: Low

m4m(64) • 37 views
ADD COMMENTlink
modified 21 days ago  • written 21 days ago by gravatar for manasahegde234 manasahegde23410
0

Let $\lambda$ be an eigen value of $A^n$.

Then there exists $X\neq0$ such that

$(A-\lambda I)X=0$

$AX-\lambda X=0$

$AX= \lambda X$ (1)

Post multiplying by A we get,

$A^2X=\lambda A X$

$A^2X=\lambda \lambda X$ from (1)

$A^2X=\lambda^2 X$

$\lambda^2$ is eigen value of $A^2$ and eigen vector corresponding to $\lambda^2$ is X.

Again pre multiplying by A,

$A^3X=\lambda^2 A X$

$A^3X=\lambda^2 \lambda X$

$A^3X=\lambda^3 X$

$\lambda^3$ is eigen value of $A^3$ and X is corresponding eigen vector. If we proceed in same manner then, we get,

$A^nX=\lambda^nX$

$\lambda^n$ is eigen value of $A^n$ and X is corresponding eigen vector.

ADD COMMENTlink
written 21 days ago by gravatar for manasahegde234 manasahegde23410
Please log in to add an answer.


Use of this site constitutes acceptance of our User Agreement and Privacy Policy.