Proof of Eigenvector Property with Simple Linear Algebra

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
Hockeystar
Messages
61
Reaction score
0

Homework Statement


Let x be an eigenvector of A with eigenvalue [tex]\lambda[/tex] and suppose x is also an eigenvector of B, corresponding to the eigenvalue [tex]\lambda[/tex]2. Let C = A + B. Show that x is an eigenvector of C. What is the corresponding eigenvalue?

to the eigenvalue 2


Homework Equations





The Attempt at a Solution



{[tex]\lambda[/tex]I - A} = {[tex]\lambda[/tex]2I - B}

C = 2{[tex]\lambda[/tex]I - A}
C is just a linear combination of the first eigenvector so it's got the same eigenvector.

Is this enough to complete the proof?

Is the corresponding eigenvalue just twice the original eigenvalue?
 
Physics news on Phys.org
Start like this: you know [tex]Ax = \lambda x[/tex] and [tex]Bx = \lambda_{2} x[/tex] from the definition of an eigenvector

Thus [tex]Cx = (A + B)x = ...[/tex] and go from there. I think it will be straightforward.