Answer:
Proof:
If λ is an eigenvalue of A then for some x≠0,
[tex]Ax=\lambda x\\\\Adding\,\,Ax\,\,on\,\,both\,\,sides\\\\Ax+Ax=\lambda x+Ax\\\\2Ax=\lambda x+\lambda x\\\\2Ax=2\lambda x[/tex]
Which shows that if λ is an eigenvalue of A, 2λ is then an eigenvalue of 2A.
