Question

Question #68ef5

Answer 1

See below.

Explanation:

If `lambda` is an eigenvalue of `B` then

`Bu = lambda u` but `B= V^-1 A V` and then

`V^-1 A V x = lambda u = lambda V^-1 Vu` or

`A(Vu)=lambda(Vu)` then `w = Vu` is an associated eigenvector to `A`

Now the eigenvalues are the values which obey

`"det"(B-lambdaI) = 0` Taking instead

`"det"(V^-1AV-lambda I) = "det"(V^-1(A-lambda I) V)`

but `det(U V) = "det"(U)"det"(V)` so `"det"(V^-1)"det"(V) = 1`

and finally

`"det"(V^-1AV-lambda I) ="det"(A-lambda I)=0` and then

`lambda` is eigenvalue for `B` and `A`