Question

: Linear Algebra: Let B = (S^-1)AS, and x be an eigenvector of B belonging to an eigenvalue L. Show that Sx is an eigenvector of A belonging to L.

Answer 1

The setup to this question indicates that
\[Bx = Lx\]
where L is a constant and B is the matrix
\[B = (S^{-1}AS)\]
Which means that
\[A = SBS^{-1}\]
It follows that
\[A(Sx) = SBS^{-1}(Sx) = SB(x) = S(Lx) = L(Sx) \]
So Sx is indeed an eigenvector of A with eigenvalue L.

Answer 2

Thank you, that reminded me to use the fact that Bx = Lx, which I forget so easily in the middle of all these hw..

Answer 3

No problem :)