# Finding eigenvector and eigenvalue for a hypothetical diagonalized matrix?

If there are some $$A, B,$$ and $$S$$ (all of which are $$n \times n$$) such that $$A = S(B)S^{-1}$$. Suppose $$A$$ has eigenvector $$v$$ and eigenvalue $$\lambda$$. How can we solve for an eigenvector of $$B$$ and find a corresponding eigenvalue?

### My Attempt

If $$B = S^{-1}(A)S$$, then the column vectors of $$S$$ give an eigenbasis for $$A$$. By that logic, the column vectors of $$S$$ in our case give an eigenbasis for $$B$$ and the diagonal entries of $$A$$ will be the eigenvalues. I don't really know where to go from here. Thank you!

• What makes you think that either $A$ or $B$ Is diagonal?
– amd
Apr 14 '20 at 2:33

My thought process so far: If $$B = S^{-1}A S$$, then the column vectors of S give an eigenbasis for A.
This only holds when $$A$$ is a diagonal matrix. Although this might look like "diagonalization", the fact that $$B = S^{-1}AS$$ doesn't necessarily have anything to do with diagonal matrices. If you want a way to think about the relationship between $$B$$ and $$A$$, then we can note that $$B$$ represents the transformation $$x \mapsto Ax$$ relative to a non-standard basis.
Setting all that aside, here is a very direct approach for your problem. We know that $$A$$ "has eigenvector $$v$$ and eigenvector $$\lambda$$". In other words, $$v$$ is a non-zero vector and $$A v = \lambda v$$. In order to try to say something about $$B$$, we can plug in: $$(S^{-1}BS)v = \lambda v$$ That doesn't look very helpful yet, but we can rewrite the equation by multiplying both sides by $$S$$ (from the left) to get $$BSv = \lambda Sv.$$ Notice that the term "$$Sv$$" appears on both sides. If we group it off, the equation reads $$B(Sv) = \lambda (Sv)$$. Since $$Sv$$ is non-zero (why?), that means that the vector $$Sv$$ is an eigenvector of $$B$$!