I’ve avoided proofs unless absolutely necessary, but the relation between the same eigenvector expressed in two different bases, is important.

Given that A_{S} is the linear transformation matrix in standard basis S, and A_{B} is its counterpart in basis B, we can write the relation between them as:

where C is the similarity transformation. We’ve seen this relation already; check here if you’ve forgotten about it.

Now, the definition of an eigenvector gives us:

x is the eigenvector in standard basis S. Now substituting A_{S} with the identity we obtained previously into the eigenvector identity, we get:

If we take x_{B} = C^{-1}.x_{S}, we get:

This tells us two important things. One, that the eigenvalue is still . Second, the eigenvector expressions in the two bases are related as:

This gives us the relation we need, and justifies the logic in this post.