Eigenvalues and eigenvectors of similar matrices.

Question

Suppose there is a transformation $T$ and let $A$ be a matrix representation of $T$ with chosen basis. If I find out the eigenvalues of matrix $A$, these eigenvalues will be the eigenvalues of the transformation $T$?

Then what about eigenvectors of $T$? As far as I know, similar matrices have same eigenvalues, so any matrix representation of $T$ with different basis has same eigenvalues, but eigenvectors corresponding to eigenvalues are dependent of matrix representation.

Then, what can I say about eigenvectors of $T$ by just looking at the eigenvectors of matrices?

Answer 1

The "new" eigenvectors are the expression of the "old" ones in the new basis. That is if $\{v_1,\ldots,v_n\}$ are the eigenvectors of $A$ (associated to the eigenvalues $\lambda_1,\ldots,\lambda_n$ respectively) and $B= UAU^{-1}$ where $U$ is an invertible matrix, then $\{Uv1,\ldots,Uv_n\}$ are the eigenvectors of $B$ since $$B(Uv_i)=(UAU^{-1})Uv_i = UAv_i = U \lambda_i v_i = \lambda_i (Uv_i) \qquad \forall 1 \leq i\leq n$$

Answer 2

Suppose $A$ is diagonalizable, $A=P D P^{-1}$ where $D$ is diagonal, and $B=C A C^{-1}$, i.e. $B$ and $A$ are similar. Then $B=CP D P^{-1} C^{-1} = (CP) D (CP)^{-1}$. This means that $B$ is diagonalizable with the eigenvector matrix $CP$ and of course the same eigenvalue matrix $D$. These are just the eigenvectors of $A$ represented in the basis given by $C$.