TITLE: Finding Eigenvalues of linear transformation
QUESTION [1 upvotes]: Let $T: V\to W $ be a linear transformation, where $B$ and $B'$ are bases of $ V$ and $W$ respectively. 
Now to find eigenvalues of this linear transformation,

we consider the matrix associated to this linear transformation wrt bases $B$ and $B'$. 
then we find the characteristic polynomial using $det(xI-A)= 0$, where $A$ is the matrix representation of linear transformation. 

But now, if we change the bases of $V$ and $W$, then we obtain a different matrix representation of the linear transformation. So, we will get different characteristic polynomial and so different eigenvalues. 
But according to definition of Eigenvalues, $T(v)=xv$, where x is the eigenvalue and hence should not depend on the bases of V and W.
Why is there such a discrepency?

REPLY [3 votes]: After changing bases, you do get a different matrix representation, but the characteristic polynomial will not change. For instance, considering $SAS^{-1}$ (where $S$ is invertible) instead of $A$ gives $\det(xI-SAS^{-1}) = \det (S(xI-A)S^{-1}) = \det(S) \det(xI-A) \det(S^{-1}) = \det(xI-A)$.

Clarification: the above explanation is for the case of changing the bases of $V$ and $W$ in the same way. If they are not the same, then the eigenvalues may not be the same. For instance, suppose $A=I$ and $V=W=\mathbb{R}^2$ with the standard basis. If we let $V$ have the standard basis scaled by $2$ instead, while keeping the original basis of $W$, then the matrix of the transformation is $2I$, whose eigenvalues are different than that of the original matrix $A=I$.
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