Eigenvalues of two matrices associated to the same endomorophism are different? I've a question on the eigenvalues of an endomorphism.
Suppose to have the endomorphism $f$ such that the matrix associated to it with respect to a basis $\mathscr{B}=(\vec{v_1},\vec{v_2},\vec{v_3})$ in the domain and in the codomain is 
$M^{\mathscr{B},\mathscr{B}}(f)=\left[ {\begin{array} \\3&0&0\\0&4&0\\0&0&4 \end{array}} \right]$
Then I know that $\vec{v_1}$ is an eigenvector with eigenvale $3$, $\vec{v_2}$ is an eigenvector with eigenvale $4$ and$\vec{v_3}$ is an eigenvector with eigenvale $4$.
But then if I make a change of basis only in the codomain to a new basis $\mathscr{C}=(3\vec{v_1},4\vec{v_2},4\vec{v_3})$ then 
$M^{\mathscr{B},\mathscr{C}}(f)=\left[ {\begin{array} \\1&0&0\\0&1&0\\0&0&1 \end{array}} \right]$
And that's a matrix that is still associated to the endomorphism $f$ since it is obtained as 
$M^{\mathscr{B},\mathscr{C}}(f)=Q^{-1} M^{\mathscr{B},\mathscr{B}}(f) P$
Where $Q=\left[ {\begin{array} \\3&0&0\\0&4&0\\0&0&4 \end{array}} \right]$ is the change of basis matrix from $\mathscr{B}$ to $\mathscr{C}$ and $P$ is the identity matrix (basis was not changed in the domain).
At this point I can't understand how can $M^{\mathscr{B},\mathscr{B}}(f)$ and $
M^{\mathscr{B},\mathscr{C}}(f)$ have different eigenvalues. 
They are not similar but they are still associated to the same linear function (an endomorphism of course). And since in theory eigenvalues are related to an endomorphism, not to a matrix, a change of basis should not influence them.
Am I missing something?
Thanks a lot for your help
 A: The theory of eigenvalues and eigenvectors relates to coordinate matrices for endomorphisms where the source and target bases are the same. 
Switching from one basis to another (both the target and source bases being the same) yields similar matrices (one coordinate matrix is the conjugate of the other).
If you mess with both bases, you are looking at a matrix equivalence. A matrix equivalence does not preserve eigenvalues for coordinate matrices. Although it does preserve other things - like rank.
Recall that for a linear endomorphism: $T:V \to V$ an eigenvector is a non-zero vector ${\bf v} \in V$ such that there is a scalar $\lambda$ such that $T({\bf v}) = \lambda{\bf v}$. 
Pick a (single) basis $\beta$ and get $[T({\bf v})]_\beta = [\lambda {\bf v}]_\beta$ so that $[T]_\beta^\beta [{\bf v}]_\beta = \lambda [{\bf v}]_\beta$.
This $\lambda$ is an eigenvalue for the coordinate matrix $[T]_\beta^\beta$ ($\beta$ being used as a basis for both domain and codomain). 
Eigenvalues for $T$ and eigenvalues for coordinate matrices $[T]_\beta^\beta$ are the same.
If you use different bases, say $\alpha$ and $\beta$ to get a coordinate matrix: $[T]_\alpha^\beta$ and find some eigenvector for this matrix, say ${\bf w}$. Then we get: $[T]_\alpha^\beta {\bf w} = \lambda {\bf w}$. Now ${\bf w} = [{\bf v}]_\alpha$ for some ${\bf v} \in V$ and ${\bf w}=[{\bf z}]_\beta$ for some ${\bf z} \in V$. Our equation becomes: $[T]_\alpha^\beta [{\bf v}]_\alpha = \lambda [{\bf z}]_\beta$ which out of coordinates says: $T({\bf v})=\lambda {\bf z}$. Unless coincidentally ${\bf v}={\bf z}$, this is no longer an eigenvector/eigenvalue for $T$! That's why matrix equivalence/using different bases is not what you want when looking at eigenvectors/eigenvalues.
