= Eigenvectors under similarity
{title2=$Av=\lambda v\ \Longrightarrow\ B(S^{-1}v)=\lambda S^{-1}v$}
Under $B=S^{-1}AS$, an <eigenvector> in old <coordinates> becomes $S^{-1}v$ in new <coordinates>. Indeed,
$$
B(S^{-1}v)=S^{-1}Av=\lambda S^{-1}v.
$$
Since $S^{-1}$ is an <invertible matrix>, the new <coordinate vector> is nonzero. The inverse conversion $S$ proves that the two <eigenspaces> have the same <dimension>, hence the same <geometric multiplicity>.
For $A=\operatorname{diag}(2,5)$, the <eigenvector> $e_2=(0,1)^T$ has <eigenvalue> $5$. In the oblique <basis> of the opening example, its <coordinate vector> is
$$
S^{-1}e_2=\begin{pmatrix}-1\\1\end{pmatrix},\qquad
B\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}-5\\5\end{pmatrix}.
$$
The <tuple> changes from $(0,1)$ to $(-1,1)$, but reconstruction gives $-b_1+b_2=e_2$: the physical vertical direction is unchanged. Calling the <eigenvectors> unchanged is accurate only for the underlying <vectors>, not for their <coordinate vectors>.
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