How to find eigenvalues and basis of the inverse of linear transformation? Let $T:\ \Bbb{R}^3\rightarrow \Bbb{R}^3$ be a linear operator that has eigenvalues $1, 2, 3$ with associated eigenvectors $(2,1,3),(1,4,0),(1,0,0)$ respectively. How to find eigenvalues of $T^{-1}$ and a basis for each eigenspace?
Since $T(x)=\lambda x$,   we have $T^{-1}(x)=\lambda^{-1} x=\frac{1}{\lambda}x$, so I think the eigenvalues are $1,\frac12, \frac13$. In terms of basis, I have no idea what to write.
 A: If $x$ is an eigenvector of $T$ with eigenvalue $\lambda$, then from your own work above,
$$Tx = \lambda x \ \Longrightarrow \ T^{-1}x = \frac{1}{\lambda}x.$$
Therefore $T^{-1}$ has eigenvalue $\frac{1}{\lambda}$ corresponding to the same eigenvector $x$.
So each eigenvalue $\frac{1}{\lambda}$ has the same eigenspace basis as $\lambda$ has for the original $T$.
For example, the basis corresponding to eigenvalue $\frac12$ is the vector $(1,4,0)$, and similarly for the other two.
A: If your trouble is to justify the result $T^{-1}x=\lambda^{-1}x$ than note that, from:
$$T(x)=\lambda x$$ if $T$ is invertible, multiplying both side by $T^{-1}$ we have
$$
T^{-1}T(x)=T^{-1}(\lambda x)
$$
and, by definition of the inverse and linearity:
$$
x=\lambda T^{-1}( x)
$$
if $\lambda \ne 0$ it has an inverse $\lambda^{-1}$ because it is an element of a field, so, multiplying both sides for $\lambda^{-1}$ ve find:
$$
\lambda^{-1} x= T^{-1}( x)
$$
and this means exactly that $x$ ( the same  vector) is an aigenvector of $T^{-1}$ for the eigenvalue $\lambda^{-1}$.
