We continue with
example.
There exists the two
eigenvectors
and
for the different
eigenvalues
and
,
so that the mapping is
diagonalizable,
due to
fact.
With respect to the
basis
, consisting of these eigenvectors, the linear mapping is described by the diagonal matrix
-
The
transformation matrix,
from the basis
to the standard basis
, consisting of
and
,
is simply
-
![{\displaystyle {}M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2b490dd41fb5f7280c4a3b4f7fae29fa0e8948d)
The
inverse matrix
is
-
![{\displaystyle {}{\frac {1}{2{\sqrt {5}}}}{\begin{pmatrix}1&{\sqrt {5}}\\-1&{\sqrt {5}}\end{pmatrix}}={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5a6d296b983d6a71f9baefb93d90f7d1b345e70)
Because of
fact,
we have the relation
![{\displaystyle {}{\begin{aligned}{\begin{pmatrix}{\sqrt {5}}&0\\0&-{\sqrt {5}}\end{pmatrix}}&={\begin{pmatrix}{\frac {1}{2}}&{\frac {\sqrt {5}}{2}}\\{\frac {1}{2}}&{\frac {-{\sqrt {5}}}{2}}\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80bac6e50d96f55fded6a4ee7aaad506733b2832)