Matrix/Eigenvalues/0510/R/Change of basis and diagonalization/Example

We continue with example. There exists the two eigenvectors ${}{\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}$ and ${}{\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}$ for the different eigenvalues ${}{\sqrt {5}}$ and ${}-{\sqrt {5}}$, so that the mapping is diagonalizable, due to fact. With respect to the basis ${}{\mathfrak {u}}$, consisting of these eigenvectors, the linear mapping is described by the diagonal matrix

${\begin{pmatrix}{\sqrt {5}}&0\\0&-{\sqrt {5}}\end{pmatrix}}.$

The transformation matrix, from the basis ${}{\mathfrak {u}}$ to the standard basis ${}{\mathfrak {v}}$, consisting of ${}e_{1}$ and ${}e_{2}$, is simply

${}M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\,.$

The inverse matrix is

${}{\frac {1}{2{\sqrt {5}}}}{\begin{pmatrix}1&{\sqrt {5}}\\-1&{\sqrt {5}}\end{pmatrix}}={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}\,.$

Because of fact, we have the relation

${}{\begin{aligned}{\begin{pmatrix}{\sqrt {5}}&0\\0&-{\sqrt {5}}\end{pmatrix}}&={\begin{pmatrix}{\frac {1}{2}}&{\frac {\sqrt {5}}{2}}\\{\frac {1}{2}}&{\frac {-{\sqrt {5}}}{2}}\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}.\end{aligned}}$