# Linear mapping/Diagonalizable/Eigentheory/Section

The restriction of a linear mapping to an eigenspace is the homothety with the corresponding eigenvalue, so this is a quite simple linear mapping. If there are many eigenvalues with high-dimensional eigenspaces, then usually the linear mapping is simple in some sense. An extreme case are the so-called diagonalizable mappings.
For a diagonal matrix
    ${\begin{pmatrix}d_{1}&0&\cdots &\cdots &0\\0&d_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1}&0\\0&\cdots &\cdots &0&d_{n}\end{pmatrix}},$
the characteristic polynomial is just
    $(X-d_{1})(X-d_{2})\cdots (X-d_{n}).$
If the number $d$ occurs $k$-times as a diagonal entry, then also the linear factor $X-d$ occurs with exponent $k$ inside the factorization of the characteristic polynomial. This is also true when we just have an upper triangular matrix. But in the case of a diagonal matrix, we can also read of immediately the eigenspaces, see example. The eigenspace for $d$ consists of all linear combinations of the standard vectors $e_{i}$, for which $d_{i}$ equals $d$. In particular, the dimension of the eigenspace equals the number how often $d$ occurs as a diagonal element. Thus, for a diagonal matrix, the algebraic and the geometric multiplicities coincide.

## Definition

Let $K$ denote a field, let $V$ denote a vector space, and let
    $\varphi \colon V\longrightarrow V$
denote a linear mapping. Then $\varphi$ is called diagonalizable, if $V$ has a basis consisting of eigenvectors
$\varphi$

## Theorem

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
denote a
* $\varphi$ is diagonalizable.
* There exists a basis ${\mathfrak {v}}$ of $V$ such that the describing matrix $M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )$ is a diagonal matrix.
* For every describing matrix $M=M_{\mathfrak {w}}^{\mathfrak {w}}(\varphi )$ with respect to a basis ${\mathfrak {w}}$, there exists an invertible matrix $B$ such that is a diagonal matrix.
    $BMB^{-1}$

### Proof

The equivalence between (1) and (2) follows from the definition, from example, and the correspondence between linear mappings and matrices. The equivalence between (2) and (3) follows from fact.
$\Box$

## Corollary

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
denote a linear mapping. Suppose that there exists $n$ different eigenvalues. Then $\varphi$ is

### Proof

Because of fact, there exist $n$ linearly independent eigenvectors. These form, due to fact, a basis.
$\Box$

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