# Endomorphism/Diagonalizability/Introduction/Section
## 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.
$\Box$
If $\varphi$ is diagonalizable and if the eigenvalues together with their geometric multiplicities are known, then we can simply write down a corresponding diagonal matrix: just take the diagonal matrix such that, in the diagonal, the eigenvalues occur as often as the geometric multiplicities. In particular, the corresponding diagonal matrix of a diagonalizable mapping is, up to the ordering of the diagonal entries, uniquely determined.

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

## 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.
$\Box$

## Lemma

Let $K$ denote a field, and let $V$ denote a $K$-vector space of finite dimension. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping. Then $\varphi$ is diagonalizable if and only if $V$ is the direct sum of the

### Proof

If $\varphi$ is diagonalizable, then there exists a basis $v_{1},\ldots ,v_{n}$ of $V$ consisting of eigenvectors. Hence,
    $\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\langle v_{i},\,{\text{the eigenvalue of }}v_{i}{\text{ is }}\lambda \rangle \,.$
Therefore,
    $V=\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{k}}{\left(\varphi \right)}\,.$
That the sum is direct follows from fact. If, the other way round,
    $V=\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{k}}{\left(\varphi \right)}\,$
holds, then we can choose in every eigenspace a basis. These bases consists of eigenvectors and yield together a basis of $V$.
$\Box$

## Example

We consider $2\times 2$-shearing matrices
    ${\begin{pmatrix}1&a\\0&1\end{pmatrix}}$
with $a\in K$. The condition for some $\lambda \in K$ to be an eigenvalue means
    ${\begin{pmatrix}1&a\\0&1\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}\,.$
This yields the equations
    $x+ay=\lambda x{\text{ and }}y=\lambda y.$
For $\lambda \neq 1$, we get $y=0$ and hence also $x=0$, that is, only $1$ can be an eigenvalue. In this case, the second equation is fulfilled, and the first equation becomes
    $x+ay=x{\text{ and }}ay=0.$
For $a\neq 0$, we get $y=0$ and thus ${\begin{pmatrix}x\\0\end{pmatrix}}$ is the eigenspace for the eigenvalue $1$, and ${\begin{pmatrix}1\\0\end{pmatrix}}$ is an eigenvector which spans this eigenspace. For $a=0$, we have the identity matrix, and the eigenspace for the eigenvalue $1$ is the total plane. For $a\neq 0$, there is a one-dimensional eigenspace, and the mapping is not diagonalizable.
The product of two diagonal matrices is again a diagonal matrix. The following example shows that the product of two diagonalizable matrices is not necessarily diagonalizable.

## Example

Let $G_{1}$ and $G_{2}$ denote two lines in $\mathbb {R} ^{2}$ through the origin, and let $\varphi _{1}$ and $\varphi _{2}$ denote the reflections at these axes. A reflection at an axis is always diagonalizable, the axis and the line orthogonal to the axis are eigenlines (with eigenvalues $1$ and $-1$). The composition
    $\psi =\varphi _{2}\circ \varphi _{1}\,$
of the reflections is a plane rotation, the angle of rotation being twice the angle between the two lines. However, a rotation is only diagonalizable if the angle of rotation is $0$ or $180$ degree. If the angle between the axes is different from $0,90$ degree, then $\psi$ does not have any eigenvector.