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

for ${}\varphi$.

Theorem

Let ${}K$ denote a field, and let ${}V$ denote a finite-dimensional vector space. Let

$\varphi \colon V\longrightarrow V$

denote a

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is diagonalizable.
2. There exists a basis ${}{\mathfrak {v}}$ of ${}V$ such that the describing matrix ${}M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )$ is a diagonal matrix.
3. 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

   $BMB^{-1}$

   is a diagonal matrix.

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

diagonalizable.

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

eigenspaces.

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 consist 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.