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

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.

$\Box$

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.

$\Box$

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

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