Generalized eigenvector

Examples

Here are some examples to illustrate the concept of generalized eigenvectors. Some of the details will be described later.

Example 1

This example is simple but clearly illustrates the point. This type of matrix is used frequently in textbooks. Suppose

${\displaystyle A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right).}$

Then there is only one eigenvalue, ${\displaystyle \lambda =1}$, and its algebraic multiplicity is ${\displaystyle m=2}$.

Notice that this matrix is in Jordan normal form but is not diagonal. Hence, this matrix is not diagonalizable. Since there is one superdiagonal entry, there will be one generalized eigenvector of rank greater than 1 (or one could note that the vector space ${\displaystyle V}$ is of dimension 2, so there can be at most one generalized eigenvector of rank greater than 1). Alternatively, one could compute the dimension of the nullspace of ${\displaystyle A-\lambda I}$ to be ${\displaystyle p=1}$, and thus there are ${\displaystyle m-p=1}$ generalized eigenvectors of rank greater than 1.

The ordinary eigenvector ${\displaystyle {\mathbf{v}}_1=\left(\begin{array}{c}1\\ 0\end{array}\right)}$ is computed as usual (see the eigenvector page for examples). Using this eigenvector, we compute the generalized eigenvector ${\displaystyle {\mathbf{v}}_2}$ by solving

${\displaystyle (A-\lambda I){\mathbf{v}}_2={\mathbf{v}}_1.}$

Writing out the values:

${\displaystyle \left(\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)-1\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right)\left(\begin{array}{c}{v}_{21}\\ v_{22}\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{v}_{21}\\ v_{22}\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right).}$

This simplifies to

${\displaystyle v_{22}=1.}$

The element ${\displaystyle v_{21}}$ has no restrictions. The generalized eigenvector of rank 2 is then ${\displaystyle {\mathbf{v}}_2=\left(\begin{array}{c}a\\ 1\end{array}\right)}$, where a can have any scalar value. The choice of a = 0 is usually the simplest.

Note that

${\displaystyle (A-\lambda I){\mathbf{v}}_2=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}a\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)={\mathbf{v}}_1,}$

so that ${\displaystyle {\mathbf{v}}_2}$ is a generalized eigenvector, because

${\displaystyle (A-\lambda I{)}^2{\mathbf{v}}_2=(A-\lambda I)[(A-\lambda I){\mathbf{v}}_2]=(A-\lambda I){\mathbf{v}}_1=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)=\mathbf{0},}$

so that ${\displaystyle {\mathbf{v}}_1}$ is an ordinary eigenvector, and that ${\displaystyle {\mathbf{v}}_1}$ and ${\displaystyle {\mathbf{v}}_2}$ are linearly independent and hence constitute a basis for the vector space ${\displaystyle V}$.

Example 2

This example is more complex than Example 1. Unfortunately, it is a little difficult to construct an interesting example of low order. The matrix

${\displaystyle A=\left(\begin{array}{ccccc}1& 0& 0& 0& 0\\ 3& 1& 0& 0& 0\\ 6& 3& 2& 0& 0\\ 10& 6& 3& 2& 0\\ 15& 10& 6& 3& 2\end{array}\right)}$

has eigenvalues ${\displaystyle {\lambda }_1=1}$ and ${\displaystyle {\lambda }_2=2}$ with algebraic multiplicities ${\displaystyle {\mu }_1=2}$ and ${\displaystyle {\mu }_2=3}$, but geometric multiplicities ${\displaystyle {\gamma }_1=1}$ and ${\displaystyle {\gamma }_2=1}$.

The generalized eigenspaces of ${\displaystyle A}$ are calculated below. ${\displaystyle {\mathbf{x}}_1}$ is the ordinary eigenvector associated with ${\displaystyle {\lambda }_1}$. ${\displaystyle {\mathbf{x}}_2}$ is a generalized eigenvector associated with ${\displaystyle {\lambda }_1}$. ${\displaystyle {\mathbf{y}}_1}$ is the ordinary eigenvector associated with ${\displaystyle {\lambda }_2}$. ${\displaystyle {\mathbf{y}}_2}$ and ${\displaystyle {\mathbf{y}}_3}$ are generalized eigenvectors associated with ${\displaystyle {\lambda }_2}$.

${\displaystyle (A-1I){\mathbf{x}}_1=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\\ 3& 0& 0& 0& 0\\ 6& 3& 1& 0& 0\\ 10& 6& 3& 1& 0\\ 15& 10& 6& 3& 1\end{array}\right)\left(\begin{array}{c}0\\ 3\\ -9\\ 9\\ -3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right)=\mathbf{0},}$

${\displaystyle (A-1I){\mathbf{x}}_2=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\\ 3& 0& 0& 0& 0\\ 6& 3& 1& 0& 0\\ 10& 6& 3& 1& 0\\ 15& 10& 6& 3& 1\end{array}\right)\left(\begin{array}{c}1\\ -15\\ 30\\ -1\\ -45\end{array}\right)=\left(\begin{array}{c}0\\ 3\\ -9\\ 9\\ -3\end{array}\right)={\mathbf{x}}_1,}$

${\displaystyle (A-2I){\mathbf{y}}_1=\left(\begin{array}{ccccc}-1& 0& 0& 0& 0\\ 3& -1& 0& 0& 0\\ 6& 3& 0& 0& 0\\ 10& 6& 3& 0& 0\\ 15& 10& 6& 3& 0\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 9\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right)=\mathbf{0},}$

${\displaystyle (A-2I){\mathbf{y}}_2=\left(\begin{array}{ccccc}-1& 0& 0& 0& 0\\ 3& -1& 0& 0& 0\\ 6& 3& 0& 0& 0\\ 10& 6& 3& 0& 0\\ 15& 10& 6& 3& 0\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 0\\ 3\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 9\end{array}\right)={\mathbf{y}}_1,}$

${\displaystyle (A-2I){\mathbf{y}}_3=\left(\begin{array}{ccccc}-1& 0& 0& 0& 0\\ 3& -1& 0& 0& 0\\ 6& 3& 0& 0& 0\\ 10& 6& 3& 0& 0\\ 15& 10& 6& 3& 0\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\\ -2\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 3\\ 0\end{array}\right)={\mathbf{y}}_2.}$

This results in a basis for each of the generalized eigenspaces of ${\displaystyle A}$. Together the two chains of generalized eigenvectors span the space of all 5-dimensional column vectors.

${\displaystyle \left\{{\mathbf{x}}_1,{\mathbf{x}}_2\right\}=\left\{\left(\begin{array}{c}0\\ 3\\ -9\\ 9\\ -3\end{array}\right),\left(\begin{array}{c}1\\ -15\\ 30\\ -1\\ -45\end{array}\right)\right\},\left\{{\mathbf{y}}_1,{\mathbf{y}}_2,{\mathbf{y}}_3\right\}=\left\{\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 9\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 0\\ 3\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\\ -2\\ 0\end{array}\right)\right\}.}$

An "almost diagonal" matrix ${\displaystyle J}$ in Jordan normal form, similar to ${\displaystyle A}$ is obtained as follows:

${\displaystyle M=\left(\begin{array}{ccccc}{\mathbf{x}}_1& {\mathbf{x}}_2& {\mathbf{y}}_1& {\mathbf{y}}_2& {\mathbf{y}}_3\end{array}\right)=\left(\begin{array}{ccccc}0& 1& 0& 0& 0\\ 3& -15& 0& 0& 0\\ -9& 30& 0& 0& 1\\ 9& -1& 0& 3& -2\\ -3& -45& 9& 0& 0\end{array}\right),}$

${\displaystyle J=\left(\begin{array}{ccccc}1& 1& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 2& 1& 0\\ 0& 0& 0& 2& 1\\ 0& 0& 0& 0& 2\end{array}\right),}$

where ${\displaystyle M}$ is a generalized modal matrix for ${\displaystyle A}$, the columns of ${\displaystyle M}$ are a canonical basis for ${\displaystyle A}$, and ${\displaystyle AM=MJ}$.

Jordan chains

Definition: Let ${\displaystyle {\mathbf{x}}_m}$ be a generalized eigenvector of rank m corresponding to the matrix ${\displaystyle A}$ and the eigenvalue ${\displaystyle \lambda }$. The chain generated by ${\displaystyle {\mathbf{x}}_m}$ is a set of vectors ${\displaystyle \left\{{\mathbf{x}}_m,{\mathbf{x}}_{m-1},\dots ,{\mathbf{x}}_1\right\}}$ given by

where ${\displaystyle {\mathbf{x}}_1}$ is always an ordinary eigenvector with a given eigenvalue ${\displaystyle \lambda }$. Thus, in general,

The vector ${\displaystyle {\mathbf{x}}_j}$, given by (2), is a generalized eigenvector of rank j corresponding to the eigenvalue ${\displaystyle \lambda }$. A chain is a linearly independent set of vectors.

Canonical basis

Definition: A set of n linearly independent generalized eigenvectors is a canonical basis if it is composed entirely of Jordan chains.

Thus, once we have determined that a generalized eigenvector of rank m is in a canonical basis, it follows that the m − 1 vectors ${\displaystyle {\mathbf{x}}_{m-1},{\mathbf{x}}_{m-2},\dots ,{\mathbf{x}}_1}$ that are in the Jordan chain generated by ${\displaystyle {\mathbf{x}}_m}$ are also in the canonical basis.

Let ${\displaystyle {\lambda }_i}$ be an eigenvalue of ${\displaystyle A}$ of algebraic multiplicity ${\displaystyle {\mu }_i}$. First, find the ranks (matrix ranks) of the matrices ${\displaystyle (A-{\lambda }_{i}I),(A-{\lambda }_{i}I{)}^2,\dots ,(A-{\lambda }_{i}I{)}^{m_i}}$. The integer ${\displaystyle m_i}$ is determined to be the first integer for which ${\displaystyle (A-{\lambda }_{i}I{)}^{m_i}}$ has rank ${\displaystyle n-{\mu }_i}$ (n being the number of rows or columns of ${\displaystyle A}$, that is, ${\displaystyle A}$ is n × n).

Now define

${\displaystyle {\rho }_k=\mathrm{rank}(A-{\lambda }_{i}I{)}^{k-1}-\mathrm{rank}(A-{\lambda }_{i}I{)}^k(k=1,2,\dots ,m_i).}$

The variable ${\displaystyle {\rho }_k}$ designates the number of linearly independent generalized eigenvectors of rank k corresponding to the eigenvalue ${\displaystyle {\lambda }_i}$ that will appear in a canonical basis for ${\displaystyle A}$. Note that

${\displaystyle \mathrm{rank}(A-{\lambda }_{i}I{)}^0=\mathrm{rank}(I)=n}$.

Computation of generalized eigenvectors

In the preceding sections we have seen techniques for obtaining the ${\displaystyle n}$ linearly independent generalized eigenvectors of a canonical basis for the vector space ${\displaystyle V}$ associated with an ${\displaystyle n\times n}$ matrix ${\displaystyle A}$. These techniques can be combined into a procedure:

Solve the characteristic equation of ${\displaystyle A}$ for eigenvalues ${\displaystyle {\lambda }_i}$ and their algebraic multiplicities ${\displaystyle {\mu }_i}$;

For each ${\displaystyle {\lambda }_i:}$

Determine ${\displaystyle n-{\mu }_i}$;

Determine ${\displaystyle m_i}$;

Determine ${\displaystyle {\rho }_k}$ for ${\displaystyle (k=1,\dots ,m_i)}$;

Determine each Jordan chain for ${\displaystyle {\lambda }_i}$;

Example 3

The matrix

${\displaystyle A=\left(\begin{array}{cccc}5& 1& -2& 4\\ 0& 5& 2& 2\\ 0& 0& 5& 3\\ 0& 0& 0& 4\end{array}\right)}$

has an eigenvalue ${\displaystyle {\lambda }_1=5}$ of algebraic multiplicity ${\displaystyle {\mu }_1=3}$ and an eigenvalue ${\displaystyle {\lambda }_2=4}$ of algebraic multiplicity ${\displaystyle {\mu }_2=1}$. We also have ${\displaystyle n=4}$. For ${\displaystyle {\lambda }_1}$ we have ${\displaystyle n-{\mu }_1=4-3=1}$.

${\displaystyle (A-5I)=\left(\begin{array}{cccc}0& 1& -2& 4\\ 0& 0& 2& 2\\ 0& 0& 0& 3\\ 0& 0& 0& -1\end{array}\right),\mathrm{rank}(A-5I)=3.}$

${\displaystyle (A-5I{)}^2=\left(\begin{array}{cccc}0& 0& 2& -8\\ 0& 0& 0& 4\\ 0& 0& 0& -3\\ 0& 0& 0& 1\end{array}\right),\mathrm{rank}(A-5I{)}^2=2.}$

${\displaystyle (A-5I{)}^3=\left(\begin{array}{cccc}0& 0& 0& 14\\ 0& 0& 0& -4\\ 0& 0& 0& 3\\ 0& 0& 0& -1\end{array}\right),\mathrm{rank}(A-5I{)}^3=1.}$

The first integer ${\displaystyle m_1}$ for which ${\displaystyle (A-5I{)}^{m_1}}$ has rank ${\displaystyle n-{\mu }_1=1}$ is ${\displaystyle m_1=3}$.

We now define

${\displaystyle {\rho }_3=\mathrm{rank}(A-5I{)}^2-\mathrm{rank}(A-5I{)}^3=2-1=1,}$

${\displaystyle {\rho }_2=\mathrm{rank}(A-5I{)}^1-\mathrm{rank}(A-5I{)}^2=3-2=1,}$

${\displaystyle {\rho }_1=\mathrm{rank}(A-5I{)}^0-\mathrm{rank}(A-5I{)}^1=4-3=1.}$

Consequently, there will be three linearly independent generalized eigenvectors; one each of ranks 3, 2 and 1. Since ${\displaystyle {\lambda }_1}$ corresponds to a single chain of three linearly independent generalized eigenvectors, we know that there is a generalized eigenvector ${\displaystyle {\mathbf{x}}_3}$ of rank 3 corresponding to ${\displaystyle {\lambda }_1}$ such that

but

Equations (3) and (4) represent linear systems that can be solved for ${\displaystyle {\mathbf{x}}_3}$. Let

${\displaystyle {\mathbf{x}}_3=\left(\begin{array}{c}{x}_{31}\\ x_{32}\\ x_{33}\\ x_{34}\end{array}\right).}$

Then

${\displaystyle (A-5I{)}^3{\mathbf{x}}_3=\left(\begin{array}{cccc}0& 0& 0& 14\\ 0& 0& 0& -4\\ 0& 0& 0& 3\\ 0& 0& 0& -1\end{array}\right)\left(\begin{array}{c}{x}_{31}\\ x_{32}\\ x_{33}\\ x_{34}\end{array}\right)=\left(\begin{array}{c}14x_{34}\\ -4x_{34}\\ 3x_{34}\\ -x_{34}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right)}$

and

${\displaystyle (A-5I{)}^2{\mathbf{x}}_3=\left(\begin{array}{cccc}0& 0& 2& -8\\ 0& 0& 0& 4\\ 0& 0& 0& -3\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{c}{x}_{31}\\ x_{32}\\ x_{33}\\ x_{34}\end{array}\right)=\left(\begin{array}{c}2x_{33}-8x_{34}\\ 4x_{34}\\ -3x_{34}\\ x_{34}\end{array}\right)\ne \left(\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right).}$

Thus, in order to satisfy the conditions (3) and (4), we must have ${\displaystyle x_{34}=0}$ and ${\displaystyle x_{33}\ne 0}$. No restrictions are placed on ${\displaystyle x_{31}}$ and ${\displaystyle x_{32}}$. By choosing ${\displaystyle x_{31}=x_{32}=x_{34}=0,x_{33}=1}$, we obtain

${\displaystyle {\mathbf{x}}_3=\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right)}$

as a generalized eigenvector of rank 3 corresponding to ${\displaystyle {\lambda }_1=5}$. Note that it is possible to obtain infinitely many other generalized eigenvectors of rank 3 by choosing different values of ${\displaystyle x_{31}}$, ${\displaystyle x_{32}}$ and ${\displaystyle x_{33}}$, with ${\displaystyle x_{33}\ne 0}$. Our first choice, however, is the simplest.

Now using equations (1), we obtain ${\displaystyle {\mathbf{x}}_2}$ and ${\displaystyle {\mathbf{x}}_1}$ as generalized eigenvectors of rank 2 and 1, respectively, where

${\displaystyle {\mathbf{x}}_2=(A-5I){\mathbf{x}}_3=\left(\begin{array}{c}-2\\ 2\\ 0\\ 0\end{array}\right),}$

and

${\displaystyle {\mathbf{x}}_1=(A-5I){\mathbf{x}}_2=\left(\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right).}$

The simple eigenvalue ${\displaystyle {\lambda }_2=4}$ can be dealt with using standard techniques and has an ordinary eigenvector

${\displaystyle {\mathbf{y}}_1=\left(\begin{array}{c}-14\\ 4\\ -3\\ 1\end{array}\right).}$

A canonical basis for ${\displaystyle A}$ is

${\displaystyle \left\{{\mathbf{x}}_3,{\mathbf{x}}_2,{\mathbf{x}}_1,{\mathbf{y}}_1\right\}=\left\{\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right)\left(\begin{array}{c}-2\\ 2\\ 0\\ 0\end{array}\right)\left(\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right)\left(\begin{array}{c}-14\\ 4\\ -3\\ 1\end{array}\right)\right\}.}$

${\displaystyle {\mathbf{x}}_1,{\mathbf{x}}_2}$ and ${\displaystyle {\mathbf{x}}_3}$ are generalized eigenvectors associated with ${\displaystyle {\lambda }_1}$, while ${\displaystyle {\mathbf{y}}_1}$ is the ordinary eigenvector associated with ${\displaystyle {\lambda }_2}$.

This is a fairly simple example. In general, the numbers ${\displaystyle {\rho }_k}$ of linearly independent generalized eigenvectors of rank ${\displaystyle k}$ will not always be equal. That is, there may be several chains of different lengths corresponding to a particular eigenvalue.

Generalized modal matrix

Let ${\displaystyle A}$ be an n × n matrix. A generalized modal matrix ${\displaystyle M}$ for ${\displaystyle A}$ is an n × n matrix whose columns, considered as vectors, form a canonical basis for ${\displaystyle A}$ and appear in ${\displaystyle M}$ according to the following rules:

• All Jordan chains consisting of one vector (that is, one vector in length) appear in the first columns of ${\displaystyle M}$.
• All vectors of one chain appear together in adjacent columns of ${\displaystyle M}$.
• Each chain appears in ${\displaystyle M}$ in order of increasing rank (that is, the generalized eigenvector of rank 1 appears before the generalized eigenvector of rank 2 of the same chain, which appears before the generalized eigenvector of rank 3 of the same chain, etc.).

Jordan normal form

Let ${\displaystyle V}$ be an n-dimensional vector space; let ${\displaystyle \varphi }$ be a linear map in L(V), the set of all linear maps from ${\displaystyle V}$ into itself; and let ${\displaystyle A}$ be the matrix representation of ${\displaystyle \varphi }$ with respect to some ordered basis. It can be shown that if the characteristic polynomial ${\displaystyle f(\lambda )}$ of ${\displaystyle A}$ factors into linear factors, so that ${\displaystyle f(\lambda )}$ has the form

${\displaystyle f(\lambda )=\pm (\lambda -{\lambda }_1{)}^{{\mu }_1}(\lambda -{\lambda }_2{)}^{{\mu }_2}\cdots (\lambda -{\lambda }_r{)}^{{\mu }_r},}$

where ${\displaystyle {\lambda }_1,{\lambda }_2,\dots ,{\lambda }_r}$ are the distinct eigenvalues of ${\displaystyle A}$, then each ${\displaystyle {\mu }_i}$ is the algebraic multiplicity of its corresponding eigenvalue ${\displaystyle {\lambda }_i}$ and ${\displaystyle A}$ is similar to a matrix ${\displaystyle J}$ in Jordan normal form, where each ${\displaystyle {\lambda }_i}$ appears ${\displaystyle {\mu }_i}$ consecutive times on the diagonal, and the entry directly above each ${\displaystyle {\lambda }_i}$ (that is, on the superdiagonal) is either 0 or 1: in each block the entry above the first occurrence of each ${\displaystyle {\lambda }_i}$ is always 0 (except in the first block); all other entries on the superdiagonal are 1. All other entries (that is, off the diagonal and superdiagonal) are 0. (But no ordering is imposed among the eigenvalues, or among the blocks for a given eigenvalue.) The matrix ${\displaystyle J}$ is as close as one can come to a diagonalization of ${\displaystyle A}$. If ${\displaystyle A}$ is diagonalizable, then all entries above the diagonal are zero. Note that some textbooks have the ones on the subdiagonal, that is, immediately below the main diagonal instead of on the superdiagonal. The eigenvalues are still on the main diagonal.

Every n × n matrix ${\displaystyle A}$ is similar to a matrix ${\displaystyle J}$ in Jordan normal form, obtained through the similarity transformation ${\displaystyle J=M^{-1}AM}$, where ${\displaystyle M}$ is a generalized modal matrix for ${\displaystyle A}$. (See Note above.)

Example 4

Find a matrix in Jordan normal form that is similar to

${\displaystyle A=\left(\begin{array}{ccc}0& 4& 2\\ -3& 8& 3\\ 4& -8& -2\end{array}\right).}$

Solution: The characteristic equation of ${\displaystyle A}$ is ${\displaystyle (\lambda -2{)}^3=0}$, hence, ${\displaystyle \lambda =2}$ is an eigenvalue of algebraic multiplicity three. Following the procedures of the previous sections, we find that

${\displaystyle \mathrm{rank}(A-2I)=1}$

and

${\displaystyle \mathrm{rank}(A-2I{)}^2=0=n-\mu .}$

Thus, ${\displaystyle {\rho }_2=1}$ and ${\displaystyle {\rho }_1=2}$, which implies that a canonical basis for ${\displaystyle A}$ will contain one linearly independent generalized eigenvector of rank 2 and two linearly independent generalized eigenvectors of rank 1, or equivalently, one chain of two vectors ${\displaystyle \left\{{\mathbf{x}}_2,{\mathbf{x}}_1\right\}}$ and one chain of one vector ${\displaystyle \left\{{\mathbf{y}}_1\right\}}$. Designating ${\displaystyle M=\left(\begin{array}{ccc}{\mathbf{y}}_1& {\mathbf{x}}_1& {\mathbf{x}}_2\end{array}\right)}$, we find that

${\displaystyle M=\left(\begin{array}{ccc}2& 2& 0\\ 1& 3& 0\\ 0& -4& 1\end{array}\right),}$

and

${\displaystyle J=\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 1\\ 0& 0& 2\end{array}\right),}$

where ${\displaystyle M}$ is a generalized modal matrix for ${\displaystyle A}$, the columns of ${\displaystyle M}$ are a canonical basis for ${\displaystyle A}$, and ${\displaystyle AM=MJ}$. Note that since generalized eigenvectors themselves are not unique, and since some of the columns of both ${\displaystyle M}$ and ${\displaystyle J}$ may be interchanged, it follows that both ${\displaystyle M}$ and ${\displaystyle J}$ are not unique.

Example 5

In Example 3, we found a canonical basis of linearly independent generalized eigenvectors for a matrix ${\displaystyle A}$. A generalized modal matrix for ${\displaystyle A}$ is

${\displaystyle M=\left(\begin{array}{cccc}{\mathbf{y}}_1& {\mathbf{x}}_1& {\mathbf{x}}_2& {\mathbf{x}}_3\end{array}\right)=\left(\begin{array}{cccc}-14& 2& -2& 0\\ 4& 0& 2& 0\\ -3& 0& 0& 1\\ 1& 0& 0& 0\end{array}\right).}$

A matrix in Jordan normal form, similar to ${\displaystyle A}$ is

${\displaystyle J=\left(\begin{array}{cccc}4& 0& 0& 0\\ 0& 5& 1& 0\\ 0& 0& 5& 1\\ 0& 0& 0& 5\end{array}\right),}$

so that ${\displaystyle AM=MJ}$.

Applications

Matrix functions

Three of the most fundamental operations which can be performed on square matrices are matrix addition, multiplication by a scalar, and matrix multiplication. These are exactly those operations necessary for defining a polynomial function of an n × n matrix ${\displaystyle A}$. If we recall from basic calculus that many functions can be written as a Maclaurin series, then we can define more general functions of matrices quite easily. If ${\displaystyle A}$ is diagonalizable, that is

${\displaystyle D=M^{-1}AM,}$

with

${\displaystyle D=\left(\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\lambda }_n\end{array}\right),}$

then

${\displaystyle D^k=\left(\begin{array}{cccc}{\lambda }_1^k& 0& \cdots & 0\\ 0& {\lambda }_2^k& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\lambda }_n^k\end{array}\right)}$

and the evaluation of the Maclaurin series for functions of ${\displaystyle A}$ is greatly simplified. For example, to obtain any power k of ${\displaystyle A}$, we need only compute ${\displaystyle D^k}$, premultiply ${\displaystyle D^k}$ by ${\displaystyle M}$, and postmultiply the result by ${\displaystyle M^{-1}}$.

Using generalized eigenvectors, we can obtain the Jordan normal form for ${\displaystyle A}$ and these results can be generalized to a straightforward method for computing functions of nondiagonalizable matrices. (See Matrix function#Jordan decomposition.)

Differential equations

Consider the problem of solving the system of linear ordinary differential equations

where

${\displaystyle \mathbf{x}=\left(\begin{array}{c}{x}_1(t)\\ x_2(t)\\ ⋮\\ x_n(t)\end{array}\right),{\mathbf{x}}^{\prime }=\left(\begin{array}{c}{x}_1^{\prime }(t)\\ x_2^{\prime }(t)\\ ⋮\\ x_n^{\prime }(t)\end{array}\right),}$ and ${\displaystyle A=(a_{ij}).}$

If the matrix ${\displaystyle A}$ is a diagonal matrix so that ${\displaystyle a_{ij}=0}$ for ${\displaystyle i\ne j}$, then the system (5) reduces to a system of n equations which take the form

In this case, the general solution is given by

${\displaystyle x_1=k_1{e}^{a_{11}t}}$

${\displaystyle x_2=k_2{e}^{a_{22}t}}$

${\displaystyle ⋮}$

${\displaystyle x_n=k_n{e}^{a_{nn}t}.}$

In the general case, we try to diagonalize ${\displaystyle A}$ and reduce the system (5) to a system like (6) as follows. If ${\displaystyle A}$ is diagonalizable, we have ${\displaystyle D=M^{-1}AM}$, where ${\displaystyle M}$ is a modal matrix for ${\displaystyle A}$. Substituting ${\displaystyle A=MD{M}^{-1}}$, equation (5) takes the form ${\displaystyle M^{-1}{\mathbf{x}}^{\prime }=D(M^{-1}\mathbf{x})}$, or

where

The solution of (7) is

${\displaystyle y_1=k_1{e}^{{\lambda }_{1}t}}$

${\displaystyle y_2=k_2{e}^{{\lambda }_{2}t}}$

${\displaystyle ⋮}$

${\displaystyle y_n=k_n{e}^{{\lambda }_{n}t}.}$

The solution ${\displaystyle \mathbf{x}}$ of (5) is then obtained using the relation (8).

On the other hand, if ${\displaystyle A}$ is not diagonalizable, we choose ${\displaystyle M}$ to be a generalized modal matrix for ${\displaystyle A}$, such that ${\displaystyle J=M^{-1}AM}$ is the Jordan normal form of ${\displaystyle A}$. The system ${\displaystyle {\mathbf{y}}^{\prime }=J\mathbf{y}}$ has the form

where the ${\displaystyle {\lambda }_i}$ are the eigenvalues from the main diagonal of ${\displaystyle J}$ and the ${\displaystyle {ϵ}_i}$ are the ones and zeros from the superdiagonal of ${\displaystyle J}$. The system (9) is often more easily solved than (5). We may solve the last equation in (9) for ${\displaystyle y_n}$, obtaining ${\displaystyle y_n=k_n{e}^{{\lambda }_{n}t}}$. We then substitute this solution for ${\displaystyle y_n}$ into the next to last equation in (9) and solve for ${\displaystyle y_{n-1}}$. Continuing this procedure, we work through (9) from the last equation to the first, solving the entire system for ${\displaystyle \mathbf{y}}$. The solution ${\displaystyle \mathbf{x}}$ is then obtained using the relation (8).

Lemma:

Given the following chain of generalized eigenvectors of length ${\displaystyle r,}$

${\displaystyle X_1=v_1{e}^{\lambda t}}$

${\displaystyle X_2=(t{v}_1+v_2)e^{\lambda t}}$

${\displaystyle X_3=\left(\frac{t^2}{2}{v}_1+t{v}_2+v_3\right)e^{\lambda t}}$

${\displaystyle ⋮}$

${\displaystyle X_r=\left(\frac{t^{r-1}}{(r-1)!}{v}_1+...+\frac{t^2}{2}{v}_{r-2}+t{v}_{r-1}+v_r\right)e^{\lambda t}}$,

these functions solve the system of equations,

${\displaystyle X^{\prime }=AX.}$

Proof:

Define

${\displaystyle v_0=0}$

${\displaystyle X_j(t)=e^{\lambda t}\sum _{i=1}^j\frac{t^{j-i}}{(j-i)!}{v}_i.}$

Then, as ${\displaystyle t^0=1}$ and ${\displaystyle 1^{\prime }=0}$,

${\displaystyle X_j^{\prime }(t)=e^{\lambda t}\sum _{i=1}^{j-1}\frac{t^{j-i-1}}{(j-i-1)!}{v}_i+e^{\lambda t}\lambda \sum _{i=1}^j\frac{t^{j-i}}{(j-i)!}{v}_i}$.

On the other hand we have, ${\displaystyle v_0=0}$ and so

${\displaystyle A{X}_j(t)=e^{\lambda t}\sum _{i=1}^j\frac{t^{j-i}}{(j-i)!}A{v}_i}$

${\displaystyle =e^{\lambda t}\sum _{i=1}^j\frac{t^{j-i}}{(j-i)!}(v_{i-1}+\lambda v_i)}$

${\displaystyle =e^{\lambda t}\sum _{i=2}^j\frac{t^{j-i}}{(j-i)!}{v}_{i-1}+e^{\lambda t}\lambda \sum _{i=1}^j\frac{t^{j-i}}{(j-i)!}{v}_i}$

${\displaystyle =e^{\lambda t}\sum _{i=1}^{j-1}\frac{t^{j-i-1}}{(j-i-1)!}{v}_i+e^{\lambda t}\lambda \sum _{i=1}^j\frac{t^{j-i}}{(j-i)!}{v}_i}$

${\displaystyle =X_j^{\prime }(t)}$

as required.