Jordan canonical form

Jordan decomposition

Let $T$ be a linear transformation on the $n$-dimensional space $V$. If the minimal polynomial of $T$ is:

$$m_A(\lambda)={(\lambda-\lambda_1)}^{r_1}{(\lambda-\lambda_2)}^{r_2}\cdots{(\lambda-\lambda_k)}^{r_k}$$

Then, by the primary decomposition theorem, the space $V$ can be decomposed as a direct sum of subspaces:

$$V=V_1\oplus V_2\oplus\cdots\oplus V_k$$

where $V_i=N\left({(A-\lambda_i I)}^{r_i}\right)$, $A$ is the matrix corresponding to $T$, and all these subspaces are invariant under $T$.

Let the transformation $T_i$ be the projection of $V$ onto the subspace $V_i$; that is, construct a polynomial $u_i(T)$ such that:

• $$T_i=u_i(T)\frac{m_A(T)}{{(T-\lambda_i T_e)}^{r_i}}$$
• $$T_1+T_2+\cdots+T_k=T_e$$

Here, $T_e$ denotes the identity transformation on $V$. Therefore:

• The transformation $T_i$ restricted to space $V_i$, namely ${T_i|}_{V_i}$, is the identity transformation on $V_i$.
• If $i$ and $j$ are distinct, transformation $T_i$ on space $V_j$, with restriction ${T_i|}_{V_j}$, is the zero transformation on $V_j$.

Thus, transformation $T_i$ maps every vector in space $V$, written $\xi$, to its component in $V_i$, namely $\xi_i$.

Construct the transformation:

$$T_D=\lambda_1 T_1+\lambda_2 T_2+\cdots+\lambda_k T_k$$

Since every transformation $T_i$ is a polynomial in $T$, the transformation $T_D$ is also a polynomial in $T$, so every subspace $V_i$ is invariant under $T_D$.

From the equations above, the restriction of transformation $T_D$ to subspace $V_i$, namely ${T_D|}_{V_i}$, is a homothety of $V_i$ with ratio $\lambda_i$. Therefore, $T_D$ is diagonalizable.

Define:

$$T_N=T-T_D$$

Then $T_N$ is also a polynomial in $T$, so every subspace $V_i$ is invariant under $T_N$. For any vector in $V_i$, written $\xi_i$, we have:

$${T_N}^{r_i}(\xi_i)={T-T_D}^{r_i}(\xi_i)={T-\lambda_i T_i}^{r_i}(\xi_i)=0$$

Let $r$ be the maximum of all the $r_i$. Then, for every vector in space $V$, written $\xi$, the power of $T_N$ of order $r$ maps $\xi$ to the zero vector. Thus $T_N$ is nilpotent.

Thus, every transformation on space $V$, denoted $T$, can be written as:

$$T=T_D+T_N$$

Here $T_D$ is diagonalizable and $T_N$ is nilpotent. Since both $T_D$ and $T_N$ are polynomials in $T$, they commute:

$$T_DT_N=T_NT_D$$

Theorem: Let $T_1$ and $T_2$ be two diagonalizable transformations of $V$ satisfying $T_1T_2=T_2T_1$. Then there is a basis with respect to which the matrices of both $T_1$ and $T_2$ are diagonal.

Theorem: Let $T$ be a linear transformation on the $n$-dimensional space $V$. Then there exist a diagonalizable transformation $T_D$ and a nilpotent transformation $T_N$ such that:

• $$T=T_D+T_N$$
• $$T_DT_N=T_NT_D$$

Both are polynomials in $T$, and they are uniquely determined by $T$.

This theorem gives a decomposition of $T$ called the Jordan decomposition of $T$. $T_D$ is called the diagonalizable part of $T$, and $T_N$ is called the nilpotent part of $T$.

Similarly, matrices have a Jordan decomposition:

Theorem: Let $A$ be an $n$-th order matrix. Then there exist a diagonalizable matrix $D$ and a nilpotent matrix $N$ such that:

• $$A=D+N$$
• $$DN=ND$$

Both are polynomials in $A$, and they are uniquely determined by $A$.

This theorem gives a decomposition of $A$ called the Jordan decomposition of $A$. $D$ is called the diagonalizable part of $A$, and $N$ is called the nilpotent part of $A$.

Lambda matrices

The next part introduces a more general kind of matrix containing the variable parameter $\lambda$, rather than merely a table of numbers. This discussion is broader than that of matrices consisting only of numbers.

For a $\lambda$ matrix, the field associated with the corresponding space becomes the field of rational functions in the variable $\lambda$.

A matrix whose entries are polynomials in $\lambda$ is called a $\lambda$ matrix, denoted by $A(\lambda)$.

Since the polynomial field contains the numerical field, a numerical matrix is a special $\lambda$ matrix, and the characteristic matrix of numerical matrix $A$, namely $\lambda I-A$, is a $\lambda$ matrix.

Elementary transformations of lambda matrices

For $\lambda$ matrices, addition, subtraction, multiplication, elementary transformations, and rank can likewise be defined. For square $\lambda$ matrices, determinants, minors, and cofactors can likewise be defined.

For $\lambda$ matrices, elementary transformations are mostly the same as for numerical matrices; only the row-addition transformation is changed to the following (using a row transformation as an example):

• Multiply one row by a polynomial in $\lambda$, namely $\varphi(\lambda)$, and add it to another row.

The scalar-multiplication transformation is not modified. This is because row addition does not change the determinant, whereas scalar multiplication does. To preserve the properties of rank over a polynomial field, the determinant may only be changed by a nonzero constant from the numerical field.

The corresponding elementary matrices are modified accordingly.

It is clear that the determinants of the three elementary matrices are all nonzero constants, so they all have full rank. Therefore, multiplying a $\lambda$ matrix on the left or right by any of them does not change its rank.

If $A(\lambda)$ is transformed into $B(\lambda)$ by finitely many elementary transformations, $A(\lambda)$ and $B(\lambda)$ are said to be equivalent.

For $\lambda$ matrices, equivalent matrices have the same rank. The converse does not hold, unlike the case for numerical matrices.

Smith normal form

Theorem: If the rank of a $\lambda$ matrix is $r$, then $A(\lambda)$ is necessarily equivalent to:

$$\begin{pmatrix} D(\lambda) & 0\\ 0 & 0\\ \end{pmatrix}$$

where:

$$D(\lambda)=\begin{pmatrix} d_1(\lambda) & & \\ & \ddots & \\ & & d_r(\lambda)\\ \end{pmatrix}$$

Each $d_i(\lambda)$ is a monic polynomial with leading coefficient $1$, and every two adjacent polynomials satisfy the divisibility relation $d_i(\lambda)|d_{i+1}(\lambda)$.

This normal form is called the Smith normal form, and the $d_i(\lambda)$ are called invariant factors.

To compute the Smith normal form, eliminate from the upper left to the lower right. At each step, make the upper-left entry the greatest common divisor of all remaining polynomials in the lower-right submatrix, and use it to eliminate every other entry in that row and column to $0$.

Theorem: $A(\lambda)$ and $B(\lambda)$ are equivalent if and only if $A(\lambda)$ and $B(\lambda)$ have exactly the same invariant factors.

Elementary divisors

By the fundamental theorem of algebra, suppose the invariant factors of $A(\lambda)$, namely $d_1(\lambda),d_2(\lambda),\cdots,d_m(\lambda)$, factor as:

$$d_i(\lambda)={(\lambda-\lambda_1)}^{e_{i1}}{(\lambda-\lambda_2)}^{e_{i2}}\cdots{(\lambda-\lambda_S)}^{e_{iS}}$$

where $\lambda_1,\cdots,\lambda_S$ are distinct. Since:

$$d_i(\lambda)|d_{i+1}(\lambda)$$

the exponents $e_{1j},e_{2j},\cdots,e_{mj}$ are nondecreasing, and every exponent in the last term $d_m(\lambda)$ is nonzero.

All factors in the expression above whose exponents are greater than zero are collectively called the elementary divisors of $A(\lambda)$.

Note that elementary divisors are counted with multiplicity. If, for some $j$, an exponent $e_{ij}$ occurs several times, the corresponding elementary divisor ${(\lambda-\lambda_j)}^{e_{ij}}$ must also occur the same number of times.

The preceding theorem states that $A(\lambda)$ and $B(\lambda)$ are equivalent if and only if they have exactly the same invariant factors. Identical invariant factors naturally imply identical elementary divisors, but the converse does not hold. In fact:

Theorem: $A(\lambda)$ and $B(\lambda)$ have exactly the same invariant factors if and only if their elementary divisors and ranks are both exactly the same.

Thus, “having exactly the same elementary divisors and rank” is also a criterion for the equivalence of $\lambda$ matrices.

When applying elementary transformations, one may first transform $A(\lambda)$ into a diagonal matrix, then find its elementary divisors and rank, and finally obtain the invariant factors and the normal form. We have the following result:

Theorem: Suppose $A(\lambda)$ is equivalent to a diagonal matrix:

$$\operatorname{diag}\{f_1(\lambda),f_2(\lambda),\cdots,f_r(\lambda),0,\cdots,0\}$$

Then the powers of the linear factors $f_1(\lambda),f_2(\lambda),\cdots,f_r(\lambda)$, namely ${(\lambda-\lambda_j)}^{e_{ij}}$, constitute the elementary divisors of $A(\lambda)$.

To construct invariant factors from the elementary divisors and the rank, first group the elementary divisors by factor and arrange the factors of each group in decreasing order in one row. Put the highest powers of the different factors in one column, pad the table with $1$ until it has $r$ columns, and take the product of each column as an invariant factor.

Application to characteristic matrices

If $A$ and $B$ are numerical matrices, their characteristic matrices are $\lambda$ matrices. We have the following result:

Theorem: Numerical matrices $A$ and $B$ are similar if and only if their characteristic matrices $\lambda I-A$ and $\lambda I-B$ are equivalent.

Since the characteristic matrix $\lambda I-A$ contains $n$ copies of $\lambda$ only on its main diagonal, its rank is $n$. By the reasoning above, characteristic matrices of numerical matrices of the same size always have the same rank, so we obtain the equivalence:

Numerical matrices $A$ and $B$ are similar if and only if the characteristic matrices $\lambda I-A$ and $\lambda I-B$ have exactly the same elementary divisors.

For the characteristic matrix $\lambda I-A$, elementary transformations preserve equivalence and therefore do not change the rank.

Considering the three elementary transformations, the only modified operation, row addition, does not change the determinant. In fact, the three elementary transformations merely multiply the resulting determinant polynomial by a constant, so they do not change its factorization or degree.

Therefore, the determinant of the characteristic matrix $\lambda I-A$ is a degree-$n$ polynomial. After elementary transformations reduce it to Smith normal form, its rank is $n$, so the determinant is the product of all invariant factors on the main diagonal, which is also the product of all elementary divisors. Hence the sum of the degrees of all elementary divisors of $\lambda I-A$ is $n$.

Jordan normal form

The matrix

$$\begin{pmatrix} \lambda & 1 & 0 & \cdots & 0 & 0\\ 0 & \lambda & 1 & \cdots & 0 & 0\\ 0 & 0 & \lambda & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & \lambda & 1\\ 0 & 0 & 0 & \cdots & 0 & \lambda\\ \end{pmatrix}$$

A matrix whose main-diagonal entries are all $\lambda$, whose entries immediately above the main diagonal are all $1$, and whose remaining entries are all $0$ is called a Jordan matrix belonging to $\lambda$, or a Jordan block.

Clearly, a nilpotent Jordan matrix is a special case of a Jordan matrix, namely the case where $\lambda$ is $0$.

Theorem: Let $T$ be a transformation of the $n$-dimensional space $V$, and let $\lambda_1,\cdots,\lambda_k$ be all distinct eigenvalues of $T$. Then there is a basis with respect to which the matrix of $T$ has the form:

$$\begin{pmatrix} B_1 & & & 0\\ & B_2 & & \\ & & \ddots & \\ 0 & & & B_k\\ \end{pmatrix}$$

where

$$B_i=\begin{pmatrix} J_{i1} & & & 0\\ & J_{i2} & & \\ & & \ddots & \\ 0 & & & J_{is_i}\\ \end{pmatrix}$$

where $J_{i1},\cdots,J_{is_i}$ are all Jordan blocks belonging to $\lambda_i$.

This follows because, first, by the minimal polynomial:

$$m_A(\lambda)={(\lambda-\lambda_1)}^{r_1}{(\lambda-\lambda_2)}^{r_2}\cdots{(\lambda-\lambda_k)}^{r_k}$$

there is a primary decomposition:

$$V=V_1\oplus V_2\oplus\cdots\oplus V_k$$

where:

$$V_i=N\left({(A-\lambda_i I)}^{r_i}\right)$$

Here $A$ is the matrix corresponding to $T$.

Let $S_i$ be the restriction of $T$ to $V_i$, namely ${T|}_{V_i}$. We next seek a Jordan decomposition for each $S_i$.

Let $T_e$ denote the identity transformation on $V$. Unlike the Jordan decomposition above, let $T_i$ denote the nilpotent part in the Jordan decomposition of $S_i$:

$$S_i=\lambda_i T_e+T_i$$

Then $T_i$ is a nilpotent transformation of subspace $V_i$; in fact, it is also the restriction of $T-\lambda_i T_e$ to $V_i$, namely ${(T-\lambda_i T_e)|}_{V_i}$.

The subspace $V_i$ can be decomposed as a direct sum of cyclic subspaces of the nilpotent transformation $T_i$:

$$V_i=W_{i1}\oplus W_{i2}\oplus\cdots\oplus W_{is_i}$$

In each cyclic subspace $W_{ij}$, take a cyclic basis and reverse its order. Together these form a basis of $V_i$, with respect to which the matrix of $T_i$ has the form:

$$N_i=\begin{pmatrix} N_{i1} & & & 0\\ & N_{i2} & & \\ & & \ddots & \\ 0 & & & N_{is_i}\\ \end{pmatrix}$$

All the $N_{ij}$ are nilpotent Jordan blocks. Thus, with respect to the basis chosen above for $V_i$, the matrix corresponding to $S_i$ is:

$$B_i=\begin{pmatrix} \lambda_i & & & 0\\ & \lambda_i & & \\ & & \ddots & \\ 0 & & & \lambda_i\\ \end{pmatrix}+\begin{pmatrix} N_{i1} & & & 0\\ & N_{i2} & & \\ & & \ddots & \\ 0 & & & N_{is_i}\\ \end{pmatrix}=\begin{pmatrix} J_{i1} & & & 0\\ & J_{i2} & & \\ & & \ddots & \\ 0 & & & J_{is_i}\\ \end{pmatrix}$$

Here $J_{i1},J_{i2},\cdots,J_{is_i}$ are all Jordan blocks belonging to $\lambda_i$.

Choose a basis for each subspace $V_i$ in the way described above and combine them into a basis of $V$. The matrix of $T$ with respect to this basis then has the form stated in the theorem.

Of the following form:

$$\begin{pmatrix} J_1 & & & 0\\ & J_2 & & \\ & & \ddots & \\ 0 & & & J_m\\ \end{pmatrix}$$

The $n$-th order matrix above, in which each $J_i$ is a Jordan block, is called a Jordan normal form.

Theorem: Every $n$-th order matrix $A$ is similar to a Jordan normal form. Apart from the ordering of the Jordan blocks, the Jordan normal form similar to $A$ is uniquely determined by $A$.

In the matrix $B_i$ constructed above, the first term is a scalar multiple of an identity matrix and therefore commutes with the second term. Thus, the first term is the diagonalizable part of the Jordan decomposition of $B_i$, and the second term is the nilpotent part of the Jordan decomposition of $B_i$.

In the Jordan normal form corresponding to a matrix, the diagonal matrix formed by the main-diagonal entries is the diagonalizable part of that Jordan normal form. Replacing the main-diagonal entries by $0$ gives its nilpotent part.

Theorem: For every Jordan block in the Jordan normal form of a matrix $A$:

$$J_i=\begin{pmatrix} \lambda_i & 1 & & & \\ & \lambda_i & 1 & & \\ & & \ddots & \ddots & \\ & & & \ddots & 1\\ & & & & \lambda_i\\ \end{pmatrix}$$

corresponds to an elementary divisor of the characteristic matrix $\lambda I-A$, namely ${(\lambda-\lambda_i)}^{n_i}$, and all elementary divisors of $\lambda I-A$ correspond to all Jordan blocks in the Jordan normal form of $A$.

This is because $A$ is similar to its Jordan normal form, so their characteristic matrices are equivalent; reducing the characteristic matrix of the Jordan normal form to Smith normal form makes the correspondence clear.

By this theorem, using the elementary divisors of the characteristic matrix $\lambda I-A$, we can write the Jordan normal form of matrix $A$.

A consequence is that $A$ is diagonalizable if and only if all elementary divisors of $\lambda I-A$ are linear.

Frobenius theorem

The discussion above shows that the rank of the Smith normal form of an $n$-th order characteristic matrix is $n$.

Theorem: Suppose the characteristic matrix of matrix $A$, namely $\lambda I-A$, has Smith normal form:

$$\operatorname{diag}\{d_1(\lambda),d_2(\lambda),\cdots,d_n(\lambda)\}$$

Then the last invariant factor $d_n(\lambda)$ is exactly the minimal polynomial of matrix $A$, denoted $m_A(\lambda)$.

Corollary: The equivalent conditions for a matrix $A$ to be diagonalizable are:

• The minimal polynomial $m_A(\lambda)$ has no repeated roots.
• The invariant factors of the characteristic matrix $\lambda I-A$ have no repeated roots.
• All elementary divisors of the characteristic matrix $\lambda I-A$ are linear.