Let T be a linear operator on a vector space V, and let $\lambda$ be a scalar. A nonzero vector x in V is called a generalized eigenvector of T corresponding to $\lambda$ if ${(\text{T}-\lambda \text{I})}^p(x)=0$ for some positive integer p.

Notice that if x is a generalized eigenvector of T corresponding to $\lambda$, and p is the smallest positive integer for which ${(\text{T}-\lambda \text{I})}^p(x)=0$, then ${(\text{T}-\lambda \text{I})}^{p-1}(x)$ is an eigenvector of T corresponding to $\lambda$. Therefore $\lambda$ is an eigenvalue of T.

In the context of Example 1, each vector in $\beta$ is a generalized eigenvector of T. In fact, $v_1,v_2,v_3$, and $v_4$ correspond to the eigenvalue 2, $v_5$ and $v_6$ correspond to the eigenvalue 3, and $v_7$ and $v_8$ correspond to the eigenvalue 0.

Just as eigenvectors lie in eigenspaces, generalized eigenvectors lie in “generalized eigenspaces.”

Let T be a linear operator on a vector space V, and let $\lambda$ be an eigenvalue of T. The generalized eigenspace of T corresponding to $\lambda$, denoted ${\text{K}}_{\lambda }$, is the subset of V defined by

Note that ${\text{K}}_{\lambda }$ consists of the zero vector and all generalized eigenvectors corresponding to $\lambda$.

Recall that a subspace W of V is T-invariant for a linear operator T if $\text{T}(\text{W})\subseteq \text{W}$. In the development that follows, we assume the results of Exercises 3 and 4 of Section 5.4. In particular, for any polynomial g(x), if W is T-invariant, then it is also g(T)-invariant. Furthermore, the range of a linear operator T is T-invariant.

(a) Clearly, $\text{0}\in {\text{K}}_{\lambda }$. Suppose that x and y are in ${\text{K}}_{\lambda }$. Then there exist positive integers p and q such that

Therefore

and hence $x+y\in {\text{K}}_{\lambda }$. The proof that ${\text{K}}_{\lambda }$ is closed under scalar multiplication is straightforward.

To show that ${\text{K}}_{\lambda }$ is T-invariant, consider any $x\in {\text{K}}_{\lambda }$. Choose a positive integer p such that ${(\text{T}-\lambda \text{I})}^p(x)=0$. Then

Therefore $\text{T}(x)\in {\text{K}}_{\lambda }$.

Finally, it is a simple observation that ${\text{K}}_{\lambda }$ is contained in ${\text{K}}_{\lambda }$.

(b) Suppose that $w\in {\text{K}}_{\lambda }\cap {\text{K}}_{\mu }$. Then there exist positive integers p and q such that

As polynomials in $t,{(t-\lambda )}^p$ and ${(t-\mu )}^q$ are relatively prime, and hence by Theorem E.2 in Appendix E there exist polymomials $q_1(t)$ and $q_2(t)$ such that

It follows that

and hence $0=w$.

(c) Let $\mu$ be any scalar such that $\mu \ne \lambda$. Since ${\text{K}}_{\lambda }$ is T-invariant, it is also $(\text{T}-\mu \text{I})$-invariant. If $(\text{T}-\mu \text{I})(w)=0$ for some $w\in {\text{K}}_{\lambda }$, then $w\in {\text{E}}_{\mu }\subset {\text{K}}_{\mu }$. Hence $w=0$ by (b), and we conclude that the restriction of $\text{T}-\mu \text{I}$ to ${\text{K}}_{\lambda }$is one-to-one.

We now show that the restriction of $\text{T}-\mu \text{I}$ to ${\text{K}}_{\lambda }$ is onto. Let $x\in {\text{K}}_{\lambda }$, p be the smallest positive integer for which ${(\text{T}-\lambda \text{I})}^p(x)=0$, and

It is easily shown that W is a T-invariant subspace of ${\text{K}}_{\lambda }$, and hence it is also $(\text{T}-\mu \text{I})$-invariant. So $\text{T}-\mu \text{I}$ maps W to W. Since $\text{T}-\mu \text{I}$ is one-to-one on ${\text{K}}_{\lambda }$ , it must be one-to-one on W. Thus, because W is finite-dimensional, $\text{T}-\mu \text{I}$ maps W onto W. So there exists a $y\in \text{W}$ such that $(\text{T}-\mu \text{I})(y)=x$. Therefore $\text{T}-\mu \text{I}$ maps ${\text{K}}_{\lambda }$ onto ${\text{K}}_{\lambda }$.

(a) Let $\text{W}={\text{K}}_{\lambda }$, and let h(t) be the characteristic polynomial of ${\text{T}}_{\text{W}}$. By Theorem 5.20 (p. 312), h(t) divides the characteristic polynomial of T, and by Theorem 7.1(b), $\lambda$ is the only eigenvalue of ${\text{T}}_{\text{W}}$. Hence $h(t)={(-1)}^d{(t-\lambda )}^d$, where $d=\dim (\text{W})$, and $d\le m$.

(b) Clearly $\text{N}({(\text{T}-\lambda \text{I})}^m)\subseteq {\text{K}}_{\lambda }$. The characteristic polynomial of T is of the form $f(t)={(t-\lambda )}^{m}g(t)$, where g(t) is a product of powers of the form $(t-\mu )$ for eigenvalues $\mu \ne \lambda$. By Theorem 7.1, g(T) is one-to one on ${\text{K}}_{\lambda }$. Since ${\text{K}}_{\lambda }$ is finite-dimensional, it is also onto. Let $x\in {\text{K}}_{\lambda }$. Then there exists $y\in {\text{K}}_{\lambda }$ such that $g(\text{T})(y)=x$. Hence

by the Cayley-Hamilton Theorem. Therefore $x\in \text{N}({(\text{T}-\lambda \text{I})}^m)$, and thus ${\text{K}}_{\lambda }\subseteq \text{N}({(\text{T}-\lambda \text{I})}^m)$.

Suppose the characteristic polynomial of T is

For $j=1,2,\dots ,k$, let

First, we prove the existence of the vectors $v_i$. Because the polynomials $f_j(t)$ are relatively prime for $j=1,2,\dots ,k$, by Theorem E.2 there exist polynomials $q_j(t)$ for $j=1,2,\dots ,k$ such that

It follows that

For $i=1,2,\dots ,k$, let $v_i=q_i(\text{T})f_i(\text{T})(v)=f_i(\text{T})q_i(\text{T})(v)$. Then for each i,

by the Cayley-Hamilton theorem. It follows that $v_i\in K_{{\lambda }_i}$ for $i=1,2,\dots ,k$.

Now we prove the uniqueness of the vectors $v_i$. Since each ${\text{K}}_{{\lambda }_j}$ is T-invariant, each ${\text{K}}_{{\lambda }_i}$ is invariant under the operator $f_j(\text{T})$ for $j=1,2,\dots ,k$. Furthermore, $f_j(\text{T})$ is one-to-one on ${\text{K}}_{{\lambda }_j}$ by Theorem 7.1(c) and maps ${\text{K}}_{{\lambda }_i}$ to {0} for $i\ne j$ by Theorem 7.2(b).

Suppose that

where $v_i,w_i\in {\text{K}}_{{\lambda }_i}$ for $i=1,2,\dots ,k$. Then for $j=1,2,\dots ,k$,

and hence $f_j(\text{T})(v_j)=f_j(\text{T})(w_j)$ since $f_j(\text{T})(v_i)=f_j(\text{T})(w_i)=0$ for $i\ne j$. Because $f_j(\text{T})$ is one-to-one on ${\text{K}}_{{\lambda }_j}$, it follows that $v_j=w_j$.

The next result extends Theorem 5.8(b) (p. 267) to all linear operators whose characteristic polynomials split. In this case, the eigenspaces are replaced by generalized eigenspaces.

(a) This is a direct consequence of Theorem 7.1(b).

(b) First, we prove that $\beta$ is linearly independent. Consider a linear combination of vectors in $\beta$ equal to 0. For each i, let $v_i$ be the sum of the terms of this linear combination involving the vectors in ${\beta }_i$. Then

and hence $v_i=0$ for all i by Theorem 7.3. Since each ${\beta }_i$ is linearly independent, it follows that all of the coefficients in the linear combination are zeros. Hence $\beta$ is linearly independent.

We now prove that $\beta$ is a spanning set for V. Consider any vector $v\in \text{V}$. By Theorem 7.3, for $i=1,2,\dots ,k$ there exist vectors $v_i\in {\text{K}}_{{\lambda }_i}$ such that $v=v_1+v_2+\cdots +v_k$. Since each $v_i$ is a linear combination of the vectors in ${\beta }_i$, v is a linear combination of the vectors in $\beta$. Hence $\beta$ spans V.

(c) Since dim(V) is equal to the degree of the characteristic polynomial of T, it follows that $\dim (\text{V})={\displaystyle {\sum }_{i=1}^k{m}_i}$. By (b), $\dim (\text{V})={\displaystyle {\sum }_{i=1}^k\dim ({\text{K}}_{{\lambda }_i})}$. Hence ${\displaystyle {\sum }_{i=1}^k(m_i-\dim ({\text{K}}_{{\lambda }_i}))=0}$. By Theorem 7.2(a), $m_i-\dim ({\text{K}}_{{\lambda }_i})\ge 0$ for all i. Consequently, $m_i-\dim ({\text{K}}_{{\lambda }_i})=0$ for all i, which is the desired result.

Combining Theorems 7.4 and 5.8(a) (p. 267), we see that T is diagonalizable if and only if $\dim ({\text{E}}_{\lambda })=\dim ({\text{K}}_{\lambda })$ for each eigenvalue $\lambda$ of T. But ${\text{E}}_{\lambda }\subseteq {\text{K}}_{\lambda }$, and hence these subspaces have the same dimension if and only if they are equal.

We now focus our attention on the problem of selecting suitable bases for the generalized eigenspaces of a linear operator so that we may use Theorem 7.4 to obtain a Jordan canonical basis for the operator. For this purpose, we consider again the basis $\beta$ of Example 1. We have seen that the first four vectors of $\beta$ lie in the generalized eigenspace ${\text{K}}_2$. Observe that the vectors in $\beta$ that determine the first Jordan block of J are of the form

Furthermore, observe that ${(\text{T}-2\text{I})}^3(v_3)=0$. The relation between these vectors is the key to finding Jordan canonical bases. This leads to the following definitions.

Let T be a linear operator on a vector space V, and let x be a generalized eigenvector of T corresponding to the eigenvalue $\lambda$. Suppose that p is the smallest positive integer for which ${(\text{T}-\lambda \text{I})}^p(x)=0$. Then the ordered set

is called a cycle of generalized eigenvectors of T corresponding to $\lambda$. The vectors ${(\text{T}-\lambda \text{I})}^{p-1}(x)$ and x are called the initial vector and the end vector of the cycle, respectively. We say that the length of the cycle is p.

Notice that the initial vector of a cycle of generalized eigenvectors of a linear operator T is the only eigenvector of T in the cycle. Also observe that if x is an eigenvector of T corresponding to the eigenvalue $\lambda$, then the set $\left\{x\right\}$ is a cycle of generalized eigenvectors of T corresponding to $\lambda$ of length 1.

In Example 1, the subsets ${\beta }_1=\{v_1,v_2,v_3\},{\beta }_2=\left\{v_4\right\},{\beta }_3=\{v_5,v_6\}$, and ${\beta }_4=\{v_7,v_8\}$ are the cycles of generalized eigenvectors of T that occur in $\beta$ . Notice that $\beta$ is a disjoint union of these cycles. Furthermore, setting ${\text{W}}_i=\text{span}({\beta }_i)$ for $i=1,2,3,4$, we see that ${\beta }_i$ is a basis for ${\text{W}}_i$ and ${[{\text{T}}_{{\text{W}}_i}]}_{{\beta }_i}$ is the ith Jordan block of the Jordan canonical form of T. This is precisely the condition that is required for a Jordan canonical basis.

(a) Suppose that $\gamma$ corresponds to the eigenvalue $\lambda ,\gamma$ has length p, and x is the end vector of $\gamma$. Then $\gamma =\{v_1,v_2,\dots ,v_p\}$, where

So

and hence $\text{T}(v_1)=\lambda v_1$. For $i>1$,

This shows that $\text{T}-\lambda \text{I}$ maps W into itself, and so T does also. Thus, by the preceding equations, we see that ${[{\text{T}}_{\text{W}}]}_{\gamma }$ is a Jordan block.

For (b), simply repeat the arguments of (a) for each cycle in $\beta$ in order to obtain ${[\text{T}]}_{\beta }$. We leave the details as an exercise.

In view of this result, we must show that, under appropriate conditions, there exist bases that are disjoint unions of cycles of generalized eigenvectors. Since the characteristic polynomial of a Jordan canonical form splits, this is a necessary condition. We will soon see that it is also sufficient. The next result moves us toward the desired existence theorem.

Exercise 5 shows that the ${\gamma }_i$’s are disjoint.

The proof that $\gamma$ is linearly independent is by mathematical induction on the number of vectors in $\gamma$. If this number is 1, then the result is clear. So assume that, for some integer $n>1$, the result is true whenever $\gamma$ has fewer than n vectors, and suppose that $\gamma$ has exactly n vectors. Let W be the subspace of V generated by $\gamma$. Clearly W is $(\text{T}-\lambda \text{I})$-invariant, and $\dim (\text{W})\le n$. Let U denote the restriction of $\text{T}-\lambda \text{I}$ to W.

For each i, let ${{\gamma }^{\prime }}_i$ denote the cycle obtained from ${\gamma }_i$ by deleting the end vector. Note that if ${\gamma }_i$ has length one, then ${{\gamma }^{\prime }}_i=\varnothing$. In the case that ${{\gamma }^{\prime }}_i\ne \varnothing$, each vector of ${{\gamma }^{\prime }}_i$ is the image under U of a vector in ${\gamma }_i$, and conversely, every nonzero image under U of a vector of ${\gamma }_i$ is contained in ${{\gamma }^{\prime }}_i$. Let ${\gamma }^{\prime }={\displaystyle \underset{i}{\cup }{{\gamma }^{\prime }}_i}$.

Then by the last statement, ${\gamma }^{\prime }$ generates R(U). Furthermore, ${\gamma }^{\prime }$ consists of $n-q$ vectors, and the initial vectors of the ${{\gamma }^{\prime }}_i$’s are also initial vectors of the ${\gamma }_i$’s. Thus we may apply the induction hypothesis to conclude that ${\gamma }^{\prime }$ is linearly independent. Therefore ${\gamma }^{\prime }$ is a basis for R(U). Hence $\dim (\text{R}(\text{U}))=n-q$. Since the q initial vectors of the ${\gamma }_i$’s form a linearly independent set and lie in N(U), we have $\dim (\text{N}(\text{U}))\ge q$. From these inequalities and the dimension theorem, we obtain

We conclude that $\dim (\text{W})=n$. Since $\gamma$ generates W and consists of n vectors, it must be a basis for W. Hence $\gamma$ is linearly independent.

The proof is by mathematical induction on $n=\dim ({\text{K}}_{\lambda })$. The result is clear for $n=1$. So suppose that for some integer $n>1$ the result is true whenever $\dim ({\text{K}}_{\lambda })<n$, and assume that $\dim ({\text{K}}_{\lambda })=n$. Let U denote the restriction of $\text{T}-\lambda \text{I}$ to ${\text{K}}_{\lambda }$. Then R(U) is a subspace of ${\text{K}}_{\lambda }$ of lesser dimension, and R(U) is the space of generalized eigenvectors corresponding to $\lambda$ for the restriction of T to R(U). Therefore, by the induction hypothesis, there exist disjoint cycles ${\gamma }_1,{\gamma }_2,\dots ,{\gamma }_q$ of generalized eigenvectors of this restriction, and hence of T itself, corresponding to $\lambda$ for which $\gamma ={\displaystyle \underset{i=1}{\overset{q}{\cup }}{\gamma }_i}$ is a basis for R(U). For $i=1,2,\dots ,q$, the end vector of ${\gamma }_i$ is the image under U of a vector $v_i\in {\text{K}}_{\lambda }$, and so we can extend each ${\gamma }_i$ to a larger cycle ${\tilde{\gamma }}_i={\gamma }_i\cup \left\{v_i\right\}$ of generalized eigenvectors of T corresponding to $\lambda$. For $i=1,2,\dots ,q$, let $w_i$ be the initial vector of ${\tilde{\gamma }}_i$ (and hence of ${\gamma }_i$). Since $\{w_1,w_2,\dots ,w_q\}$ is a linearly independent subset of ${\text{E}}_{\lambda }$, this set can be extended to a basis $\{w_1,w_2,\dots ,w_q,u_1,u_2,\dots ,u_s\}$ for ${\text{E}}_{\lambda }$. Then ${\tilde{\gamma }}_1,{\tilde{\gamma }}_2,\dots ,{\tilde{\gamma }}_q,\left\{u_1\right\},\left\{u_2\right\},\dots ,\left\{u_s\right\}$ are disjoint cycles of generalized eigenvectors of T corresponding to $\lambda$ such that the initial vectors of these cycles are linearly independent. Therefore their union $\tilde{\gamma }$ is a linearly independent subset of ${\text{K}}_{\lambda }$ by Theorem 7.6.

Finally, we show that $\tilde{\gamma }$ is a basis for ${\text{K}}_{\lambda }$. Suppose that $\gamma$ consists of $r=\text{rank}(\text{U})$ vectors. Then $\tilde{\gamma }$ consists of $r+q+s$ vectors. Furthermore, since $\{w_1,w_2,\dots ,w_q,u_1,u_2,\dots ,u_s\}$ is a basis for ${\text{E}}_{\lambda }=\text{N}(\text{U})$, it follows that $\text{nullity}(\text{U})=q+s$. Therefore

So $\tilde{\gamma }$ is a linearly independent subset of ${\text{K}}_{\lambda }$ containing $\dim ({\text{K}}_{\lambda })$ vectors, and thus $\tilde{\gamma }$ is a basis for ${\text{K}}_{\lambda }$.

Let ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k$ be the distinct eigenvalues of T. By Theorem 7.7, for each i there is an ordered basis ${\beta }_i$ consisting of a disjoint union of cycles of generalized eigenvectors corresponding to ${\lambda }_i$. Let $\beta ={\beta }_1\cup {\beta }_2\cup \cdots \cup {\beta }_k$. Then, by Theorem 7.5(b), $\beta$ is a Jordan canonical basis for V.

The Jordan canonical form also can be studied from the viewpoint of matrices.

Let $A\in {\text{M}}_{n\times n}(F)$ be such that the characteristic polynomial of A (and hence of ${\text{L}}_A$) splits. Then the Jordan canonical form of A is defined to be the Jordan canonical form of the linear operator ${\text{L}}_A$ on ${\text{F}}^n$.

The next result is an immediate consequence of this definition and Corollary 1.

Exercise.

We can now compute the Jordan canonical forms of matrices and linear operators in some simple cases, as is illustrated in the next two examples. The tools necessary for computing the Jordan canonical forms in general are developed in the next section.