In linear algebra, cycles are strings of linearly independent vectors obtained by applying increasing powers of an operator to a single vector; cyclic subspaces are the subspaces spanned by cycles.

In this lecture we focus on the cycles and cyclic subspaces obtained by applying nilpotent operators.

Let be a vector space.

Remember that a linear operator is said to be nilpotent of index if and only iffor any and for at least one vector .

Clearly,and where denotes the null space of an operator and denotes strict inclusion.

Here is a formal definition of cycle.

Definition Let be a vector space and a nilpotent linear operator. Let be a non-zero vector. Let be the largest non-negative integer such that . Then, the setis called the cycle generated by . The vector is called the initial vector of the cycle. The subspace spanned by the cycle, that is,is called the cyclic subspace generated by .

Note that and the cycle has elements. The number of elements of the cycle is called the length of the cycle.

Also note that the assumption that is nilpotent guarantees that exists (because can at most be equal to the index of ). As a consequence, cycles are well-defined.

In what follows, we call a cycle obtained from a nilpotent operator an -cycle if we want to emphasize the specific operator that was used to obtain the cycle.

How do we find the largest integer such that ?

In other words, how do we determine the length of a cycle?

The answer is simple: we start from and we compute increasingly larger powers: , , .... The first time that we get a zero vector, we stop. In fact, if thenfor any integer .

In other words, once we encounter the first zero vector, then we know that it will be followed by all zero vectors. This also implies that a cycle contains no zero vectors (proved formally in the next section).

We could also change the definition and say that the length of the cycle is the smallest non-negative integer such that .

Example Let be the space of vectors. Consider the nilpotent mapping defined by where The mapping is nilpotent becauseConsider the vector Then, we haveThus, the cycle of length generated by isNow, consider another vector Then, we haveThus, the cycle of length generated by is

As an exercise, we recommend trying to rigorously prove the following simple result (before reading the proof).

Proposition All the vectors of a cycle are non-zero.

Proof

The proof is by contradiction. Suppose that for (where is the length of the cycle). Then,since any linear operator maps the zero vector into itself. This contradicts the assumption that the cycle has length (hence ).

The first important result concerns the linear independence of cycles.

Proposition Let be a vector space, a nilpotent linear operator, a non-zero vector and a cycle of length generated by . Then, is a linearly independent set.

Proof

Since the length of the cycle is , then andfor any integer . Suppose that there exist scalars such thatBy applying to both sides of the equation, we obtain which implies (since ). Thus, the equation becomesBy applying to both sides of the equation, we obtainwhich implies . Hence, the equation becomesWe proceed in this way until we have demonstrated that , which proves that is a linearly independent set.

When working with multiple cycles together, it is often convenient to use so-called cycle tableaux.

Suppose that we have three different vectors , and , which generate three cycles having lengths , and respectively.

A linear combination involving all the vectors of the three cycles iswhere the first subscript of the coefficients is used to index the three different cycles and the second one is used to index the powers of the nilpotent operator.

A cycle tableau is a table that contains all the summands of the linear combinationwhere each cycle is written on a different row, always starting from the leftmost column.

When we apply to all the elements of the tableau, all the elements in the first columns on the left become equal to zero.

For example, if we apply to the elements of the tableau, we get

But, by the definition of length of a cycle (see also the proof in the previous section), we have

Thus, the tableau becomesor, simply,

If we instead apply to the original tableau, we get

Another useful result about linear independence follows.

Proposition Let be a vector space and a nilpotent linear operator. Let be non-zero vectors, generating the cycles of lengths . If the setis linearly independent, then also the setis linearly independent.

Proof

Suppose that there exist scalars ( and ) such thatNote that the linear combination in (1) includes all the vectors of the set . Arrange all the summands of the linear combination in a cycle tableau. Suppose that there are columns in the tableau (which implies that the length of the longest chain is equal to ). Then, we apply to equation (1) and all the elements of the tableau, so that we are left with a single non-zero column. By the assumed linear independence, all the scalars in that column must be equal to zero and that column can be dropped from the original tableau. Thus, we have a new tableau with columns (the rightmost one has been dropped). Then, we iteratively apply . At each iteration, we obtain that some coefficients of the linear combination are zero and we can drop another column. In the end, we obtain that all the coefficients of the linear combination are zero, which implies that the set is linearly independent.

The iterations outlined in the proof are fully shown in the next example.

Example Let us continue with the same example shown in the section above on cycle tableaux. We need to show that implies that all the coefficients of the linear combination are equal to zero, provided that , and are linearly independent. The starting tableau isThe length of the longest cycle is . So, to start with, we apply to all the summands. As a result, we obtainorwhich implies that (because ). Thus, we can drop the last column and the tableau becomesWe now apply to all the summands. As a result, we obtainorwhich implies that . We can drop another column:By applying , we getwhich implies by the assumed linear independence. The final tableau isWe leave it unchanged (or, which is the same, we apply the identity operator ). Thus,which implies by the assumed linear independence. This is the final step and it proves that the union of the cycles is a linearly independent set.

When the union of two or more cycles forms a linearly independent set, we call it a non-overlapping union of cycles.

Definition Let be a vector space and a nilpotent linear operator. Let be non-zero vectors, generating the cycles . The setis called a non-overlapping union of cycles if and only if it is linearly independent.

The next proposition is the key reason why the theory of cycles is so important.

Proposition Let be a finite-dimensional vector space and a nilpotent linear operator. Then, there exists a non-overlapping union of -cycles that is a basis for .

Proof

The proof is by induction on the dimension of , which is equal to the number of elements of any one of its bases. For , is the zero mapping (as proved in the lecture on nilpotent matrices). Thus, we can choose any non-zero , and is both a basis for and a cycle of length one (because ), which trivially constitutes a union of non-overlapping cycles. Now, suppose that the proposition holds for all the spaces of dimension or less. Let us consider a space such that . There are two possible cases: 1) ; 2) . In case 1) is the zero operator (which maps every vector to zero) and any basis of is a non-overlapping union of cycles because for any , so that any constitutes a cycle by itself. In case 2), the restriction of to is a nilpotent operator (remember that ranges are invariant; thus, the restriction is indeed an operator). Moreover, because a nilpotent operator cannot have full rank. As a consequence, there is a non-overlapping union of -cycles that spans . Suppose that the union iswhere . By the definition of , there exist vectors such thatWe can use the vectors to lengthen the cycles and define a new union of cycles which is non-overlapping by the previous proposition on the linear independence of unions of cycles. Suppose that the lengths of the cycles are . The vectors are linearly independent and belong to (being the final vectors in their respective cycles). If they do not form a basis for , we can find linearly independent vectorsbelonging to that complete the basis. Moreover, for . As a consequence, the newly found vectors form, by themselves, cycles of length . They are also linearly independent from the final vectors of the cycles found previously. Therefore,is a union of non-overlapping cycles. The number of vectors belonging to isby the rank-nullity theorem. Therefore, contains linearly independent vectors belonging to . As a consequence, is a basis for .

Now that we know about cycles, we can further hone our understanding of minimal polynomials.

Remember that the minimal polynomial of a nilpotent matrix of index iswhere is the nilpotency index of the matrix, as well as the smallest power after which the null spaces of the powers of stop growing.

We can now provide another characterization.

Proposition Let be the space of vectors. Let be a nilpotent matrix of index . Then, is the length of the longest cycle generated by the nilpotent operator defined by .

Proof

Since , no cycle can be longer than . However, there must be at least one cycle of length . To prove it, suppose to the contrary that all cycles have length less than . Then, it must be that for any . But this implies that cannot be nilpotent of index .

Below you can find some exercises with explained solutions.

Let be a vector space and a nilpotent linear operator.

Suppose that four different vectors , , and generate four cycles having lengths , , and respectively.

Consider a linear combination of all the vectors belonging to the four cycles (where the coefficient of a generic summand is ).

Use the cycle tableau to represent the linear combination. What happens to the tableau when we apply to the linear combination?

Solution

A linear combination involving all the vectors of the four cycles isWritten in a cycle tableau, the combination isBy applying to all the elements of the tableau, we get By considering the length of the various cycle represented in the tableau, we obtain Hence, the tableau becomesor, by pruning the zero column,

Let be the space of vectors. Consider the nilpotent mapping defined by where Find the length of the cycle generated by

Solution

The first vector of the cycle is itself. The second vector isThe third vector isThe next one isTherefore, the length of the cycle is .

Please cite as:

Taboga, Marco (2017). "Cyclic subspace", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/cyclic-subspace.

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