This subsection is optional, and requires material from the optional Direct Sum subsection.
The prior subsection shows that as j increases, the dimensions of the 's fall while the dimensions of the 's rise, in such a way that this rank and nullity split the dimension of V. Can we say more; do the two split a basis— is ?
The answer is yes for the smallest power j = 0 since . The answer is also yes at the other extreme.
Where is a linear transformation, the space is the direct sum . That is, both and .
We will verify the second sentence, which is equivalent to the first. The first clause, that the dimension n of the domain of t^{n} equals the rank of t^{n} plus the nullity of t^{n}, holds for any transformation and so we need only verify the second clause.
Assume that , to prove that is . Because is in the nullspace, . On the other hand, because , the map is a dimensionpreserving homomorphism and therefore is onetoone. A composition of onetoone maps is onetoone, and so is onetoone. But now— because only is sent by a onetoone linear map to — the fact that implies that .
Technically we should distinguish the map from the map because the domains or codomains might differ. The second one is said to be the restriction^{[1]} of t to . We shall use later a point from that proof about the restriction map, namely that it is nonsingular.
In contrast to the j = 0 and j = n cases, for intermediate powers the space V might not be the direct sum of and . The next example shows that the two can have a nontrivial intersection.
Consider the transformation of defined by this action on the elements of the standard basis.
The vector
is in both the rangespace and nullspace. Another way to depict this map's action is with a string.^{[Index]}
A map whose action on is given by the string
has equal to the span , has , and has . The matrix representation is all zeros except for some subdiagonal ones.
Transformations can act via more than one string. A transformation t acting on a basis by
is represented by a matrix that is all zeros except for blocks of subdiagonal ones
(the lines just visually organize the blocks).
In those three examples all vectors are eventually transformed to zero.
^{[Index]}A nilpotent transformation^{[Index]}^{[Index]} is one with a power that is the zero map. A nilpotent matrix^{[Index]}^{[Index]} is one with a power that is the zero matrix. In either case, the least such power is the index of nilpotency.^{[Index]}^{[Index]}
In Example 2.3 the index of nilpotency is two. In Example 2.4 it is four. In Example 2.5 it is three.
The differentiation map is nilpotent of index three since the third derivative of any quadratic polynomial is zero. This map's action is described by the string and taking the basis gives this representation.
Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones.
With the matrix from Example 2.4, and this fourvector basis
a change of basis operation produces this representation with respect to D,D.
The new matrix is nilpotent; it's fourth power is the zero matrix since
and is the zero matrix.
The goal of this subsection is Theorem 2.13, which shows that the prior example is prototypical in that every nilpotent matrix is similar to one that is all zeros except for blocks of subdiagonal ones.
Let t be a nilpotent transformation on V. A tstring generated by^{[Index]} is a sequence . This sequence has length k. A tstring basis^{[Index]}^{[Index]} is a basis that is a concatenation of tstrings.
In Example 2.5, the tstrings and , of length three and two, can be concatenated to make a basis for the domain of t.
If a space has a tstring basis then the longest string in it has length equal to the index of nilpotency of t.
Suppose not. Those strings cannot be longer; if the index is k then t^{k} sends any vector— including those starting the string— to . So suppose instead that there is a transformation t of index k on some space, such that the space has a tstring basis where all of the strings are shorter than length k. Because t has index k, there is a vector such that . Represent as a linear combination of basis elements and apply t^{k − 1}. We are supposing that t^{k − 1} sends each basis element to but that it does not send to . That is impossible.
We shall show that every nilpotent map has an associated string basis. Then our goal theorem, that every nilpotent matrix is similar to one that is all zeros except for blocks of subdiagonal ones, is immediate, as in Example 2.5.
Looking for a counterexample, a nilpotent map without an associated string basis that is disjoint, will suggest the idea for the proof. Consider the map with this action.
Even after ommitting the zero vector, these three strings aren't disjoint, but that doesn't end hope of finding a tstring basis. It only means that will not do for the string basis.
To find a basis that will do, we first find the number and lengths of its strings. Since t's index of nilpotency is two, Lemma 2.12 says that at least one string in the basis has length two. Thus the map must act on a string basis in one of these two ways.
Now, the key point. A transformation with the lefthand action has a nullspace of dimension three since that's how many basis vectors are sent to zero. A transformation with the righthand action has a nullspace of dimension four. Using the matrix representation above, calculation of t's nullspace
shows that it is threedimensional, meaning that we want the lefthand action.
To produce a string basis, first pick and from
(other choices are possible, just be sure that is linearly independent). For pick a vector from that is not in the span of .
Finally, take and such that and .
Now, with respect to , the matrix of t is as desired.
Any nilpotent transformation t is associated with a tstring basis. While the basis is not unique, the number and the length of the strings is determined by t.
This illustrates the proof. Basis vectors are categorized into kind 1, kind 2, and kind 3. They are also shown as squares or circles, according to whether they are in the nullspace or not.
Fix a vector space V; we will argue by induction on the index of nilpotency of . If that index is 1 then t is the zero map and any basis is a string basis , ..., . For the inductive step, assume that the theorem holds for any transformation with an index of nilpotency between 1 and k − 1 and consider the index k case.
First observe that the restriction to the rangespace is also nilpotent, of index k − 1. Apply the inductive hypothesis to get a string basis for , where the number and length of the strings is determined by t.
(In the illustration these are the basis vectors of kind 1, so there are i strings shown with this kind of basis vector.)
Second, note that taking the final nonzero vector in each string gives a basis for . (These are illustrated with 1's in squares.) For, a member of is mapped to zero if and only if it is a linear combination of those basis vectors that are mapped to zero. Extend C to a basis for all of .
(The 's are the vectors of kind 2 so that is the set of squares.) While many choices are possible for the 's, their number p is determined by the map t as it is the dimension of minus the dimension of .
Finally, is a basis for because any sum of something in the rangespace with something in the nullspace can be represented using elements of B for the rangespace part and elements of for the part from the nullspace. Note that
and so can be extended to a basis for all of V by the addition of i more vectors. Specifically, remember that each of is in , and extend with vectors such that . (In the illustration, these are the 3's.) The check that linear independence is preserved by this extension is Problem 13.
^{[Index]}^{[Index]}^{[Index]} Every nilpotent matrix is similar to a matrix that is all zeros except for blocks of subdiagonal ones. That is, every nilpotent map is represented with respect to some basis by such a matrix.
This form is unique in the sense that if a nilpotent matrix is similar to two such matrices then those two simply have their blocks ordered differently. Thus this is a canonical form for the similarity classes of nilpotent matrices provided that we order the blocks, say, from longest to shortest.
The matrix
has an index of nilpotency of two, as this calculation shows.
The calculation also describes how a map m represented by M must act on any string basis. With one map application the nullspace has dimension one and so one vector of the basis is sent to zero. On a second application, the nullspace has dimension two and so the other basis vector is sent to zero. Thus, the action of the map is and the canonical form of the matrix is this.
We can exhibit such a mstring basis and the change of basis matrices witnessing the matrix similarity. For the basis, take M to represent m with respect to the standard bases, pick a and also pick a so that .
(If we take M to be a representative with respect to some nonstandard bases then this picking step is just more messy.) Recall the similarity diagram.
The canonical form equals Rep_{B,B}(m) = PMP ^{− 1}, where
and the verification of the matrix calculation is routine.
The matrix
is nilpotent. These calculations show the nullspaces growing.
That table shows that any string basis must satisfy: the nullspace after one map application has dimension two so two basis vectors are sent directly to zero, the nullspace after the second application has dimension four so two additional basis vectors are sent to zero by the second iteration, and the nullspace after three applications is of dimension five so the final basis vector is sent to zero in three hops.
To produce such a basis, first pick two independent vectors from
then add such that and
and finish by adding ) such that .
What is the index of nilpotency of the leftshift operator, here acting on the space of triples of reals?
Three. It is at least three because . It is at most three because .
For each string basis state the index of nilpotency and give the dimension of the rangespace and nullspace of each iteration of the nilpotent map.
Also give the canonical form of the matrix.
Decide which of these matrices are nilpotent.
By Lemma 1.3 the nullity has grown as large as possible by the nth iteration where n is the dimension of the domain. Thus, for the matrices, we need only check whether the square is the zero matrix. For the matrices, we need only check the cube.
Another way to see that the second and fourth matrices are not nilpotent is to note that they are nonsingular.
Find the canonical form of this matrix.
The table os calculations
gives these requirements of the string basis: three basis vectors are sent directly to zero, one more basis vector is sent to zero by a second application, and the final basis vector is sent to zero by a third application. Thus, the string basis has this form.
From that the canonical form is immediate.
Consider the matrix from Example 2.16.
Each of these matrices is nilpotent.
Put each in canonical form.
shows that any map represented by the matrix must act on the string basis in this way
because the nullspace after one application has dimension one and exactly one basis vector, , is sent to zero. Therefore, this representation with respect to is the canonical form.
The table shows that the string basis is of the form
because the nullspace after one application of the map has dimension two— and are both sent to zero— and one more iteration results in the additional vector being brought to zero.
shows that any map represented by this basis must act on a string basis in this way.
Therefore, this is the canonical form.
Describe the effect of left or right multiplication by a matrix that is in the canonical form for nilpotent matrices.
A couple of examples
suggest that left multiplication by a block of subdiagonal ones shifts the rows of a matrix downward. Distinct blocks
act to shift down distinct parts of the matrix.
Right multiplication does an analgous thing to columns. See Problem 1.
Is nilpotence invariant under similarity? That is, must a matrix similar to a nilpotent matrix also be nilpotent? If so, with the same index?
Yes. Generalize the last sentence in Example 2.9. As to the index, that same last sentence shows that the index of the new matrix is less than or equal to the index of , and reversing the roles of the two matrices gives inequality in the other direction.
Another answer to this question is to show that a matrix is nilpotent if and only if any associated map is nilpotent, and with the same index. Then, because similar matrices represent the same map, the conclusion follows. This is Exercise 14 below.
Show that the only eigenvalue of a nilpotent matrix is zero.
Observe that a canonical form nilpotent matrix has only zero eigenvalues; e.g., the determinant of this lowertriangular matrix
is ( − x)^{3}, the only root of which is zero. But similar matrices have the same eigenvalues and every nilpotent matrix is similar to one in canonical form.
Another way to see this is to observe that a nilpotent matrix sends all vectors to zero after some number of iterations, but that conflicts with an action on an eigenspace unless λ is zero.
Is there a nilpotent transformation of index three on a twodimensional space?
No, by Lemma 1.3 for a map on a twodimensional space, the nullity has grown as large as possible by the second iteration.
In the proof of Theorem 2.13, why isn't the proof's base case that the index of nilpotency is zero?
The index of nilpotency of a transformation can be zero only when the vector starting the string must be , that is, only when V is a trivial space.
Let be a linear transformation and suppose is such that but . Consider the tstring .
Finish the proof of Theorem 2.13.
We must check that is linearly independent where B is a tstring basis for , where is a basis for , and where . Write
and apply t.
Conclude that the coefficients are all zero as is a basis. Substitute back into the first displayed equation to conclude that the remaining coefficients are zero also.
Show that the terms "nilpotent transformation" and "nilpotent matrix", as given in Definition 2.6, fit with each other: a map is nilpotent if and only if it is represented by a nilpotent matrix. (Is it that a transformation is nilpotent if an only if there is a basis such that the map's representation with respect to that basis is a nilpotent matrix, or that any representation is a nilpotent matrix?)
For any basis B, a transformation n is nilpotent if and only if N = Rep_{B,B}(n) is a nilpotent matrix. This is because only the zero matrix represents the zero map and so n^{j} is the zero map if and only if N^{j} is the zero matrix.
Let T be nilpotent of index four. How big can the rangespace of T^{3} be?
It can be of any size greater than or equal to one. To have a transformation that is nilpotent of index four, whose cube has rangespace of dimension k, take a vector space, a basis for that space, and a transformation that acts on that basis in this way.
possibly other, shorter, strings
So the dimension of the rangespace of T^{3} can be as large as desired. The smallest that it can be is one— there must be at least one string or else the map's index of nilpotency would not be four.
Recall that similar matrices have the same eigenvalues. Show that the converse does not hold.
These two have only zero for eigenvalues
but are not similar (they have different canonical representatives, namely, themselves).
Prove a nilpotent matrix is similar to one that is all zeros except for blocks of superdiagonal ones.
A simple reordering of the string basis will do. For instance, a map that is assoicated with this string basis
is represented with respect to by this matrix
but is represented with respect to in this way.
Prove that if a transformation has the same rangespace as nullspace. then the dimension of its domain is even.
Let be the transformation. If then the equation shows that dim(V) is even.
Prove that if two nilpotent matrices commute then their product and sum are also nilpotent.
For the matrices to be nilpotent they must be square. For them to commute they must be the same size. Thus their product and sum are defined.
Call the matrices A and B. To see that AB is nilpotent, multiply (AB)^{2} = ABAB = AABB = A^{2}B^{2}, and (AB)^{3} = A^{3}B^{3}, etc., and, as A is nilpotent, that product is eventually zero.
The sum is similar; use the Binomial Theorem.
Consider the transformation of given by t_{S}(T) = ST − TS where S is an matrix. Prove that if S is nilpotent then so is t_{S}.
Some experimentation gives the idea for the proof. Expansion of the second power
the third power
and the fourth power
suggest that the expansions follow the Binomial Theorem. Verifying this by induction on the power of t_{S} is routine. This answers the question because, where the index of nilpotency of S is k, in the expansion of
for any i at least one of the S^{i} and S^{2k − i} has a power higher than k, and so the term gives the zero matrix.
Show that if N is nilpotent then I − N is invertible. Is that "only if" also?
Use the geometric series: . If N^{k + 1} is the zero matrix then we have a right inverse for I − N. It is also a left inverse.
This statement is not "only if" since
is invertible.
^{[Index]}
