This subsection is optional, and requires material from the optional Direct Sum subsection.
The prior subsection shows that as 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 .
Can we say more;
do the two split a basis— is
The answer is yes for the smallest power since
The answer is also yes at the other extreme.
- Lemma 2.1
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 of the domain of equals
the rank of plus the nullity of , 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 dimension-preserving homomorphism and therefore is one-to-one. A composition of one-to-one maps is one-to-one, and so is one-to-one. But now— because only is sent by a one-to-one linear map to — the fact that implies that .
- Note 2.2
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 of to . We shall use later a point from that proof about the restriction map, namely that it is nonsingular.
In contrast to the and cases, for intermediate powers
the space might not be the direct sum of
The next example shows that the two can have a nontrivial intersection.
- Example 2.3
Consider the transformation of
defined by this action on the elements of the standard basis.
is in both the rangespace and nullspace. Another way to depict this map's action is with a string.
- Example 2.4
whose action on is given by
equal to the
and has .
The matrix representation is all zeros except for
some subdiagonal ones.
- Example 2.5
Transformations can act via more than one string.
A transformation acting on a basis
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
- Definition 2.6
A nilpotent transformation is one with a power that is the zero map. A nilpotent matrix is one with a power that is the zero matrix. In either case, the least such power is the index of nilpotency.
- Example 2.8
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
- Example 2.9
With the matrix from Example 2.4,
and this four-vector basis
a change of basis operation
produces this representation with respect to .
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.
- Definition 2.10
Let be a nilpotent transformation on . A -string generated by is a sequence . This sequence has length . A -string basis is a basis that is a concatenation of -strings.
- Example 2.11
In Example 2.5, the -strings and , of length three and two, can be concatenated to make a basis for the domain of .
- Lemma 2.12
If a space has a -string basis then the longest string in it has length equal to the index of nilpotency of .
Suppose not. Those strings cannot be longer; if the index is then sends any vector— including those starting the string— to . So suppose instead that there is a transformation of index on some space, such that the space has a -string basis where all of the strings are shorter than length . Because has index , there is a vector such that . Represent as a linear combination of basis elements and apply . We are supposing that 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 omitting the zero vector, these three
strings aren't disjoint, but that doesn't end hope of finding a
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 '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 left-hand action has a
nullspace of dimension three since that's how many basis vectors are
sent to zero.
A transformation with the right-hand action has a nullspace of
Using the matrix representation above, calculation of 's nullspace
shows that it is three-dimensional,
meaning that we want the left-hand 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
Now, with respect to ,
the matrix of is as desired.
- Theorem 2.13
Any nilpotent transformation is associated with a -string basis. While the basis is not unique, the number and the length of the strings is determined by .
This illustrates the proof.
Basis vectors are categorized into kind , kind , and
They are also shown as squares or circles,
according to whether they are in the nullspace or not.
Fix a vector space ; we will argue by induction on the index of nilpotency
If that index is then 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 and
and consider the index case.
First observe that the restriction to
is also nilpotent, of index .
Apply the inductive hypothesis to get a string basis for
where the number and length of the strings
is determined by .
(In the illustration these are the basis vectors of kind ,
so there are strings shown with this kind of basis vector.)
Second, note that taking the final nonzero vector in each string
gives a basis
(These are illustrated with '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
Extend to a basis for all of .
(The 's are the vectors of kind
so that is the set of squares.)
While many choices are possible for the 's,
their number is determined by the map as it is the dimension of
minus the dimension of
is a basis for
because any sum of something in the rangespace with something in the nullspace
can be represented using elements of for the rangespace
part and elements of for the part from the nullspace.
and so can be extended to a basis for all of by the addition of more vectors. Specifically, remember that each of is in , and extend with vectors such that . (In the illustration, these are the 's.) The check that linear independence is preserved by this extension is Problem 13.
- Corollary 2.14
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.
- Example 2.15
has an index of nilpotency of two, as this calculation shows.
The calculation also describes how a map represented by must act on any
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 -string basis
and the change of basis matrices witnessing the matrix similarity.
For the basis,
take to represent with respect to the standard bases,
and also pick a
so that .
(If we take 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 , where
and the verification of the matrix calculation is routine.
- Example 2.16
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
such that and
and finish by adding )
such that .