A square matrix is said to be nilpotent if, by rasing it to a sufficiently high integer power, we get the zero matrix as a result.
We start with a definition.
Definition Let be a matrix. We say that is nilpotent of index if and only ifandfor .
The index is often called the index of nilpotency (or nilpotency index) of the matrix.
Let be the space of all vectors.
Remember that the null space of isand, as we demonstrated in the lecture on matrix powers, the larger the power of a matrix is, the larger its null space is.
Proposition Let be the space of all vectors. Let be a matrix. Then, is nilpotent of index if and only ifandwhere denotes strict inclusion.
Let us prove the "only if" part, starting from the assumption that is nilpotent of index . Then, for any ,which implies that . Moreover, which implies that at least one column of is different from zero. Suppose it is the -th column. Let be the vector of the canonical basis such that its -th entry is equal to and all its other entries are equal to . Then, , being equal to the -th column of . As a consequence, and . Let us now prove the "if" part, starting from the assumption that and . For any , we haveand, in particular,for every vector of the canonical basis , which implies that all the columns of are equal to zero. Hence, . Since , there exists such thatwhich is possible only if at least one column of is different from zero. Hence, .
An important fact is that the range (or column space) of a nilpotent matrix cannot coincide with .
Proposition Let be the space of all vectors. Let be a matrix and denote its range by . Then, is nilpotent only ifwhere denotes strict inclusion.
If , then is full-rank and is full-rank (because the product of two full-rank matrices is full-rank). Hence, for any .
The following proposition characterizes nilpotent matrices in terms of their eigenvalues.
Proposition A matrix is nilpotent if and only if all its eigenvalues are equal to zero.
Let us prove the "only if" part, starting from the hypothesis that is nilpotent of index . Let be an eigenvalue of with associated eigenvector , that is,By pre-multiplying both sides by , we getSince , the previous equation becomesBut , which implies . Since was an arbitrary eigenvalue, all the eigenvalues of must be equal to zero. Let us prove the "if" part, starting from the hypothesis that all the eigenvalues of are zero. The proof is by contradiction. Suppose that has an eigenvalue . Then, we can construct a Schur decompositionin which is a unitary matrix and is an upper triangular matrix such that . In other words, the first column of isThe matrix is unitary, hence invertible. As a consequence, we can find a vector such thatwhere is the first vector of the canonical basis (its first entry is equal to and all its other entries are zero). Then,which implies , a contradiction. Hence, cannot have any non-zero eigenvalue.
In other words, nilpotent matrices are those that have a single eigenvalue equal to zero, with algebraic multiplicity equal to .
We now prove a simple fact about geometric multiplicity.
Proposition Let be a nilpotent matrix. Then, is non-defective (i.e., the geometric multiplicity of the eigenvalue is ) if and only if .
Let be the space of all vectors. Let us prove the "only if" part, starting from the hypothesis that in non-defective. Then, the eigenvectors of corresponding to the eigenvalue span . In other words, is the eigenspace of . Then, for any , we havewhich implies . Let us now prove the "if" part, starting from the hypothesis that . Then, for any , we havewhich implies that the eigenspace of is . Hence, is non-defective.
Since a matrix is diagonalizable if and only if it is non-defective, the previous proposition implies that the only nilpotent diagonalizable matrix is the zero matrix.
Remember that the minimal polynomial of a matrix is the lowest-degree monic polynomial such that .
The minimal polynomial of a matrix having distinct eigenvalues with algebraic multiplicities can be written aswhere for .
Thus, when is nilpotent, its minimal polynomial becomeswhere .
Proposition Let be a matrix. Then, is nilpotent of index if and only if the minimal polynomial of is
Let us prove the "only if" part, starting from the assumption that is nilpotent of index . Then, and if . This implies that the exponent of the minimal polynomial must be equal to (if it was larger, the minimal polynomial would not be the lowest-degree monic annihilating polynomial). Let us now prove the "if" part, starting from the assumption that the minimal polynomial is . Then, since the minimal polynomial is annihilating, Moreover,otherwise would not be minimal.
With the previous proposition, although extremely simple and intuitive, we are at a wow moment!
Three different concepts we previously defined turn out to be the same thing when we confine our attention to nilpotent matrices:
the nilpotency index;
the exponent of the linear factor in the minimal polynomial;
the index of the matrix (i.e., the smallest power after which null spaces stop growing).
We highly recommend revising the lecture on the minimal polynomial while having the previous proposition in mind.
The concept of a nilpotent matrix can be generalized to that of a nilpotent operator.
Definition Let be a vector space. Let be a linear operator. We say that is nilpotent of index if and only if, for any ,and there exists a non-zero such that
In other words, a linear operator is nilpotent of index if and only ifandwhere denotes strict inclusion and denotes the null space (or kernel) of the operators.
When is finite-dimensional, we can as usual represent:
the elements as coordinate vectors with respect to a basis ;
with a square matrix (see matrix of a linear map) that transforms coordinates by multiplication (i.e., ).
By the proposition above on null spaces, is a nilpotent mapping if and only if is a nilpotent matrix for any basis .
We end this section with a simple but useful fact.
Proposition Let be a vector space and a nilpotent operator. If , then is the zero mapping.
Choose a basis , which is made up of a single vector since . Therefore, the matrix of the operator is a scalar. Suppose that the operator is nilpotent of index . Then, the conditioncan be true only if .
Let us consider again the space of all vectors.
Let be a matrix, one of its eigenvalues, a strictly positive integer and a generalized eigenspace associated to :
We can consider the restricted linear operator defined, for any , by
For the definition to make sense, we need to check that the range of is included in , that is, whenever .
But implies that
If we multiply both sides of the last equation by , we getwhich implies that . Thus, is indeed an operator because we can consider its codomain to be equal to its domain.
Moreover, is a nilpotent operator: by the very definition of the domain , we havefor every . Note that is not necessarily idempotent of index , as there might be a smaller integer such thatfor every .
Please cite as:
Taboga, Marco (2017). "Nilpotent matrix", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/nilpotent-matrix.
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