Section NLT  Nilpotent Linear Transformations

From A First Course in Linear Algebra
Version 0.94
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

This Section Under Construction

We have seen that some matrices are diagonalizable and some are not. Some authors refer to a non-diagonalizable matrix as defective, but we will study them carefully anyway. Examples of such matrices include Example EMMS4, Example HMEM5, and Example CEMS6. Each of these matrices has at least one eigenvalue with geometric multiplicity strictly less than its geometric multiplicity, and therefore Theorem DMFE tells us these matrices are not diagonalizable.

Given a square matrix $A$, it is likely similar to many, many other matrices. Of all these possibilities, which is the best? “Best” is a subjective term, but we might agree that a diagonal matrix is certainly a very nice choice. Unfortunately, as we have seen, this will not always be possible. What form of a matrix is “next-best”? Our goal, which will take us several sections to reach, is to show that every matrix is similar to a matrix that is “nearly-diagonal.” More precisely, every matrix is similar to a matrix with elements on the diagonal, and zeros and ones on the diagonal just above the main diagonal (the “super diagonal”), with zeros everywhere else. In the language of equivalence relations (see Theorem SER), we are determining a systematic representative for each equivalence class. Such a representative for a set of similar matrices is called a canonical form.

We have just discussed the determination of a canonical form as a question about matrices. However, we know that every square matrix creates a natural linear transformation (Theorem MBLT) and every linear transformation with identical domain and codomain has a square matrix representation for each choice of a basis, with a change of basis creating a similarity transformation (Theorem SCB). So we will state, and prove, theorems using the language of linear transformations on abstract vector spaces, while most of our examples will work with square matrices. You can, and should, mentally translate between the two settings frequently and easily.

Subsection NLT: Nilpotent Linear Transformations

We will discover that nilpotent linear transformations are the essential obstacle in a non-diagonalizable linear transformation. So we will study them carefully first, both as an object of inherent mathematical interest, but also as the object at the heart of the argument that leads to a pleasing canonical form for any linear transformation. Once we understand these linear transformations thoroughly, we will be able to easily analyze the structure of any linear transformation.

Definition NLT
Nilpotent Linear Transformation
Suppose that $T:V\mapsto V$ is a linear transformation such that there is an integer $p>0$ such that $T^p\left(v\right)=0$ for every $v\in V$. The smallest $p$ for which this condition is met is called the $index$ of $T$. $△$

Of course, the linear transformation $T$ defined by $T\left(v\right)=0$ will qualify as nilpotent of index 1. But are there others?

Example NM64
Nilpotent matrix, size 6, index 4
Recall that our definitions and theorems are being stated for linear transformations on abstract vector spaces, while our examples will work with square matrices (and use the same terms interchangeably). In this case, to demonstrate the existence of nontrivial nilpotent linear transformations, we desire a matrix such that some power of the matrix is the zero matrix. Consider

Thus we can say that $A$ is nilpotent of index 4.

Because it will presage some upcoming theorems, we will record some extra information about the eigenvalues and eigenvectors of $A$ here. $A$ has just one eigenvalue, $\lambda =0$, with algebraic multiplicity $6$ and geometric multiplicity $2$. The eigenspace for this eigenvalue is

If there were degrees of singularity, we might say this matrix was very singular, since zero is an eigenvalue with maximum algebraic multiplicity (Theorem SMZE, Theorem ME). Notice too that $A$ is “far” from being diagonalizable (Theorem DMFE). $⊠$

Another example.

Example NM62
Nilpotent matrix, size 6, index 2
Consider the matrix

So $B$ is nilpotent of index 2. Again, the only eigenvalue of $B$ is zero, with algebraic multiplicity $6$. The geometric multiplicity of the eigenvalue is $3$, as seen in the eigenspace,

Again, Theorem DMFE tells us that $B$ is far from being diagonalizable. $⊠$

On a first encounter with the definition of a nilpotent matrix, you might wonder if such a thing was possible at all. That a high power of a nonzero object could be zero is so very different from our experience with scalars that it seems very unnatural. Hopefully the two previous examples were somewhat surprising. But we have seen that matrix algebra does not always behave the way we expect (Example MMNC), and we also now recognize matrix products not just as arithmetic, but as function composition (Theorem MRCLT). We will now turn to some examples of nilpotent matrices which might be more transparent.

Definition JB
Jordan Block
Given the scalar $\lambda \in ℂ$, the Jordan block $J_n\left(\lambda \right)$ is the $n\times n$ matrix defined by

(This definition contains Notation JB.) $△$

Example JB4
Jordan block, size 4
A simple example of a Jordan block,

We will return to general Jordan blocks later, but in this section we are just interested in Jordan blocks where $\lambda =0$. Here’s an example of why we are specializing in these matrices now.

Example acronym
Nilpotent Jordan block, size 5
Consider

So $J_5\left(0\right)$ is nilpotent of index 5. As before, we record some information about the eigenvalues and eigenvectors of this matrix. The only eigenvalue is zero, with algebraic multiplicity 5, the maximum possible (Theorem ME). The geometric multiplicity of this eigenvalue is just 1, the minimum possible (Theorem ME), as seen in the eigenspace,

There should not be any real surprises in this example. We can watch the ones in the powers of $J_5\left(0\right)$ slowly march off to the upper-right hand corner of the powers. In some vague way, the eigenvalues and eigenvectors of this matrix are equally extreme. $⊠$

We can form combinations of Jordan blocks to build a variety of nilpotent matrices. Simply place Jordan blocks on the diagonal of a matrix with zeros everywhere else, to create a block diagonal matrix.

Example NM83
Nilpotent matrix, size 8, index 3
Consider the matrix

So $C$ is nilpotent of index 3. You should notice how block diagonal matrices behave in products (much like diagonal matrices) and that it was the largest Jordan block that determined the index of this combination. All eight eigenvalues are zero, and each of the three Jordan blocks contributes one eigenvector to a basis for the eigenspace, resulting in zero having a geometric multiplicity of 3. $⊠$

It would appear that nilpotent matrices only have zero as an eigenvalue, so the algebraic multiplicity will be the maximum possible. However, by creating block diagonal matrices with Jordan blocks on the diagonal you should be able to attain any desired geometric multiplicity for this lone eigenvalue. Likewise, the size of the largest Jordan block employed will determine the index of the matrix. So nilpotent matrices with various combinations of index and geometric multiplicities are easy to manufacture. The predictable properties of block diagonal matrices in matrix products and eigenvector computations, along with the next theorem, make this possible.

Subsection PNLT: Properties of Nilpotent Linear Transformations

In this subsection we collect some basic properties of nilpotent linear transformations. After studying the examples in the previous section, some of these will be no surprise.

Theorem ENLT
Eigenvalues of Nilpotent Linear Transformations
Suppose that $T:V\mapsto V$ is a linear transformation and $\lambda$ is an eigenvalue of $T$. Then $\lambda =0$. $\square$

Proof   Let $x$ be an eigenvector of $T$ for the eigenvalue $\lambda$, and suppose that $T$ is nilpotent with index $p$. Then

Because $x$ is an eigenvector, it is nonzero, and therefore Theorem SMEZV tells us that ${\lambda }^p=0$ and so $\lambda =0$. $■$

Paraphrasing, all of the eigenvalues of a nilpotent linear transformation are zero. So in particular, the characteristic polynomial of a nilpotent linear transformation, $T$, on a vector space of dimension $n$, is simply $p_T\left(x\right)=x^n$.

The next theorem is not critical for what follows, but it will explain our interest in nilpotent linear transformations. More specifically, it is the first step in backing up the assertion that nilpotent linear transformations are the essential obstacle in a non-diagonalizable linear transformation.

Theorem DNLT
Diagonalizable Nilpotent Linear Transformations
Suppose the linear transformation $T:V\mapsto V$ is nilpotent. Then $T$ is diagonalizable if and only $T$ is the zero linear transformation. $\square$

Proof   We start with the easy direction. Let $n=dim\left(V\right)$.

($\Leftarrow$) The linear transformation $Z:V\mapsto V$ defined by $Z\left(v\right)=0$ for all $v\in V$ is nilpotent of index $p=1$ and a matrix repesentation relative to any basis of $V$ is the $n\times n$ zero matrix, $\mathcal{O}$. Quite obviously, the zero matrix is a diagonal matrix (Definition DIM) and hence $Z$ is diagonalizable (Definition DZM).

($\Rightarrow$) Assume now that $T$ is diagonalizable, so ${\gamma }_T\left(\lambda \right)={\alpha }_T\left(\lambda \right)$ for every eigenvalue $\lambda$ (Theorem DMFE). By Theorem ENLT, $T$ has only one eigenvalue (zero), which therefore must have algebraic multiplicity $n$ (Theorem NEM). So the geometric multiplicity of zero will be $n$ as well, ${\gamma }_T\left(0\right)=n$.

Let $B$ be a basis for the eigenspace ${\mathcal{ℰ}}_T\left(0\right)$. Then $B$ is a linearly independent subset of $V$ of size $n$, and by Theorem G will be a basis for $V$. For any $x\in B$ we have

So $T$ is identically zero on a basis for $B$, and since the action of a linear transformation on a basis determines all of the values of the linear transformation (Theorem LTDB), it is easy to see that $T\left(v\right)=0$ for every $v\in V$. $■$

So, other than one trivial case (the zero matrix), every nilpotent linear transformation is not diagonalizable. It remains to see what is so “essential” about this broad class of non-diagonalizable linear transformations. For this we now turn to a discussion of kernels of powers of nilpotent linear transformations, beginning with a result about general linear transformations that may not necessarily be nilpotent.

Theorem KPLT
Kernels of Powers of Linear Transformations
Suppose $T:V\mapsto V$ is a linear transformation, where $dim\left(V\right)=n$. Then there is an integer $m$, $0\le m\le n$, such that

Proof   There are several items to verify in the conclusion as stated. First, we show that $\mathcal{K}\left(T^k\right)\subseteq \mathcal{K}\left(T^{k+1}\right)$ for any $k$. Choose $z\in \mathcal{K}\left(T^k\right)$. Then

So by Definition KLT, $z\in \mathcal{K}\left(T^{k+1}\right)$ and by Definition SSET we have $\mathcal{K}\left(T^k\right)\subseteq \mathcal{K}\left(T^{k+1}\right)$.

Second, we demonstrate the existence of a power $m$ where consecutive powers result in equal kernels. A by-product will be the condition that $m$ can be chosen so that $m\le n$. To the contrary, suppose that

Since $\mathcal{K}\left(T^k\right)⊊\mathcal{K}\left(T^{k+1}\right)$, Theorem PSSD implies that $dim\left(\mathcal{K}\left(T^{k+1}\right)\right)\ge dim\left(\mathcal{K}\left(T^k\right)\right)+1$. Repeated application of this observation yields

Thus, $\mathcal{K}\left(T^{n+1}\right)$ has a basis of size at least $n+1$, which is a linearly independent set of size greater than $n$ in the vector space $V$ of dimension $n$. This contradicts Theorem G.

This contradiction yields the existence of an integer $k$ such that $\mathcal{K}\left(T^k\right)=\mathcal{K}\left(T^{k+1}\right)$, so we can define $m$ to be smallest such integer with this property. From the argument above about dimensions resulting from a strictly increasing chain of subspaces, it should be clear that $m\le n$.

It remains to show that once two consecutive kernels are equal, then all of the remaining kernels are equal. More formally, if $\mathcal{K}\left(T^m\right)=\mathcal{K}\left(T^{m+1}\right)$, then $\mathcal{K}\left(T^m\right)=\mathcal{K}\left(T^{m+j}\right)$ for all $j\ge 1$. We will give a proof by induction on $j$ (Technique I). The base case ($j=1$) is precisely our defining property for $m$.

In the induction step, we assume that $\mathcal{K}\left(T^m\right)=\mathcal{K}\left(T^{m+j}\right)$ and endeavor to show that $\mathcal{K}\left(T^m\right)=\mathcal{K}\left(T^{m+j+1}\right)$. At the outset of this proof we established that $\mathcal{K}\left(T^m\right)\subseteq \mathcal{K}\left(T^{m+j+1}\right)$. So Definition SE requires only that we establish the subset inclusion in the opposite direction. To wit, choose $z\in \mathcal{K}\left(T^{m+j+1}\right)$. Then

So by Definition KLT, $z\in \mathcal{K}\left(T^m\right)$ as desired. $■$

We now specialize Theorem KPLT to the case of nilpotent linear transformations, which buys us just a bit more precision in the conclusion.

Theorem KPNLT
Kernels of Powers of Nilpotent Linear Transformations
Suppose $T:V\mapsto V$ is a nilpotent linear transformation with index $p$ and $dim\left(V\right)=n$. Then $0\le p\le n$ and

Proof   Since $T^p=0$ it follows that $T^{p+j}=0$ for all $j\ge 0$ and thus $\mathcal{K}\left(T^{p+j}\right)=V$ for $j\ge 0$. So the value of $m$ guaranteed by Theorem KPLT is at most $p$. The only remaining aspect of our conclusion that does not follow from Theorem KPLT is that $m=p$. To see this we must show that $\mathcal{K}\left(T^k\right)⊊\mathcal{K}\left(T^{k+1}\right)$ for $0\le k\le p-1$. If $\mathcal{K}\left(T^k\right)=\mathcal{K}\left(T^{k+1}\right)$ for some $k<p$, then $\mathcal{K}\left(T^k\right)=\mathcal{K}\left(T^p\right)=V$. This implies that $T^k=0$, violating the fact that $T$ has index $p$. So the smallest value of $m$ is indeed $p$, and we learn that $p<n$. $■$

The structure of the kernels of powers of nilpotent linear transformations will be crucial to what follows. But immediately we can see a practical benefit. Suppose we are confronted with the question of whether or not an $n\times n$ matrix, $A$, is nilpotent or not. If we don’t quickly find a low power that equals the zero matrix, when do we stop trying higher and higher powers? Theorem KPNLT gives us the answer: if we don’t see a zero matrix by the time we finish computing $A^n$, then it is not going to ever happen. We’ll now take a look at one example of Theorem KPNLT in action.

Example KPNLT
Kernels of Powers of a Nilpotent Linear Transformation
We will recycle the nilpotent matrix $A$ of index 4 from Example NM64. We now know that would have only needed to look at the first 6 powers of $A$ if the matrix had not been nilpotent. We list bases for the null spaces of the powers of $A$. (Notice how we are using null spaces for matrices interchangeably with kernels of linear transformations, see Theorem KNSI for justification.)

With the exception of some convenience scaling of the basis vectors in $\mathcal{N}\left(A^2\right)$ these are exactly the basis vectors described in Theorem BNS. We can see that the dimension of $\mathcal{N}\left(A\right)$ equals the geometric multiplicity of the zero eigenvalue. Why is this not an accident? We can see the dimensions of the kernels consistently increasing, and we can see that $\mathcal{N}\left(A^4\right)={ℂ}^6$. But Theorem KPNLT says a little more. Each successive kernel should be a superset of the previous one. We ought to be able to begin with a basis of $\mathcal{N}\left(A\right)$ and extend it to a basis of $\mathcal{N}\left(A^2\right)$. Then we should be able to extend a basis of $\mathcal{N}\left(A^2\right)$ into a basis of $\mathcal{N}\left(A^3\right)$, all with repeated applications of Theorem ELIS. Verify the following,

Do not be concerned at the moment about how these bases were constructed since we are not describing the applications of Theorem ELIS here. Do verify carefully for each alleged basis that, (1) it is a superset of the basis for the previous kernel, (2) the basis vectors really are members of the kernel of the right power of $A$, (3) the basis is a linearly independent set, (4) the size of the basis is equal to the size of the basis found previously for each kernel. With these verifications, Theorem G will tell us that we have successfully demonstrated what Theorem KPNLT guarantees. $⊠$

Subsection CFNLT: Canonical Form for Nilpotent Linear Transformations

Our main purpose in this section is to find a basis so that a nilpotent linear transformation will have a pleasing, nearly-diagonal matrix representation. Of course, we will not have a definition for “pleasing,” nor for “nearly-diagonal.” But the short answer is that our preferred matrix representation will be built up from Jordan blocks, $J_n\left(0\right)$. Here’s the theorem.

TALK UP EXAMPLE HERE............

Theorem CFNLT
Canonical Form for Nilpotent Linear Transformations
Suppose that $T:V\mapsto V$ is a nilpotent linear transformation of index $d$. Then there is a basis for $V$ so that the matrix representation, $M_{B,B}^T$, is block diagonal with each block being a Jordan block, $J_n\left(0\right)$. The size of the largest block is the index $d$, and the total number of blocks is the nullity of $T$, $n\left(T\right)$. $\square$

Proof   We will explicitly construct the desired basis, so the proof is constructive (Technique C), and can be used in practice. As we begin, the basis vectors will not be in the proper order, but we will rearrange them at the end of the proof. For convenience, define $n_i=n\left(T^i\right)$, so for example, $n_0=0$, $n_1=n\left(T\right)$ and $n_d=n\left(T^d\right)=dim\left(V\right)$.

The proof is in essence the following claim. For $2\le k\le d$, let $Z_k=\left\{z_{n_{k-1}+1},z_{n_{k-1}+2},z_{n_{k-1}+3},\dots ,z_{n_k}\right\}$ be any linearly independent set of $n_k-n_{k-1}$ vectors such that $z_i\in ker{T}^k$ and $z_i\notin ker{T}^{k-1}$, for all $n_{k-1}+1\le i\ne n_k$. Then there exists a set of vectors $\left\{z_1,z_2,z_3,\dots ,z_{n_{k-1}}\right\}$ such that

1. $\left\{z_1,z_2,z_3,\dots ,z_{n_{k-2}}\right\}\subseteq ker{T}^{k-2}$
2. $z_i\in ker{T}^{k-1}$ and $z_i\notin ker{T}^{k-2}$, for all $n_{k-2}+1\le i\le n_{k-1}$
3. $T\left(z_{n_{k-1}+i}\right)=vect{z}_{n_{k-2}+i}$ for $1\le i\le n_k-n_{k-1}$
4. $\left\{z_1,z_2,z_3,\dots ,z_k\right\}$ is a basis of $\mathcal{K}\left(T^k\right)$.