Linear Algebra/Strings

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 $ℛ(t^j)$'s fall while the dimensions of the $𝒩(t^j)$'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 $V=ℛ(t^j)\oplus 𝒩(t^j)$?

The answer is yes for the smallest power $j=0$ since $V=ℛ(t^0)\oplus 𝒩(t^0)=V\oplus \{\overrightarrow{0}\}$. The answer is also yes at the other extreme.

In contrast to the $j=0$ and $j=n$ cases, for intermediate powers the space $V$ might not be the direct sum of $ℛ(t^j)$ and $𝒩(t^j)$. The next example shows that the two can have a nontrivial intersection.

In those three examples all vectors are eventually transformed to zero.

Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones.

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.

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 $t:{ℂ}^5\to {ℂ}^5$ with this action.

                 ${\mathrm{R}\mathrm{e}\mathrm{p}}_{{ℰ}_5,{ℰ}_5}(t)=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 1& 1& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0\end{array}\right)$

Even after omitting the zero vector, these three strings aren't disjoint, but that doesn't end hope of finding a $t$-string basis. It only means that ${ℰ}_5$ 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.

$\begin{array}{ccccc}{\overrightarrow{\beta }}_1& \mapsto & {\overrightarrow{\beta }}_2& \mapsto & \overrightarrow{0}\\ {\overrightarrow{\beta }}_3& \mapsto & {\overrightarrow{\beta }}_4& \mapsto & \overrightarrow{0}\\ {\overrightarrow{\beta }}_5& \mapsto & \overrightarrow{0}\end{array}$                

$\begin{array}{ccccc}{\overrightarrow{\beta }}_1& \mapsto & {\overrightarrow{\beta }}_2& \mapsto & \overrightarrow{0}\\ {\overrightarrow{\beta }}_3& \mapsto & \overrightarrow{0}\\ {\overrightarrow{\beta }}_4& \mapsto & \overrightarrow{0}\\ {\overrightarrow{\beta }}_5& \mapsto & \overrightarrow{0}\end{array}$

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 dimension four. Using the matrix representation above, calculation of $t$'s nullspace

$𝒩(t)=\{\left(\begin{array}{c}x\\ -x\\ z\\ 0\\ r\end{array}\right)|x,z,r\in ℂ\}$

shows that it is three-dimensional, meaning that we want the left-hand action.

To produce a string basis, first pick ${\overrightarrow{\beta }}_2$ and ${\overrightarrow{\beta }}_4$ from $ℛ(t)\cap 𝒩(t)$

${\overrightarrow{\beta }}_2=\left(\begin{array}{c}0\\ 0\\ 1\\ 0\\ 0\end{array}\right){\overrightarrow{\beta }}_4=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 1\end{array}\right)$

(other choices are possible, just be sure that $\{{\overrightarrow{\beta }}_2,{\overrightarrow{\beta }}_4\}$ is linearly independent). For ${\overrightarrow{\beta }}_5$ pick a vector from $𝒩(t)$ that is not in the span of $\{{\overrightarrow{\beta }}_2,{\overrightarrow{\beta }}_4\}$.

${\overrightarrow{\beta }}_5=\left(\begin{array}{c}1\\ -1\\ 0\\ 0\\ 0\end{array}\right)$

Finally, take ${\overrightarrow{\beta }}_1$ and ${\overrightarrow{\beta }}_3$ such that $t({\overrightarrow{\beta }}_1)={\overrightarrow{\beta }}_2$ and $t({\overrightarrow{\beta }}_3)={\overrightarrow{\beta }}_4$.

${\overrightarrow{\beta }}_1=\left(\begin{array}{c}0\\ 1\\ 0\\ 0\\ 0\end{array}\right){\overrightarrow{\beta }}_3=\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\end{array}\right)$

Now, with respect to $B=⟨{\overrightarrow{\beta }}_1,\dots ,{\overrightarrow{\beta }}_5⟩$, the matrix of $t$ is as desired.

${\mathrm{R}\mathrm{e}\mathrm{p}}_{B,B}(t)=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right)$

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.

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.

Exercises

/Solutions/