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Section 32.6 Theory

Here, we are most interested in the theory of cyclic subspaces relative to a nilpotent matrix, but we will start with some general facts.

Proof.
We have already outlined the justification for this fact in Subsection 32.4.4, and we leave the details of a formal proof to you, the reader.

Proof.
A proper proof would require an induction-like argument, but we will argue more informally.
The nonzero generating vector \(\uvec{w}\) is linearly independent all by itself. On the other hand, the collection
\begin{equation*} \uvec{w}, A \uvec{w}, A^2 \uvec{w} , \dotsc, A^n \uvec{w} \end{equation*}
must be linearly dependent because it is a collection of \(n+1\) vectors in a \(n\)-dimensional space (Lemma 18.5.7). So there must be a “transition” exponent \(k \le n\) where
\begin{equation*} \uvec{w}, A \uvec{w}, A^2 \uvec{w}, \dotsc, A^{k-1} \uvec{w} \end{equation*}
is independent but
\begin{equation*} \uvec{w}, A \uvec{w}, A^2 \uvec{w}, \dotsc, A^k \uvec{w} \end{equation*}
is dependent (where we take \(A^0 \uvec{w}\) to mean \(\uvec{w}\) even if \(A\) is not invertible). We claim that this transition value \(k\) is the one required by the statement of the lemma.
Since we already have independence, to verify this claim we only need to demonstrate that
\begin{equation*} W' = \Span \{ \uvec{w}, A \uvec{w}, A^2 \uvec{w}, \dotsc, A^{k-1} \uvec{w} \} \end{equation*}
is the same space as the full cyclic subspace
\begin{equation*} W = \Span \{ \uvec{w}, A \uvec{w}, A^2 \uvec{w}, \dotsc \} \text{.} \end{equation*}
And to do that, it is enough to verify that each of
\begin{equation*} A^k \uvec{w}, A^{k+1} \uvec{w}, A^{k+2} \uvec{w}, \dotsc \end{equation*}
is in \(W'\) (Proposition 17.5.6).
Now, we already know that \(A^k \uvec{w}\) must be in \(W'\) by its dependence with the finite spanning set for \(W'\text{.}\) If we write
\begin{equation*} A^k \uvec{w} = a_0 \uvec{w} + a_1 A \uvec{w} + a_2 A^2 \uvec{w} + \dotsb + a_{k-1} A^{k-1} \uvec{w} \text{,} \end{equation*}
then multiplying through by \(A\) gives
\begin{equation*} A^{k+1} \uvec{w} = a_0 A \uvec{w} + a_1 A^2 \uvec{w} + a_2 A^3 \uvec{w} + \dotsb + a_{k-1} A^k \uvec{w} \text{.} \end{equation*}
This expression for \(A^{k+1} \uvec{w}\) is a linear combination of vectors in \(W'\text{,}\) and so must be in \(W'\) as well (Proposition 17.5.2). Similarly, we can multiply that expression for \(A^{k+1} \uvec{w}\) through by \(A\) to get an expression for \(A^{k+2} \uvec{w}\) as a linear combination of vectors in \(W'\text{.}\) And similarly for \(A^{k+3} \uvec{w}\text{,}\) and so on.
Even when matrix \(A\) is not nilpotent, if an \(A\)-cyclic spanning set terminates in \(\zerovec\text{,}\) the preceding spanning vectors must be a basis for that cyclic space.

Proof.
First, we’ll use the Test for Linear Dependence/Independence (Procedure 18.3.1) to verify that
\begin{equation*} \{ \uvec{w}, A\uvec{w}, A^2 \uvec{w}, \dotsc, A^{\ell - 1} \uvec{w} \} \end{equation*}
is a linearly independent set. So assume
\begin{equation*} c_0 \uvec{w} + c_1 A \uvec{w} + c_2 A^2 \uvec{w} + \dotsb + c_{k-1} A^{k-1} \uvec{w} = \zerovec \end{equation*}
for some unknown scalars \(c_j\text{.}\) Since we have assumed \(A^\ell \uvec{w} = \zerovec\text{,}\) we can also say that \(A^j \uvec{w} = \zerovec\) for \(j \ge \ell\text{.}\) Therefore, multiplying the homogeneous vector equation above through by \(A^{\ell - 1}\) yields
\begin{equation*} c_0 A^{\ell - 1} \uvec{w} = \zerovec \text{.} \end{equation*}
But we have also assumed that \(A^{\ell - 1} \uvec{w} \neq \zerovec\text{,}\) so we must have \(c_0 = 0\) (Rule 1.e of Proposition 16.6.2), and our initial homogeneous vector equation above becomes
\begin{equation*} c_1 A \uvec{w} + c_2 A^2 \uvec{w} + \dotsb + c_{k-1} A^{k-1} \uvec{w} = \zerovec \text{.} \end{equation*}
We can repeat the argument, multiplying this new equation through by \(A^{\ell - 2}\) to obtain
\begin{equation*} c_1 A^{\ell - 1} \uvec{w} = \zerovec \text{,} \end{equation*}
from which we can conclude that \(c_1 = 0\text{.}\) And so on, until all the coefficients \(c_j\) have been shown to be zero, which completes the verification of independence.
Finally, we know from Lemma 32.6.2 that there is a positive exponent value \(k\) for which
\begin{equation*} \uvec{w}, A\uvec{w}, A^2 \uvec{w}, \dotsc, A^{k - 1} \uvec{w} \end{equation*}
is independent and
\begin{equation*} \uvec{w}, A\uvec{w}, A^2 \uvec{w}, \dotsc, A^{k - 1} \uvec{w}, A^k \uvec{w} \end{equation*}
is dependent. Since \(A^\ell \uvec{w} = \zerovec\) and we have already shown
\begin{equation*} \uvec{w}, A\uvec{w}, A^2 \uvec{w}, \dotsc, A^{\ell - 1} \uvec{w} \end{equation*}
to be independent, we can conclude that \(\ell = k\) and
\begin{equation*} \uvec{w}, A\uvec{w}, A^2 \uvec{w}, \dotsc, A^{\ell - 1} \uvec{w} \end{equation*}
is a basis for the \(A\)-cyclic subspace generated by \(\uvec{w}\text{,}\) as in Lemma 32.6.2.
We can specialize this theorem to the case of a nilpotent matrix.

Proof.
By the definition of degree of nilpotency, we have \(N^{k-1} \ne \zerovec\) and \(N^k = \zerovec\text{.}\) We claim there exists a vector \(\uvec{w}\) such that \(N^{k-1} \uvec{w} \ne \zerovec\text{.}\) For example, we could take \(\uvec{w} = \uvec{e}_j\text{,}\) the \(\nth[j]\) standard basis vector, if the \(\nth[j]\) column of \(N^{k-1}\) is nonzero. However, we also have \(N^k \uvec{w} = \zerovec \uvec{w} = \zerovec\text{.}\) Therefore, we can apply Theorem 32.6.3 to conclude that
\begin{equation*} \{ \uvec{w}, N \uvec{w}, \dotsc, N^{k-1} \uvec{w} \} \end{equation*}
is a basis for the \(N\)-cyclic subspace generated by \(\uvec{w}\text{.}\) Since this basis has \(k\) vectors in it, the cyclic space that it spans has dimension \(k\text{.}\)
If \(\uvec{u}\) is any other nonzero \(n\)-dimensional vector, then \(N^k = \zerovec\) implies that \(N^k \uvec{u} = \zerovec\text{,}\) and so
\begin{align*} \Span \{ \uvec{u}, N\uvec{u}, N^2 \uvec{u}, \dotsc \} \amp = \Span \{ \uvec{u}, N \uvec{u}, N^2 \uvec{u}, \dotsc, N^{k-1} \uvec{u}, \zerovec, \zerovec, \dotsc \}\\ \amp = \Span \{ \uvec{u}, N \uvec{u}, N^2 \uvec{u}, \dotsc, N^{k-1} \uvec{u} \}\text{.} \end{align*}
Therefore, the \(N\)-cyclic space generated by \(\uvec{u}\) can be spanned by a collection of \(k\) vectors, which means that the dimension of this space is less than or equal to \(k\) (Proposition 19.5.1). This proves that no other \(N\)-cyclic subspace can have dimension greater than \(k\text{.}\)
In this chapter, we are most interested in the case where \(k = n\text{.}\)

Proof.
Suppose \(N\) has maximum degree of nilpotency \(n\text{.}\) By Corollary 32.6.4, there exists an \(N\)-cyclic subspace of dimension \(n\text{.}\) However, the only subspace of \(\R^n\) of dimension \(n\) is \(\R^n\text{,}\) hence \(\R^n\) itself is an \(N\)-cyclic space, as required. (Or similarly for \(\C^n\) if we are working with complex scalars and vectors.)
The statements about the nullity and rank of \(N\) follow from the fact that any cyclic basis for \(\R^n\) can be used as the columns of a matrix \(P\) to obtain
\begin{equation*} \inv{P} N P = \begin{bmatrix} 0 \\ 1 \amp 0 \\ \amp \ddots \amp \ddots \\ \amp \amp 1 \amp 0 \end{bmatrix}\text{.} \end{equation*}
Similar matrices have the same rank and nullity (Corollary 26.5.7), so
\begin{align*} \rank N \amp = \rank (\inv{P}NP) = n-1 \text{,} \amp \nullity N \amp = \nullity (\inv{P}NP) = 1 \text{.} \end{align*}

Remark 32.6.6.

The facts in this section guarantee that, in the case of a nilpotent \(N\) with \(N^{n-1} \ne \zerovec\text{,}\) we will be able to successfully carry out Procedure 32.4.2, and that the transition matrix \(P\) we construct in that procedure will be invertible, as its columns will be linearly independent.
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