Suppose that $N$ is a nilpotent linear operator on the finite-dimensional space $V$. Let us look at the cyclic decomposition for $N$ which we obtain from Theorem 3. We have a positive integer $r$ and $r$ non-zero vectors $\alpha_1,…,\alpha_r$ in $V$ with $N$-annihilators $p_1,…,p_r$ such that $$V=Z(\alpha_1 ;N)\oplus \cdots \oplus Z(\alpha_r ;N)$$ and $p_{i+1}$ divides $p_i$ for $i=1,…,r-1$. Since $N$ is nilpotent, the minimal polynomial is $x^k$ for some $k\leq n$. Thus each $p_i$ is of the form $p_i = x^{k_i}$, and the divisibility condition simply says that $k_1\geq …\geq k_r$. Of course, $k_1=k$ and $k_r\geq 1$. The companion matrix of $x^{k_i}$ is the $k_i\times k_i$ $$A_i=\begin{bmatrix} & & & \\1& & & \\ & \ddots & & \\ & & 1& \\ \end{bmatrix}.$$ Thus Theorem 3 gives us an ordered basis for $V$ in which the matrix of $N$ is the direct sum of the elementary nilpotent matrices, the sizes of which decrease as $i$ increases.
Here is one thing we should like to point out about the nilpotent operator $N$ above. The positive integer $r$ is precisely the nullity of $N$; in fact, the null space has as a basis the $r$ vectors $$N^{k_i-1}(\alpha_i).$$
For, let $\alpha$ be in the null space of $N$. We write $\alpha$ in the form $$\alpha =f_1(\alpha_1)+…+f_r(\alpha_r)$$
where $f_i$ is a polynomial, the degree of which we may assume is less than $k_i$. Since $N(\alpha)=0$, for each $i$ we have $$\begin{align}0&= N(f_i(\alpha_i))\\ &= Nf_i(N)(\alpha_i)\\ &=(xf_i)(N)(\alpha_i)\end{align}$$
Thus $xf_i$ is divisible by $x^{k_i}$, and since $\deg (f_i)\gt k_i$ this means that $$f_i=c_ix^{k_i-1}$$
where $c_i$ is some scalar. But then $$\alpha = c_1x^{k_1-1}(\alpha_1)+…+c_rx^{k_r-1}(\alpha_r)$$ which shows us that $\{N^{k_1-1}(\alpha_1),…,N^{k_r-1}(\alpha_r)\}$ form a basis for the null space of $N$.
Que: I don’t understand “since $N(\alpha)=0$, for each $i$ we have $0= N(f_i(\alpha_i))= Nf_i(N)(\alpha_i)=(xf_i)(N)(\alpha_i)$. Thus $xf_i$ is divisible by $x^{k_i}$, and since $\deg (f_i)\gt k_i$ this means that $f_i=c_ix^{k_i-1}$ where $c_i$ is some scalar”.
In particular, I can’t prove $N(f_i(\alpha_i))= Nf_i(N)(\alpha_i)$. Why $\deg (f_i)\gt k_i$? We assumed $\deg (f_i)\leq k_i-1$. Since $x^{k_i}|xf_i$, we have $\exists g_i\in F[x]$ such that $xf_i=g_ix^{k_i}$. So $$\deg (xf_i)=\deg (g_ix^{k_i})=1+\deg (f_i)=\deg (g_i)+k_i\leq 1+k_i-1=k_i.$$ Which implies $\deg (g_i)\leq 0$. So $g_i$ is a scalar polynomial, let say $g_i=c_i$. Since $xf_i=c_ix^{k_i}$, we have $f_i=c_ix^{k_i-1}$. Therefore I suppose $\deg (f_i)\gt k_i$ is typo. It should be $\deg (f_i)\lt k_i$.
My attempt: Let $B=\{\alpha_1,…,N^{k_1-1}(\alpha_1),…,\alpha_r,…,N^{k_r-1}(\alpha_r)\}$ be basis of $V$. Let $\alpha$ be in null space of $N$. Then $$\alpha =a_{1,1}\alpha_1+…+a_{1,k_1}N^{k_1-1}(\alpha_1)+…+a_{r,1}\alpha_r+…+a_{r,k_r}N^{k_r-1}(\alpha_r).$$ From $A_i$, we see $N(N^{k_i-1}(\alpha_i))=N^{k_i}(\alpha_i)=0$. So
$$N(\alpha)= a_{1,1}N(\alpha_1)+…+a_{1,k_1-1}N^{k_1-1}(\alpha_1)+…+a_{r,1}N(\alpha_r)+…+a_{r,k_r-1}N^{k_r-1}(\alpha_r)=0.$$
Since $B$ is independent, we have $a_{ij}=0$, $\forall 1\leq j\leq k_i-1$. So $\alpha=a_{1,k_1}N^{k_1-1}(\alpha_1)+…+a_{r,k_r}N^{k_r-1}(\alpha_r)$. Thus $\alpha \in \text{span}(\{N^{k_i-1}(\alpha_i)\mid 1\leq i\leq r\})$. Clearly $\text{span}(\{N^{k_i-1}(\alpha_i)\mid 1\leq i\leq r\})$ is subset of null space of $N$. Since $B$ is independent, we have $\{N^{k_i-1}(\alpha_i)\mid 1\leq i\leq r\}$ is independent. Hence $\{N^{k_i-1}(\alpha_i)\mid 1\leq i\leq r\}$ is basis of null space of $N$ and its dimension is $r$. Is my proof correct?
Best Answer
Actually the notations (where author proves r is the nullity of N) is not clear.
That's the reason why beginners may feel confused.
For,let $\alpha$ be in the null space of N.We write $\alpha$ in the form
$\begin{align} \alpha=f_1(N)\alpha_1+\cdots+f_r(N)\alpha_r \end{align}$
where $f_i$ is a polynomial,the degree of which we may assume is less than $k_i$.
Because if we have $x^i$ which degree is equal to or more than $k_i$ ,then $N^i(\alpha_i)=0.$
It cannot do any influence.
$\\$
Since $N\alpha=0$,for each i we have
$0=N\alpha=Nf_i(N)\alpha_i =(xf_i)(N)\alpha_i$
Thus $xf_i$ is dibisible by $x^{k_i}$ which means that $deg(xf_i)\geq k_i.$
So $deg(f_i)=k_i-1$,we have
$f_i=c_ix^{k_i-1}$
where $c_i$ is some scalar.