3.24 Diagonal and nilpotent
Recall
In particular
Let's look at the particular case \(\deg p_{i}(x)=1,\forall i,\,1\leq i\leq r\), that is, all the eigenvalues of \(f\) belong to \(k\)
Let's write \(p_i(x) = x - c_i, \quad c_i \in k, \quad c_i \neq c_j \text{ if } i \neq j.\)
Define \(D: V \to V\) by \(D = c_1 E_1 + c_2 E_2 + \dots + c_r E_r\) (the \(E_i\)'s are those of the Thm).
Let \(B_i\) be a ordered basis of \(W_i\), \(\forall i\), \(B = \bigcup_{i=1}^{r}B_{i}\) (ordered union).
So \(D\) is a diagonalizable \(k\)-linear map.
Let \(N = f - D : V \to V\) \(k\) -linear map
\(D = c_1 E_1 + \dots + c_r E_r\)
\(f=f\circ\text{ Id}_{V}=f\circ(E_{1}+\dots+E_{r})=f\circ E_{1}+f\circ E_{2}+\dots +f\circ E_{r}\)
\(N=f-D=(f-c_{1}\text{Id}_{V})\circ E_{1}+(f-c_{2}\text{Id}_{V})\circ E_{2}+\dots+ (f-c_{r}\text{Id}_{V})\circ E_{r}\)
\(N^{2}=\sum_{i=1}^{r}(f-c_{i}\text{Id}_{V})^{2}\circ E_{i}\) since \(E_{i}^{2}=E_{i}\quad\forall i\) and \(E_{i}\circ E_{j}=0,\quad\forall i\neq j\)
In general, prove by induction Exercise \(\forall \ell \in \mathbb{N}\) : \(N^{\ell}=\sum_{i=1}^{r}(f-c_{i}\text{Id}_{V})^{\ell}\circ E_{i}=\sum_{i=1}^{r}P_{i} (f)^{\ell}\circ E_{i}\) where \(p_i(x)=x-c_i\)
Recall \(E_{i}=h_{i}(f)=\frac{m_{f}(x)}{(x-c_{i})^{m_{i}}}(f)\)
For \(\ell\geq\max_{1\leq i\leq r}\left\lbrace m_{i}\right\rbrace\), \(\text{we get that }(f-c_{i}\text{Id})^{\ell}\circ h_{i}(f)=0\) since we can cancel the denominator and use minimal polynomial
So for \(\ell\geq\max_{1\leq i\leq r}\left\lbrace m_{i}\right\rbrace\), we get \(N^\ell = 0\)
That is \(N\) is a nilpotent \(k\) -linear map.
Definition
Let \(t\) be a linear map \(t : V \to V\) is nilpotent if \(\exists s \in \mathbb{N}\) such that \(t^s = 0\) .
The smallest \(s\) such that \(t^s = 0\) is called the index of nilpotency of \(t\) .
Thus, \(f = D + N\) where \(D\) is diagonalizable and \(N\) is nilpotent.
Example
\(f : \mathbb{R}^3 \to \mathbb{R}^3\) such that \(m_f (x) = \chi_f (x) = (x - 2)^2 (x - 3)\)
\(p_1 (x) = x - 2, \quad c_1 = 2, \quad m_1 = 2\)
\(p_2 (x) = x - 3, \quad c_2 = 3, \quad m_2 = 1\)
We know \(f\) is not diagonalizable.
\(W_1 = \ker (f - 2 \text{Id})^2 = \text{Im} E_1\) and \(W_2 = \ker (f - 3 \text{Id}) = \text{Im} E_2\)
\(m_{f_1} (x) = (x - 2)^2\) and \(m_{f_2} (x) = (x - 3)\)
\(\text{Id}_{\mathbb{R}^3}=E_{1}+E_{2},\quad f=f\circ E_{1}+f\circ E_{2}\)
Let \(D = c_1 E_1 + c_2 E_2, \quad \text{it is diagonalizable.}\)
Let \(B_1 = \{ v_1, v_2 \}\) be an ordered basis of \(W_1\) ,
\(B_2 = \{ v_3 \}\) be an ordered basis of \(W_2\) where \(v_3\) is eigenvector of eigenvalue 3
Set \(B = \{ v_1, v_2, v_3 \}\), then 
\(N=f-D=(f-2\text{Id})\circ E_{1}+(f-3\text{Id})\circ E_{2}\)
\(v_{1},v_{2}\in\) \(\ker(f-2\text{Id})^{2}\supsetneq\ker(f-2\text{Id})\) since \(m_{f}(x)=(x-2)^{2}\neq(x-2)\)
Let us choose \(w_{1}\in\ker(f-2\text{Id})^{2}\setminus\ker(f-2\text{Id})\)
\(0_{V}\neq(f-2\text{Id})(w_{1})=w_{2}\) (we set this)
Thus, \(B_{1}' = \{ w_{1}, w_{2} \}\)is a basis of \(W_1\) .
Prove that \(\{ w_1, w_2 \}\) is linearly independent
Let \(a, b \in \mathbb{R}, \quad 0 = a w_1 + b w_2\) \(\Rightarrow - a w_1 = b w_2\)
If \(a \neq 0\) , then \(w_2 = \frac{b}{a} w_1\), then \(w_1\) is also eigenvector, contradiction
If \(b\neq 0\), similarly, contradiction. Thus \(a=b=0\)
Let \(B' = \{ w_1, w_2, w_3 \}\), then 
So, \([N]_{B'} = [f]_{B'} - [D]_{B'}\) 
\(N^2 = 0\) since \(f(w_2) = 2w_2\) and \(w_2 = (f - 2I)(w_1) = f(w_1) - 2w_1\) since
So, \(f(w_1) = 2w_1 + w_2\)
Suppose \(f=D^{\prime}+N^{\prime}\) where \(D'\) is diagonalizable and \(N'\) is nilpotent.
Remark: \(N\circ D=D\circ N\)
Since \(N\circ D=D\circ N\iff (f-D)\circ D=D\circ (f-D)\iff f\circ D-D^2=D\circ f-D^2\iff D\circ f=f\circ D\)
Since \(D=c_1E_1+...+c_rE_r\) and \(E_{i}=h_{i}(f)\), then \(E_i \circ f = f \circ E_i\) for all \(i\)
Theorem
Given \(\dim V < \infty\) , \(f: V \to V\) is \(k\) -linear, \(m_{f}(x)=\prod_{i=1}^{r}(x-c_{i})^{n_{i}},c_{i}\in k,c_{i}\neq c_{j}\text{ if }i \neq j\)
Then
-
\(f = D + N\)
-
\(N\circ D=D\circ N\)
-
If \(f = D' + N'\) such that \(D'\) is diagonalizable, \(N'\) is nilpotent and \(N^{\prime}\circ D^{\prime}=D^{\prime}\circ N^{\prime}\), then \(D=D^{\prime}\text{ and }N=N^{\prime}\)
Proof
\(f = D + N\), \(f = D' + N'\) \(\Rightarrow D + N = D' + N'\) \(\Rightarrow (D - D') = (N' - N)\) (nilpotent) and \(D-D'\) is diagonalizable since \(D\circ D'=D'\circ D\), then they are diagonalizable in the same basis
\(N^{\prime}\circ D^{\prime}=D^{\prime}\circ N^{\prime}\iff D^{\prime}\circ f=f\circ D^{\prime}\Rightarrow f\circ N^{\prime}=N'\circ f\), so \(N'\) commutes with any polynomial in \(f\).
And we know \(N\) is a polynomial in \(f\), then \(N\circ N^{\prime}=N^{\prime}\circ N\)
Similarly, \(N\circ D=D\circ N\iff D\circ f=f\circ D\)
Why is \(N\) a polynomial in \(f\)?
\(N = f - D = f - (c_1 E_1 + \dots + c_r E_r)\) \(=f-c_{1}h_{1}(f)-c_{2}h_{2}(f)-\dots-c_{r}h_{r}(f)\)
For given \(l \in \mathbb{N}\), \((N^{\prime}-N)^{l}=\sum_{j=0}^{l}\binom{l}{j}(-1)^{j}(N^{\prime})^{l-j}N^{j}\) since \(N^{\prime}\circ N=N\circ N^{\prime}\)
Choose \(l \geq\) index of nilpotency of \(N\) + index of nilpotency of \(N'\). Then \((N^{\prime}-N)^{l}=0\)
Addition: Since \(D-D'=\)
, then \((D-D')^l=\)
, then it equals to \(0\) since \((N'-N)^l=0\)
Thus we have \([N'-N]^{l}_{\mathcal{B}}=0=[D-D']^{l}_{\mathcal{B}}\), then \(= \begin{pmatrix} a_{1} & \placeholder{} & \placeholder{} \\ \placeholder{} & \ddots & \placeholder{} \\ \placeholder{} & \placeholder{} & a_{n} \end{pmatrix}^{l}= \begin{pmatrix} a_{1}^{l} & \placeholder{} & \placeholder{} \\ \placeholder{} & \ddots & \placeholder{} \\ \placeholder{} & \placeholder{} & a_{n}^{l} \end{pmatrix}\)
Then \(a_i^l=0,\forall i,1\leq i\leq n\), then \(a_{i}=0,\forall i,1\leq i\leq n\), then \(D-D^{\prime}=0\), then \(D=D',N=N'\)