4.24 Jordan Form

Find all natural number for which there exists a nilpotent matrix \(A\in \mathbb{C}^{10\times 10}\) such that satisfies:

\(\text{rank}(A^{6})=0,\text{rank}(A^{5})=1,\dim\ker A^4=8,\text{rank}(A^3)=3,\text{rank}(A^2)=10-n\) and \(\dim\ker A=3\)

Solution

\(A\) \(A^2\) \(A^3\) \(A^4\) \(A^5\) \(A^6\)

rank 7 10-n 3 2 1 0

dim Ker 3 n 7 8 9 10

We know \(\{0\} \subsetneq \text{Ker } A \subsetneq \text{Ker } A^2 \subsetneq \text{Ker } A^3 \subsetneq \text{Ker } A^4 \subsetneq \text{Ker } A^5 \subsetneq \text{Ker } A^6 = \mathbb{C}^{10}\)

Since \(\text{dim Ker } A = 3 < \underline{\text{dim Ker } A^2} < 7 = \text{dim Ker } A^3\), we have \(n \in \{4, 5, 6\}\)

Since \(\text{dim Ker } A = 3 \implies J_A\) has 3 blocks.

‍

\(\text{Ker } A \subsetneq \text{Ker } A^2 \subsetneq \text{Ker } A^3 \subsetneq \text{Ker } A^4 \subsetneq \text{Ker } A^5 \subsetneq \text{Ker } A^6\)
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If \(n=4\) :
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Impossible

If \(n=5\):
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Yes

If \(n=6\):
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Yes
General Jordan Form.

Def: Let \(\lambda \in \mathbb{K}\). We call Jordan block relative to the eigenvalue \(\lambda\) of size n the matrix.

If \(V\) is a \(f.d. v.s.\) of dim \(m\) and \(f: V \to V\) is l. op. st. \(m_f(x) = (x-\lambda i d)^k\), then \((f-\lambda id)^k = 0\) and \((f-\lambda id)^{k-1} \neq 0\).
Thus, if \(g = f-\lambda id\), \(g\) is a nilpotent l. op. of index \(k\).
Therefore, there exists a basis \(\mathcal{B}\) for \(V\) s.t. \([f-\lambda id]_\mathcal{B} = [g]_\mathcal{B} = \begin{bmatrix} J_1 \\ & J_2 \\ & & \ddots \\ & & & J_r \end{bmatrix}\) (Nilpotent Jordan form).
So, \([f]_{\mathcal{B}}=[f-\lambda id+\lambda id]_{\mathcal{B}}=[f-\lambda id]_{\mathcal{B}} +[\lambda id]_{\mathcal{B}}= \begin{bmatrix} J_{1}+\lambda I_{n_1} & \placeholder{} & \placeholder{} & \placeholder{} \\ & J_{2}+\lambda I_{n_2} & \placeholder{} & \placeholder{} \\ & & \ddots & \placeholder{} \\ & & & J_{r}+\lambda I_{n_{r}} \end{bmatrix}\)

If \(f: V \to V\) s.t. \(m_f(x) = \prod_{i=1}^k (x-\lambda_i)^{d_i}\) where \(\lambda_i \neq \lambda_j\) (\(\forall i \neq j\)).
then there exists a basis \(\mathcal{B}\) for \(V\) s.t. \([f]_\mathcal{B} = \begin{bmatrix} J_1 \\ & J_2 \\ & & \ddots \\ & & & J_k \end{bmatrix}\) where each \(J_i\) consists of Jordan blocks relative to the eigenvalue \(\lambda_i\).
\(\implies\) Jordan form of \(f\) and \(\mathcal{B}\) is a Jordan basis for \(V\).
Find the general Jordan form and a Jordan basis for \(A\in \R^{6\times 6}\) where \(A= \begin{pmatrix} \begin{array}{ccc|ccc}0 & 0 & 1 & 0 & -1 & 1 \\ 1 & 2 & -1 & 1 & 1 & -2 \\ 0 & 1 & 1 & 1 & 1 & -2 \\\hline & \placeholder{} & \placeholder{} & 1 & 0 & 1 \\ & & \placeholder{} & 0 & 2 & 0 \\ & & \placeholder{} & 0 & 0 & 2\end{array} \end{pmatrix}\)
\(i)\chi_{A}(X)=\det(XI_{A}-A)\) \(= (X-2)^2 (X-1) (X-1)^3 = (X-2)^2 (X-1)^4\)

\(R^6 = Ker(A-I)^4 \oplus Ker(A-2I)^2\)
ii) Final basis for \(Ker(A-I)^{4}, Ker(A-2I)^{2}\) with \(1 \leq j \leq 4, 1 \leq j \leq 2\).
\(Ker(A-I) = span \{e_1+e_3, e_2-e_4\}\) and dim \(Ker(A-I) = 2 \neq 4\).
\(Ker(A-I)^2 = span \{ e_1+e_3, e_2+e_3, e_4+e_3 \}\). and dim \(Ker(A-I)^2=3 \neq 4\).
\(Ker(A-I)^3 = span \{e_1, e_2, e_3, e_4 \}\) and dim \(Ker(A-I)^3 = 4 = mult(1, \chi_A)\).
Then consider \(\{0\} \subsetneq Ker(A-I) \subsetneq Ker(A-I)^2 \subsetneq Ker(A-I)^3\)
\(Ker(A-I)^3 = span \{ e_1, e_2, e_3, e_4 \}\).
\(Ker(A-I)^2 = span \{ e_1+e_3, e_2+e_3, e_4+e_3 \}\).
\(Ker(A-I) = span \{ e_1+e_3, e_2-e_4 \}\).
We take \(v \in Ker(A-I)^3 \setminus Ker(A-I)^2\).
If \(v = e_1\), then \((A-I) e_1 = -e_1 + e_2 \in Ker(A-I)^2\). \((A-I)^2 e_1 = (A-I)(-e_1+e_2) = e_1 + e_3\)

\(e_1 + e_3 \leftarrow -e_1 + e_2 \leftarrow e_1\)
\(e_2 - e_4\)
\(B_1 = \{e_1, -e_1+e_2, e_1+e_3, e_2-e_4 \}\)

b)\(Ker(A-2I) = span \{ e_3+e_5, e_4+e_5+e_6 \}\)
Thus \(B_2 = \{ e_3+e_5, e_4+e_5+e_6 \}\).
Let \(B = \{ e_1, -e_1+e_2, e_1+e_3, e_2-e_4, e_3+e_5, e_4+e_5+e_6 \}\), then
Find \(A^n\) with A in Ex. 2.
\(A = P J_A P^{-1} \Rightarrow A^n = (P J_A P^{-1})^n = P J_A^n P^{-1}\), \(J_A = D+N\).
Remark: If \(M = \begin{bmatrix} M_1 & & \\ & \ddots & \\ & & M_k \end{bmatrix} \Rightarrow M^l = \begin{bmatrix} M_1^l & & \\ & \ddots & \\ & & M_k^l \end{bmatrix}\).
If \(M_i = \lambda I + N_i\) then \((M_i)^l = (\lambda I + N_i)^l = \sum_{j=0}^l \binom{l}{j} (\lambda I)^{l-j} N_i^j\).
Thus \(\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}^n = \sum_{j=0}^n \binom{n}{j} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}^{n-j} \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}^j = \begin{cases} I & \text{if } n=0 \\ I + \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} & \text{if } n=1 \\ I + \binom{n}{1} \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} + \binom{n}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} & \text{if } n \geq 2. \end{cases}\)
Then \(\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}^{n}= \begin{bmatrix} 1 & 0 & 0 \\ \binom{n}{1} & 1 & 0 \\ \binom{n}{2} & \binom{n}{1} & 1 \end{bmatrix}\), therefore \(A^{n}=P \begin{pmatrix} 1 & \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} & \\ \binom{n}{1} & 1 & \placeholder{} & \placeholder{} & \placeholder{} & \\ \binom{n}{2} & \binom{n}{1} & 1 & \placeholder{} & \placeholder{} & \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 & \placeholder{} & \\ \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} & 2^{n} & \\ \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} & \placeholder{} & 2^{n} \end{pmatrix}P^{-1}\checkmark\)
Compute \(A^8\) with

\(A^8 = (D+N)^8 = \sum_{j=0}^8 \binom{8}{j} D^{8-j} N^j\) \(= D^8 + \binom{8}{1} D^7 N^1 + \binom{8}{2} D^6 N^2 + \binom{8}{3} D^5 N^3 + 0\)

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	\(A\)	\(A^2\)	\(A^3\)	\(A^4\)	\(A^5\)	\(A^6\)
rank	7	10-n	3	2	1	0
dim Ker	3	n	7	8	9	10