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4.8 Jordan Block and Basis

Jordan Block

\(\dim V=n,f:V\to V\) be \(k\) -linear. Suppose \(\chi_f(x)=\prod^l_{i=1}(x-c_1)^{d_i}\) with \(c_i\neq c_j\) if \(i\neq j\) and \(d_i\geq 1,\forall i,\sum^l_{i=1}d_i=n\)

And \(m_f(x)=\prod^l_{i=1}(x-c_i)^{r_i}\) where \(1\leq r_i\leq d_i\)

Let \(W_i=\ker(f-c_iId)^{r_i}\) where \(1\leq i\leq l\)

But the Primary Decomposition Theorem: \(V=W_1\oplus W_2\oplus...\oplus W_l\) and each \(W_i\) is \(f\)-invariant

Let \(f_i=f|_{w_i}\) for all \(i\) where \(f_i:W_i\to W_i\), also \(m_{f_i}(x)=(x-c_{i})^{r_i},\forall i,1\leq i\leq l\)

Given \(i\), \(1\leq i\leq l\), set \(N_{i}=f_{i}-c_{i}Id:\underbrace{W_{i}}_{\ker\underbrace{(f-c_iId)^{r_i}}_{(f_i-c_iId)^{r_i}}}\to W_{i}\) where \(N_i\) is nilpotent with index of nilpotency \(r_i\)

There exists a basis \(\mathcal{B}_i\) of \(W_i\) such that \([N_{i}]_{\mathcal{B_i}}=\) image this is index not power.

Where \(\forall j,A_j^i=\)image​

This comes form the Theorem(Cyclic Decomposition Theorem)

By theorem: \(\exists z_{1}^{i},...,z_{s_{i}}^{i}:W_{i}=\langle z_{1}^{i}\rangle_{N_{i}}\oplus \cdots\oplus\langle z_{s_{i}}^{i}\rangle_{N_{i}}\) where \(m_{N_{i},z_{j}}(x)\mid m_{N_{i},z_{j-1}^{i}}(x),\forall i,2\leq j\leq s_{i}\) and \(m_{N_i,z_1^i}(x)=x^{r_i}\)

The size of \([N_{i}]_{\mathcal{B_i}}\) of the blocks is not increasing and the first one has size \(r_i\times r_i\), also \(s_i=\dim\ker N_i\)

So \([f_{i}]_{\mathcal{B_{i}}}=[N_{i}]_{\mathcal{B_{i}}}+c_{i}Id\) since \(f_i=N_i+c_iId\)

Thenimage and

Let \(B=B_1\cup B_2\cup ...\cup B_l\), thenimage. The blocks of type \(\tilde{A_j^i}\) is called the elementary Jordan block with eigenvalue \(c_i\)

Then we call \([f_{i}]_{\mathcal{B}_{i}}\) the Jordan block corresponding to the eigenvalue \(c_i\)

Thenimage​

Remark

  1. Except for the main diagonal and the diagonal that is below it, all the other coefficients of the matrix are all zero

  2. For each eigenvalue \(c_i\), the first elementary Jordan block has size \(r_i\times r_i\) and the size of \([f_{i}]_{\mathcal{B}_i}\) is \(d_i\times d_i\) since these matrices are all lower triangular

  3. Given \(A,A'\in M_{n}(k)\) such that their Jordan form are same, then \(\exists C,C'\in Gl_n(k)\) (invertible) such that \(A=CJ_{A}C^{-1}\) and \(A^{\prime}=C^{\prime}J_{A^{\prime}}C^{\prime -1}\) and \(J_A=J_A'\)​

    Then \(A=(CC'^{-1})A'(CC'^{-1})\), then \(A\sim A'\)​

Examples

  1. \(A\in M_5(\R)\) such that \(\chi_A(x)=(x-2)^3(x+7)^2\) and \(m_A(x)=(x-2)^2(x+7)\)​

    image​

    Remark: If \(m_f(x)=\prod^l_{i=1}(x-c_i),c_i\neq c_j,\) if \(i\neq j\), then \(J_f\) is a diagonal matrix \(W_{i}=\ker(f_{i}-c_{i}Id)=\left\lbrace\text{ eigenvectors of eigenvalue }c_{i}\cup \{0_{V}\}\right\rbrace\)​

  2. Let \(N_1\) and \(N_2\) be \(3\times 3\) nilpotent matrices. Prove that \(N_1\sim N_2\iff m_{N_1}(x)=m_{N_2}(x)\)​

    Possible Jordan forms of a nilpotent matrix \(3\times 3\)image​

    If \(N_1,N_2\) are \(4\times 4\) nilpotent matrices.

    image​

    \(\chi_{N_1}(x)=x^4\) and \(m_{N_1}(x)=x^2\)

    image​

    \(\chi_{N_2}(x)=x^{4}\) and \(m_{N_2}(x)=x^{2}\)

    But \(N_1\not\sim N_2\) because they have different Jordan forms

    Remark: Suppose \(A_1,A_2\in M_n(k)\) and \(A_1\sim A_2\), then \(A_{1}=CJ_{A_1}C^{-1}=\tilde CA_2\tilde C^{-1}\)

    Then \(A_{2}=(\tilde C^{-1}C)J_{A_1}(\tilde C^{-1}C)^{-1}\), then Jordan form of \(A_2\) is \(J_{A_1}\)

  3. Let \(A\in M_6(\R)\) such that \(\chi_A(x)=(x-2)^5(x+1)\). Which are all the possible Jordan forms of \(A\)?

    Then \(m_{A}(x)= \begin{cases} (x-2)(x+1) \\ (x-2)^{2}(x+1) \\ (x-2)^{3}(x+1) \\ (x-2)^{2}(x+1) \\ (x-2)^{4}(x+1) \\ (x-2)^{5}(x+1) \end{cases}\)​

    Then we have 5 cases

    image​

  4. Given \(A= \begin{pmatrix} 2 & 0 & 0 & 0 & -2 \\ -5 & 2 & -1 & 1 & 2 \\ -5 & 1 & 5 & -1 & 2 \\ -15 & 1 & 3 & 2 & 6 \\ 10 & 0 & 0 & 0 & -2 \end{pmatrix}\in \R^{5\times 5}\) with \(\chi_A(x)=(x-3)^4(x-2)\) and \(m_A(x)=(x-3)^3(x-2)\)​

    image​

  5. \(f:\R^7\to \R^7\) is \(\R\)-linear and \(\chi_f(x)=(x-1)^4x^3\) and \(m_f(x)=(x-1)^3x^2\)​

    image​

Problem

Given \(f:V\to V\), k-linear map such that \(\chi_f(x)=\prod^l_{i=1}(x-c_i)^{d_i}\) and \(c_i\neq c_j\) if \(i\neq j\)

How to find a basis \(B\) of \(V\) such that \([f]_B=J_f\)

Lemma

Suppose \(f\) is nilpotent and \(m_f(x)=x^s\). Then \(\{0_{V}\}\subsetneq \ker f\subsetneq \ker f^2\subsetneq...\subsetneq\ker f^s=V\)

Proof

Suppose \(\exists i,1\leq i<s\) and \(\ker f^i=\ker f^{i+1}\) (Hypothesis). Then \(\ker f^{i+1}=\ker f^{i+2}\) (NTP)

We know that \(\ker f^{i+1}\subseteq \ker f^{i+2}\), suppose \(\ker f^{i+1}\subsetneq \ker f^{i+2}\).

Let \(v\in \ker^{i+2}\setminus \ker f^{i+1}\), then \(0=f^{i+2}(v)=f^{i+1}(f(v))\Rightarrow f(v)\in \ker f^{i+1}=\ker f^i\) by hypothesis

So \(v\in \ker f^{i+1}\), which is a contradiction. So \(\ker f^{i+1}=\ker f^{i+2}\)

So let \(v\in \ker f^s\setminus \ker f^{s-1}\). Consider \(\{v,f(v),f^{2}(v),...,f^{s-1}(v)\in\ker f\}\) is linear independent why?

Suppose \(\exists a_{0},...,a_{s-1}\in k:0=a_{0}v+a_{1}f(v)+...+a_{s-1}f^{s-1}(v)\)

Then \(f^{s-1}(0)=a_{0}f^{s-1}(v)+a_{1}f^{s}(v)+...+a_{s-1}f^{2s-1}(v)\), then \(0=a_{0}f^{s-1}(v)\) and since \(f^{s-1}(v)\neq 0\), then \(a_0=0\)

By using \(f^{s-j},1\leq j\leq s-1\), we get that \(a_i=0,\forall i\)

Thus it is linear independent

Particular case

Suppose \(\dim V=n\), \(\chi_f(x)=x^n\). Suppose \(m_f(x)=x^n\), by lemma \(\{0_{V}\}\subsetneq \ker f\subsetneq \ker f^2\subsetneq...\subsetneq\ker f^s=V\)

Here we have \(n\) subspaces and for \(i:1\leq i\leq s-1\), \(\dim\ker f^i<\dim \ker f^{i+1}\)

Then the \(\dim\) of those kernel is increasing exact by 1

General case

\(f\) nilpotent and \(\chi_f(x)=x^n,m_f(x)=x^s\) where \(1\leq s\leq n\). By lemma: \(\{0_{V}\}\subsetneq \ker f\subsetneq \ker f^2\subsetneq...\subsetneq\ker f^s=V\)

We will construct bases of \(\ker f^{s},\ker f^{s-1},...,\ker f\)

Let's start with an any basis \(B_{s-1}\) of \(\ker f^{s-1}\). Let \(C_s\) be such that \(B_{s-1}\cup C_s\) is a basis of \(\ker f^s=V\)

Where \(C_{s}=\{v_{1}^{\left(s\right)},v_{2}^{\left(s\right)},\ldots,v_{r_{s}}^{\left(s\right)} \rbrace\) this is index not power. So \(\ker f^{s-1}\oplus \lang C_{s}\rang=\ker f^{s}=V\)

Now, Let prove \(f(C_s)\subseteq \ker f^{s-1}\) is linearly independent

Suppose \(0_{V}=\sum_{j=1}^{r_{s}}\lambda_{j}f\left(v_{j}^{(s)}\right)=f\left(\sum_{j=1}^{r_{s}} \lambda_{j}v_{j}^{(s)}\right)\) with \(\lambda_{j}\in k\)

So \(\sum_{j=1}^{r_{s}}\lambda_{j}v_{j}^{(s)}\in\ker f\subseteq\underbrace{\ker f^{s-1}} _{\text{has basis }B_{s-1}}\), so \(\sum_{j=1}^{r_{s}}\lambda_{j}v_{j}^{(s)}\) is a linear combination of the vectors in \(B_{s-1}\)

If \(\exists j:\lambda_j\neq 0\), then we will have a non-zero linear combination vectors of \(C_s\) in \(B_{s-1}\)

And since \(\ker f^{s-1}\oplus\langle C_{s}\rangle\), then \(\ker f^{s-1}\cap\langle C_{s}\rangle=\left\lbrace0_{V}\right\rbrace\), then that linear combination should be \(0_V\)

And since \(v_{1}^{\left(s\right)},v_{2}^{\left(s\right)},\ldots,v_{r_{s}}^{\left(s\right)}\) are all not zero, then \(\lambda_j\) has to be zero. Contradiction, thus \(\lambda_j=0,\forall j\)

Remark

\(\lang f(C_s)\rang\cap \ker f^{s-2}=\{0_V\}\)

Exercise

Since \(v_{i}^{(s)}\in C_s\) and \(v_i^{(s)}\in \ker f^{s}\setminus\ker f^{s-1}\), then \(f\left(v_i^{(s)}\right)\in \ker f^{s-1}\setminus \ker f^{s-2}\)​