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4.28 Jordan basis and form

Let \(A\in M_n(k)\) and \(J_A\) is Jordan form of \(A\), \(\exists P\in Gln(k)\) such that \(A=P\cdot J_A\cdot P^{-1}\) where \(J_A=D+N\)

For example if image​

\(A^{2}=PJ_{A}P^{-1}PJ_{A}P^{-1}=PJ_{A}^{2}P^{-1}\). In general by induction \(A^{r}=PJ_{A}^{r}P^{-1}=P\left(D+N\right)^{r}P^{-1}\)

Moreover we know \(DN=ND\), then \((D+N)^{r}\overset{\text{since }DN=ND}{=}\sum^{r}_{i=0}\binom{r}{i}D^{r-i}N^i\)

image​

In the example, \(A^{2}=PJ_{A}^{2}P^{-1}=P(D+N)^{2}P^{-1}=P(D^2+2DN+N^2)P^{-1}\)

\(=\)image​

Then \(A^{3}=P(D^{3}+3D^{2}N+3DN^{2}+N^{3})P^{-1}=P(D^{3}+3D^{2}N)P^{-1}\)

Example

Let \(f:\R^6\to \R^6\) and \(\R\)-linear be such that

  1. \(\ker (f-3Id)^2\subsetneq \ker (f-3Id)^3\)​

  2. \(\dim\ker (f-Id)^2=1\)​

  3. \((f-3Id)^4(f-Id)^2=0\)​

Find all possible Jordan forms of \(f\)

\(\chi_{f}(x)=(x-3)^{4}(x-1)^{2}\), then \(m_{f}(x)=(x-3)^{3}(x-1)\) or \((x-3)^4(x-1)\)

Notes on board:

From 1, we get \(\left\lbrace0\right\rbrace\subsetneq\ker(f-Id)\subsetneq\ker(f-3Id)^{2}\subsetneq \ker(f-3Id)^{3}\), then \(3\) is an eigenvalue and \(m_f(x)\) is divisible by \((x-3)^3\)

From 2, we get \(\{0\}\subseteq\ker(f-Id)\subseteq \underbrace{\ker(f-Id)^2}_{ \dim =1}\), then \(\{0\}\subsetneq \ker(f-Id)\)

Then \(\dim\ker(f-Id)\geq 1\) and \(\dim\ker(f-Id)\leq 1\), then \(\dim\ker(f-Id)=1\)

Then \(2\) is an eigenvalue

From 3, \(m_f(x)\mid (x-3)^4(x-1)^2\), so the only eigenvalue is \(3\) and \(1\)

Then \(m_{f}(x)=(x-3)^{3}(x-1)\) or \((x-3)^4(x-1)\)

Then \(\chi_f(x)=(x-3)^i(x-1)^j\) where \(i\geq 3\) and \(i+j=6\) or \(\chi_{f}(x)=(x-3)^{i}(x-1)^{j}\) where \(i\geq4\) and \(i+j=6\)

Since \(\dim(\ker f-Id)=1\), we have that there is only one \(1\) in the diagonal, so \(j=1\) and this coincides with the algebraic multiplicity

image​

Example

Given \(A\in M_5(\mathbb{C})\) such that

  1. \(x^{2}-6x+9\mid m_A(x)\)​

  2. \(\text{rank} (A+Id)=3\)​

  3. \((A-3Id)^3(A+Id)^3(A-Id)=0\)​

Find all possible Jordan forms of \(A\)

Solution

Since 1, then \(3\) is an eigenvalue of \(A\) and we have an elementary Jordan block eigen.3 if size \(k\times k\) with \(k\geq 2\)

Then \(m_A(x)=(x-3)^j\) where \(j\geq 2\)

Since 2, \(5=\dim\mathbb{C}^{5}=\dim\ker(A+Id)+\dim\text{Im}(A+Id)=\dim\ker(A+Id)+\text{rank} (A+Id)=3\), \(\dim\ker (A+Id)=2\)

So there are 2 elementary Jordan blocks with eigenvalue \(-1\). Then \(m_A(x)=(x-3)^j(x+1)^i\) where \(j\geq 2\) and \(i\geq 1\)

Since 3, we get that \(m_A(x)\mid (x-3)^3(x+1)^3(x-1)\), so the possible eigenvalue is \(3,-1,1\)

So \(m_A(x)=(x-3)^i(x+1)^j(x-1)^k\) where \(i\geq 2, i\geq 1,k\geq 0\)

\(\chi_{f}(x)\mid(x-3)^{3}(x+1)^{3}(x-1)\)​

  1. \(\chi_{f}(x)=(x-3)^{2}(x+1)^{2}(x-1)\) and \(m_f(x)=(x-3)^2(x+1)(x-1)\)​

  2. \(\chi_{f}(x)=(x-3)^{2}(x+1)^{3}\) and \(m_{f}(x)=(x-3)^{2}(x+1)^{2}\)

  3. \(\chi_{f}(x)=(x-3)^{3}(x+1)^{2}\) and \(m_{f}(x)=(x-3)^{2}(x+1)\)

  4. \(\chi_{f}(x)=(x-3)^{3}(x+1)^{2}\) and \(m_{f}(x)=(x-3)^{3}(x+1)\)

Example

\(A= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{pmatrix}\), find \(J_A\) and Jordan basis

\(\chi_{A}(x)=\det \begin{pmatrix} x & -1 & 0 & 0 \\ 0 & x & -1 & 0 \\ 0 & 0 & x & -1 \\ -1 & 0 & 0 & x \end{pmatrix}=x\det \begin{pmatrix} x & -1 & 0 \\ 0 & x & -1 \\ 0 & 0 & x \end{pmatrix}+\det \begin{pmatrix} -1 & 0 & 0 \\ x & -1 & 0 \\ 0 & x & -1 \end{pmatrix}=x^{4}-1=\left(x+1\right)\left(x-1\right)\left(x+i\right)\left(x-i\right )\) which is diagonalizable

\(E_{-1}=\ker \begin{pmatrix} -1 & -1 & 0 & 0 \\ 0 & -1 & -1 & 0 \\ 0 & 0 & -1 & -1 \\ -1 & 0 & 0 & -1 \end{pmatrix}=\ker \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix}=\text{span}\{\left(-1,1,-1,1\right)\}\)​

\(E_{1}=\ker \begin{pmatrix} 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 \\ 1 & 0 & 0 & -1 \end{pmatrix}=\)....

Example

\(A= \begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{pmatrix}\), find \(J_A\) and Jordan basis

\(\chi_A(x)=(x-1)^4\) and \(V_1=\lang e_1\rang\)

image​