References
Jordan canonical form exercises (Math 122 homework 8, Spring 2016)
Computing a Jordan canonical form: an example
Jordan canonical form and the minimal and characteristic polynomials
Let \(F\) be an algebraically closed field, \(A\in \Mat_n(F)\), and \(\operatorname{Spec}(A)\) the set of eigenvalues of \(A\). Write \(J(d,\lambda)\) for the Jordan block of size \(d\) with eigenvalue \(\lambda\), and call it a Jordan \(\lambda\)-block.
- For \(\lambda\in F\), \(\lambda\in\operatorname{Spec}(A)\) if and only if the Jordan canonical form of \(A\) contains at least one Jordan \(\lambda\)-block.
- The characteristic polynomial is \(\chi_A(x)=\prod_{\lambda\in\operatorname{Spec}(A)}(x-\lambda)^{s_\lambda}\), where \(s_\lambda\) is the sum of the sizes of all Jordan \(\lambda\)-blocks.
- The minimal polynomial is \(\mu_A(x)=\prod_{\lambda\in\operatorname{Spec}(A)}(x-\lambda)^{m_\lambda}\), where \(m_\lambda\) is the maximum size of a Jordan \(\lambda\)-block.
For \(\lambda\in F\) and \(k\geq 1\), the number \(n_A(k,\lambda)\) of Jordan \(\lambda\)-blocks of size at least \(k\) is \[ n_A(k,\lambda)=\rank\big((A-\lambda I)^{k-1}\big)-\rank\big((A-\lambda I)^{k}\big). \]
Let \(T\) be a nilpotent operator on an \(n\)-dimensional vector space \(V\) with \(T^{n-1}\neq 0\).
- For \(0\leq k\leq n\), \(\im T^k=\ker T^{n-k}\).
- If \(v_{n-1}\notin \im T=\ker T^{n-1}\) and \(v_i=Tv_{i+1}\) for \(n-2\geq i\geq 0\), then \(\{v_0,\dots,v_{n-1}\}\) is a Jordan basis for \(T\).
Practice
Let \[ A=\begin{bmatrix}1&1&0\\0&1&2\\0&0&3\end{bmatrix}. \] The characteristic polynomial is \((3-\lambda)(1-\lambda)^2\), so the eigenvalues are \(\lambda_1=3\) and \(\lambda_2=1\), with eigenvectors \(\mathbf v_1=(1,2,2)\) and \(\mathbf v_2=(1,0,0)\). A generalized eigenvector for \(\lambda_2\) is a vector \(\mathbf v_3\neq\mathbf 0\) with \((A-\lambda_2I)^2\mathbf v_3=\mathbf 0\) and \((A-\lambda_2 I)\mathbf v_3\neq\mathbf 0\); take \(\mathbf v_3=(0,1,0)\), for which \((A-\lambda_2I)\mathbf v_3=\mathbf v_2\).
Let \[ A=\begin{bmatrix}1&2&0\\1&1&2\\0&-1&1\end{bmatrix}. \] The only eigenvalue is \(\lambda=1\), with the single eigenvector \(\mathbf v_1=(-2,0,1)\) up to scaling. Since \[ (A-I)^2=\begin{bmatrix}2&0&4\\0&0&0\\-1&0&-2\end{bmatrix}, \] \(\mathbf v_2=(0,1,0)\) is a generalized eigenvector, and since \((A-I)^3=\mathbf 0\), \(\mathbf v_3=(1,0,0)\) completes a basis of generalized eigenvectors.
Let \[ A=\begin{pmatrix}1&1\\0&1\end{pmatrix}. \] The only eigenvalue is \(\lambda=1\), of algebraic multiplicity \(m=2\), and \(A\) is in Jordan normal form but not diagonal, so it is not diagonalizable. The null space of \(A-\lambda I\) has dimension \(p=1\), so there are \(m-p=1\) generalized eigenvectors of rank greater than \(1\). The eigenvector is \(\mathbf v_1=(1,0)^T\). Solving \((A-\lambda I)\mathbf v_2=\mathbf v_1\), that is, \[ \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}v_{21}\\v_{22}\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}, \] gives \(v_{22}=1\) with \(v_{21}\) unrestricted, so the generalized eigenvectors of rank \(2\) are \(\mathbf v_2=(a,1)^T\) for scalars \(a\).