Useful reference: https://mattbaker.blog/2015/07/31/the-jordan-canonical-form/
- Let \(f:V\to V\), say \(f\) is decomposable iff \(V\) splits into a direct sum of \(f{\hbox{-}}\)invariant subspaces.
- In this case, \(f\) has a block form and restricts to an indecomposable map on the invariant subspaces, so only consider those.
- For any indecomposable \(h\), there is an integer \(m\geq 1\) such that \(V = \ker h^m \oplus \operatorname{im}h^m\).
- Find an eigenvalue \(fv = \lambda v\) and apply this fact to \(h\coloneqq f - \lambda I\).
- Since \(V\) is irreducible and \(v\in \ker h\), this forces \(V = \ker h^m\), making \(h\) nilpotent of some minimal degree \(k\leq m\), so \(h^k = 0\).
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Then one can find a cyclic vector \(w\) where \(h^kw = 0\) but \(h^{k-1}w \neq 0\), so \(\left\{{w, hw, \cdots, h^{k-1}w}\right\}\) is a basis for \(V\).
- That \(k=\dim V\) follows because any \(k< \dim v\) would yield a proper \(f{\hbox{-}}\)invariant subspace.
- The matrix of any cyclic vector is exactly a Jordan block.
Facts
The JCF corresponds to elementary divisors.
\todo[inline]{Make more precise..}
The following algorithm always works for computing \(\operatorname{JCF}(A)\):
- Compute and factor the characteristic polynomial as \(\chi_A(x) = \prod_{i} (x-\lambda_i)^{m_i}\).
- For each \(\lambda_i\), find the constant \(\ell_i\) such that \begin{align*} \cdots \operatorname{rank}(A-\lambda_i I)^{\ell_i - 1} > \operatorname{rank}(A-\lambda_i I)^{\ell_i} {\color{red} = } \operatorname{rank}(A-\lambda_i I)^{\ell_i+1} {\color{red} = } \operatorname{rank}(A-\lambda_i I)^{\ell_i+1} {\color{red} = } \cdots .\end{align*}
- Find as many usual eigenvectors \(\mathbf{v}_i\) as you can. The number of eigenvectors you find will be \(\dim E_{\lambda_i}\). Suppose you just get one, \(\mathbf{v}\).
- Solve the systems: \begin{align*} (A - \lambda_i I)\mathbf{v}_1 = \mathbf{v} &\implies \mathbf{v}_1 = ? \\ (A - \lambda_i I)^2\mathbf{v}_2 = \mathbf{v}_1 &\implies \mathbf{v}_2 = ? \\ &\vdots \\ ,\end{align*} which can be solved by putting the \(\mathbf{v}_i\) in an augmented matrix and computing the RREF.
- This terminates in at most \(\ell_i\) steps, and these vectors correspond to a single Jordan block.
- If there are other eigenvectors \(\mathbf{w}, \cdots\) for \(\lambda_i\), repeating this process yields a Jordan block for each of them. Assemble \(P\) by placing these \(\mathbf{v}_i\) in the appropriate columns.
Writing \(\operatorname{Spec}(A) = \left\{{(\lambda_i, m_i)}\right\}\), \begin{align*} \min_A(x) &= \prod_i (x- \lambda_i)^{\ell_i} \\ \chi_A(x) &= \prod (x- \lambda_i)^{m_i} \\ E_{\lambda_i} &= \dim(A - \lambda_i I) \end{align*}
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The roots both polynomials are precisely the eigenvalues \(\lambda_i\) of \(A\).
- \(m_i\) are the algebraic multiplicities.
- \(\dim E_{\lambda_i}\) are the geometric multiplicities.
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\(\ell_i \leq m_i\) by Cayley-Hamilton.
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\(\ell_i\) is
- The size of the largest Jordan block associated to \(\lambda_i\) 1 , and
- The “stabilizing constant”.
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\(m_i\), associated to the characteristic polynomial, is
- The sum of sizes of all Jordan blocks associated to \(\lambda_i\),
- The number of times \(\lambda_i\) appears on the diagonal of \(JCF(A)\),
- The dimension of the generalized eigenspace \(V^{\lambda_i}\).
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\(\dim E_{\lambda_i}\) is
- The number of Jordan blocks associated to \(\lambda_i\)
- The number of (usual) eigenvector associated to \(\lambda_i\), i.e. the dimension of their span.
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\(A\) is diagonalizable iff \(\dim E_{\lambda_i} = m_i\) for all \(i\).
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Eigenvectors for distinct Jordan blocks are linearly independent.
Suppose \(A\) is \(5\times 5\) with \begin{align*} \min_A(t) &= (t-4)^2(t+6) \\ \chi_A(t) &= (t-4)^3(t+6)^2 .\end{align*}
Some deductions:
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For \(\lambda = 4\):
- The total size of all blocks is 3
- The largest block is size 2
- So this yields \(J_1 \oplus J_2\).
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For \(\lambda = -6\):
- The total size of all blocks is 2
- The largest block is size 1
- So this must be \(J_1 \oplus J_1\)
The data of \(\min_A(t), \chi_A(t)\) is not enough to uniquely determine \(\operatorname{JCF}(A)\). Counterexample: there are two distinct possibilities for \(4\times 4\) matrices with \(\min_A(t) = t^2\) and \(\chi_A(t) = t^4\), namely \(J_2 \oplus J_2\) and \(J_2 \oplus J_1 \oplus J_1\).
The elementary divisors of \(A\) are the minimal polynomials of the Jordan blocks.
Writing \(\operatorname{Ann}(\mathbf{v})\) as the annihilator of \(\mathbf{v}\), a generalized eigenvector for the pair \((\lambda_i, \mathbf{v})\) for a matrix \(A\) is any operator in the space \(\sqrt{\operatorname{Ann}(\mathbf{v})}\), where we view \(V\) as a \(k[x]{\hbox{-}}\)module using \(p(x) \curvearrowright\mathbf{v} \coloneqq p(A)(\mathbf{v})\). So \begin{align*} \operatorname{Ann}(\mathbf{v}) \coloneqq\left\{{ q(x) \in k[x] {~\mathrel{\Big\vert}~}q(x) \curvearrowright\mathbf{v} = 0}\right\} = \left\{{q(x) \in k[x] {~\mathrel{\Big\vert}~}q(A)(\mathbf{v}) = 0}\right\} .\end{align*} Now use that \(\mathbf{w}\) is an eigenvector for \(A\) with eigenvalue \(\lambda_i \iff A-\lambda_i I \in \operatorname{Ann}(\mathbf{w})\), and is a generalized eigenvector iff \begin{align*} (A-\lambda_i I)^k\in \operatorname{Ann}(\mathbf{w}) &\text{ for some }k \iff A-\lambda_i I \in \sqrt{\operatorname{Ann}(\mathbf{w})} .\end{align*}
We can then write \begin{align*} V^{\lambda_i} &\coloneqq\left\{{\mathbf{v}\in V {~\mathrel{\Big\vert}~}(A-\lambda_i I)^n \mathbf{v} = 0 \text{ for some }n }\right\} \\ &= \left\{{\mathbf{v}\in V {~\mathrel{\Big\vert}~}(A-\lambda_i I)^n \in \operatorname{Ann}(\mathbf{v}) }\right\} \\ &= \left\{{\mathbf{v}\in V {~\mathrel{\Big\vert}~}A-\lambda_i I \in \sqrt{\operatorname{Ann}(\mathbf{v})} }\right\} ,\end{align*} and the theorem is that \(V \cong \bigoplus_i V^{\lambda_i}\). It also turns out that \(V^{\lambda_i} = \ker (A-\lambda_i I)^n\) for \(n\coloneqq\dim V\).
- Suppose \(\chi_A(x) = \prod (x-\lambda_i)^{n_i}\).
- Define \(V^{j} \coloneqq\ker (A-\lambda_i I)^n\) as the generalized eigenspace for each \(i\).
- Fix \(j\) and define \(h_j(x) = \prod_{i\neq j}(x-\lambda_i)^{n_i}\), the characteristic polynomial with the \(\lambda_j\) term deleted.
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Define \(W^j \coloneqq\operatorname{im}(h_j(A))\), then the claim is \(W^j \subseteq V^j\)
- This follows because \(0 = \chi_A(A) = (A-\lambda_j I)^{n_j} h_j(A)\), so in fact \(W^j \subseteq \ker (A - \lambda_j)^{n_j}\).
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Claim: \(\sum V^j = V\):
- Let \(\mathbf{v}\in V\) be arbitrary, then by Euclid’s algorithm write \(\sum_i f_i h_i = 1\) since the \(h_i\) are coprime.
- Thus \(\sum f_i(A) h_i(A) = I \implies \qty{\sum f_i(A) h_i(A)}(\mathbf{v}) = \mathbf{v} \implies \mathbf{v} \in \sum W^j\)
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Claim: the sum is direct.
- It suffices to show \(0=\sum w_i\) with \(w_i \in W^i\) implies \(w_i =0\) for all \(i\).
- Use that \(h_j(w_i) = 0\) for \(i\neq j\) since \(w_i \in W^i \coloneqq\ker(A-\lambda_I I)^{n_i}\).
- Write \(\mathbf{w}_i = \sum f_j(A) h_j(A) \mathbf{w}_i\), which collapses to \(f_i(A) h_i(A) \mathbf{w}_i\).
- So \(f_i(A) h_i(A) \qty{\sum w_i} = 0 \implies w_i = 0\).
\todo[inline]{Messy indexing.}
Exercises
Prove Cayley-Hamilton using the JCF.
Prove the rank-nullity theorem using JCF.
Compute \(\operatorname{JCF}(A)\) for \begin{align*} A \coloneqq { \begin{bmatrix} {1} & {-1} & {0} \\ {-1} & {4} & {-1} \\ {-4} & {13} & {-3} \end{bmatrix} } .\end{align*}
- \(\operatorname{det}(A) = 0\)
- \({\mathrm{tr}}(A) = 2\)
- \({\mathrm{tr}}(\bigwedge\nolimits^2 A) = 1\)
- \(\chi_A(t) = t^3 - 2t^2 + t\)
- \(e_1 = {\left[ {1,1,3} \right]}\)
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\(e_2 = {\left[ {1,0,-1} \right]}\)
- \(e_{2, 1} = {\left[ {-3,-1, 0} \right]}\).
Determine \(\operatorname{JCF}(B)\) for \begin{align*} B \coloneqq \left(\begin{array}{cccc} 5 & -1 & 0 & 0 \\ 9 & -1 & 0 & 0 \\ 0 & 0 & 7 & -2 \\ 0 & 0 & 12 & -3 \end{array}\right) .\end{align*}