Let \(A\) be an \(n\times n\) matrix over a field \(k\). The Jordan canonical form of \(A\) exists if and only if the characteristic polynomial \(\chi_A\) splits over \(k\); the rational canonical form exists over every field.
Make \(V=k^n\) a \(k[x]\)-module with \(x\) acting by \(A\). If \(\chi_A\) splits, the elementary divisors of \(V\) have the form \((x-\lambda)^e\), and the Jordan form of \(A\) has one Jordan block \(J_e(\lambda)\) for each elementary divisor \((x-\lambda)^e\).
Computing the Jordan canonical form and a change of basis via Jordan chains (Tag PR-5A2W4)
Let \(k\) be a field and \(A \in \Mat_{n\times n}(k)\), and suppose that the characteristic polynomial factors as \(\chi_A(x) = \pm\prod_{i=1}^r (x-\lambda_i)^{m_i}\) with \(\lambda_1, \ldots, \lambda_r \in k\) distinct. For each \(i\), put \(N_i \coloneqq A - \lambda_i I\). The following procedure produces \(P \in \GL_n(k)\) such that \(\inverseof{P} A P = \JCF(A)\) is a Jordan canonical form of \(A\).
Let \(\ell_i\) be the least integer \(\ell \geq 1\) with \(\rank N_i^{\ell} = \rank N_i^{\ell+1}\). Then \[ n = \rank N_i^0 > \rank N_i > \cdots > \rank N_i^{\ell_i} = \rank N_i^{\ell_i+1} = \rank N_i^{\ell_i+2} = \cdots, \] \(\dim_k \ker N_i^{\ell_i} = m_i\), and \(\ell_i\) is the size of the largest Jordan block with eigenvalue \(\lambda_i\).
The eigenspace \(E_{\lambda_i} = \ker N_i\) has dimension \(n - \rank N_i\), the number of Jordan blocks with eigenvalue \(\lambda_i\). For \(0 \leq j \leq \ell_i\), put \(K_{i,j} \coloneqq \ker N_i \cap \im N_i^{j}\), so that \[ \ker N_i = K_{i,0} \supseteq K_{i,1} \supseteq \cdots \supseteq K_{i,\ell_i} = 0 \quad\text{and}\quad \dim_k K_{i,j} = \rank N_i^{j} - \rank N_i^{j+1}. \]
Choose a basis of \(K_{i,\ell_i - 1}\) and extend it successively to bases of \(K_{i,\ell_i-2}, \ldots, K_{i,0}\). Each basis vector \(v\) chosen at the stage of \(K_{i,s-1}\) is an eigenvector lying in \(\im N_i^{s-1}\).
For each such \(v\), solve \(N_i^{s-1} w = v\), for instance by row reducing the augmented matrix \([\,N_i^{s-1} \mid v\,]\); the system is consistent because \(v \in \im N_i^{s-1}\). The Jordan chain of \(v\) is \(N_i^{s-1}w = v,\ N_i^{s-2}w,\ \ldots,\ N_i w,\ w\).
Let \(P\) be the matrix whose columns are the Jordan chains of all the vectors chosen in step (3), for all \(i\), each chain listed consecutively in the order of step (4). A chain of length \(s\) for \(\lambda_i\) contributes a Jordan block of size \(s\) with eigenvalue \(\lambda_i\) to \(\inverseof{P} A P\).
Jordan form data from the minimal and characteristic polynomials (Tag L-CD6QT)
Let \(k\) be a field and let \(A\in\Mat_n(k)\) have characteristic polynomial \(\chi_A(x)=\det(xI-A)\) that splits over \(k\), with distinct eigenvalues \(\lambda_1,\ldots,\lambda_r\). Write \[ \chi_A(x)=\prod_{i=1}^r(x-\lambda_i)^{m_i},\qquad\min_A(x)=\prod_{i=1}^r(x-\lambda_i)^{\ell_i}, \] where \(\min_A\) is the minimal polynomial, and let \(E_{\lambda_i}=\ker(A-\lambda_iI)\) be the eigenspace and \(V^{\lambda_i}=\ker(A-\lambda_iI)^{m_i}\) the generalized eigenspace of \(\lambda_i\). The exponent \(m_i\) is the algebraic multiplicity and \(\dim E_{\lambda_i}\) the geometric multiplicity of \(\lambda_i\). In the Jordan canonical form \(\JCF(A)\):
The roots of \(\chi_A\) and of \(\min_A\) are exactly the eigenvalues \(\lambda_i\), and \(1\le\ell_i\le m_i\).
\(\ell_i\) is the size of the largest Jordan block with eigenvalue \(\lambda_i\), and it is the least \(j\ge1\) with \(\ker(A-\lambda_iI)^j=\ker(A-\lambda_iI)^{j+1}\).
\(m_i\) is the sum of the sizes of the Jordan blocks with eigenvalue \(\lambda_i\), which is the number of times \(\lambda_i\) appears on the diagonal of \(\JCF(A)\), and \(m_i=\dim V^{\lambda_i}\).
\(\dim E_{\lambda_i}\) is the number of Jordan blocks with eigenvalue \(\lambda_i\).
\(A\) is diagonalizable over \(k\) if and only if \(\dim E_{\lambda_i}=m_i\) for every \(i\), if and only if \(\ell_i=1\) for every \(i\).
Choose a basis in which \(A\) is the block diagonal matrix \(\JCF(A)\), and let \(J=J_s(\lambda)\) be one of its Jordan blocks, of size \(s\). Then \(J-\lambda I\) sends the \(j\)th basis vector of the block to the \((j-1)\)st and the first to \(0\), so \(\dim\ker(J-\lambda I)^j=\min(j,s)\), while \(J-\mu I\) is invertible for \(\mu\ne\lambda\). In particular \(\chi_J(x)=(x-\lambda)^s\), and \((x-\lambda)^j\) kills \(J\) exactly when \(j\ge s\), so \(\min_J(x)=(x-\lambda)^s\).
The characteristic polynomial of a block diagonal matrix is the product of those of the blocks, which gives 3 and the first part of 1; the minimal polynomial is their least common multiple, which gives the first part of 2. The inequality \(\ell_i\le m_i\) is Cayley–Hamilton: \(\min_A\) divides \(\chi_A\). Summing \(\dim\ker(J-\lambda_iI)^j=\min(j,s)\) over the blocks with eigenvalue \(\lambda_i\) shows that \(\dim\ker(A-\lambda_iI)^j\) increases strictly for \(j<\ell_i\) and is constant for \(j\ge\ell_i\), and that \(\dim V^{\lambda_i}=m_i\); taking \(j=1\) gives 4, since the first basis vectors of the blocks with eigenvalue \(\lambda_i\) form a basis of \(E_{\lambda_i}\). Finally, \(A\) is diagonalizable if and only if every Jordan block has size \(1\), which by 2, 3, and 4 is each of the two conditions in 5.
Jordan blocks from the minimal and characteristic polynomials
(From the two polynomials)
Suppose \(A\) is \(5\times 5\) with \[ \begin{aligned} \min_A(t) &= (t-4)^2(t+6), \\ \chi_A(t) &= (t-4)^3(t+6)^2. \end{aligned} \]
For \(\lambda = 4\), the block sizes sum to \(3\) and the largest is \(2\), so the blocks are \(J_2(4) \oplus J_1(4)\). For \(\lambda = -6\), the block sizes sum to \(2\) and the largest is \(1\), so the blocks are \(J_1(-6)\oplus J_1(-6)\).
(The two polynomials do not determine the form)
For \(4\times 4\) matrices with \(\min_A(t) = t^2\) and \(\chi_A(t) = t^4\) there are two similarity classes, \(J_2(0)\oplus J_2(0)\) and \(J_2(0)\oplus J_1(0)\oplus J_1(0)\). They are distinguished by \(\dim\ker A\), which is \(2\) and \(3\) respectively. In general, \(\dim\ker(A-\lambda I)^k-\dim\ker(A-\lambda I)^{k-1}\) is the number of Jordan blocks for \(\lambda\) of size at least \(k\).
Existence
(Sketch of existence)
Assume \(\chi_A\) splits over \(k\), and let \(f\colon V\to V\) be the linear map given by \(A\).
\(V\) is a direct sum of \(f\)-invariant subspaces on each of which \(f\) is indecomposable, and \(f\) is block diagonal with respect to such a decomposition, so it suffices to treat indecomposable \(f\).
By Fitting’s lemma, for every linear map \(h\colon V\to V\) there is \(m\geq 1\) with \(V = \ker h^m \oplus \im h^m\), and both summands are \(h\)-invariant.
Let \(\lambda\) be an eigenvalue of \(f\) with eigenvector \(v\), and apply Fitting’s lemma to \(h\da f-\lambda I\). Since \(v\in\ker h^m\) and \(f\) is indecomposable, \(V = \ker h^m\), so \(h\) is nilpotent; let \(k\) be its nilpotency index.
Choose \(w\) with \(h^{k-1}w \neq 0\). The span \(W\) of \(w, hw, \ldots, h^{k-1}w\) is \(h\)-invariant, these vectors are linearly independent, and \(W\) has an \(h\)-invariant complement; since \(f\) is indecomposable, \(W = V\) and \(k=\dim V\).
In the basis \(h^{k-1}w,\ldots,hw,w\), the matrix of \(f = \lambda I + h\) is the Jordan block \(J_k(\lambda)\).
Generalized eigenspaces
Elementary divisors are the minimal polynomials of the Jordan blocks (Tag L-W5S2W)
Let \(k\) be a field and let \(A\in\Mat_n(k)\) have characteristic polynomial that splits over \(k\), so that \(A\) is similar to its Jordan canonical form \(\JCF(A)\). Then the elementary divisors of \(A\), the prime-power factors in the decomposition of the \(k[x]\)-module \(k^n\) with \(x\) acting as \(A\), are the minimal polynomials \((x-\lambda)^s\) of the Jordan blocks \(J_s(\lambda)\) of \(\JCF(A)\), listed with multiplicity.
A Jordan block \(J_s(\lambda)\) is the matrix of multiplication by \(x\) on \(k[x]/((x-\lambda)^s)\) in the basis \((x-\lambda)^{s-1},\ldots,(x-\lambda),1\), and its minimal polynomial is \((x-\lambda)^s\). So the block decomposition of \(\JCF(A)\) is a decomposition \(k^n\cong\bigoplus k[x]/((x-\lambda)^s)\), one summand for each block, and the elementary divisors are unique up to order.
(The module structure)
Make \(V\) a \(k[x]\)-module by \(p(x)\actson \vector v \da p(A)\vector v\), and for \(\vector v\in V\) let \[ \Ann(\vector v) \da \ts{ q(x) \in k[x] \st q(A)\vector v = 0}. \] A nonzero \(\vector w\) is an eigenvector with eigenvalue \(\lambda_i\) if and only if \(x - \lambda_i \in \Ann(\vector w)\), and a generalized eigenvector for \(\lambda_i\) if and only if \[ (x-\lambda_i)^k\in \Ann(\vector w) \text{ for some } k \iff x-\lambda_i \in \sqrt{\Ann(\vector w)}. \] The generalized eigenspace for \(\lambda_i\) is \[ \begin{aligned} V^{\lambda_i} &\da \ts{\vector v\in V \st (A-\lambda_i I)^m \vector v = 0 \text{ for some }m } \\ &= \ts{\vector v\in V \st x-\lambda_i \in \sqrt{\Ann(\vector v)} }, \end{aligned} \] and \(V^{\lambda_i} = \ker (A-\lambda_i I)^n\) for \(n \da \dim V\). If \(\chi_A\) splits, then \(V = \bigoplus_i V^{\lambda_i}\).
(of the generalized eigenspace decomposition)
Write \(\chi_A(x) = \prod_i (x-\lambda_i)^{n_i}\) with the \(\lambda_i\) distinct, and set \(V^{j} \da \ker (A-\lambda_j I)^n\).
For each \(j\), let \(h_j(x) = \prod_{i\neq j}(x-\lambda_i)^{n_i}\) and \(W^j \da \im(h_j(A))\).
\(W^j \subseteq \ker (A - \lambda_j I)^{n_j}\subseteq V^j\), since \((A-\lambda_j I)^{n_j} h_j(A) = \chi_A(A) = 0\) by Cayley–Hamilton.
\(\sum_j W^j = V\): the \(h_j\) have no common factor, so there are \(f_j\in k[x]\) with \(\sum_j f_j h_j = 1\), hence \(\sum_j f_j(A)h_j(A) = I\) and \(\vector v = \sum_j f_j(A)h_j(A)\vector v\in\sum_j W^j\) for every \(\vector v\).
The sum is direct: \(h_j(A)W^i = 0\) for \(i\neq j\), because \((x-\lambda_i)^{n_i}\) divides \(h_j\). Hence for \(\vector w\in W^i\), \(\vector w = \sum_j f_j(A)h_j(A)\vector w = f_i(A)h_i(A)\vector w\), and applying \(f_i(A)h_i(A)\) to a relation \(0 = \sum_j \vector w_j\) with \(\vector w_j\in W^j\) gives \(\vector w_i = 0\).
\(V^j = W^j\): for \(i\neq j\), \(A-\lambda_j I = (A-\lambda_i I) + (\lambda_i-\lambda_j)I\) is the sum of a nilpotent operator and a nonzero scalar on the invariant subspace \(W^i\), hence invertible there. If \(\vector v = \sum_i \vector w_i\in V^j\), then \(0=(A-\lambda_j I)^n\vector v = \sum_i (A-\lambda_j I)^n\vector w_i\) with each term in \(W^i\), so \(\vector w_i = 0\) for \(i\neq j\) and \(\vector v\in W^j\).
Exercises
Jordan form of $\begin{pmatrix}1&-1&0\\-1&4&-1\\-4&13&-3\end{pmatrix}$ (Tag E-23SLE)
Compute \(\JCF(A)\) for \[ A \da \mattt{1}{-1}{0}{-1}{4}{-1}{-4}{13}{-3}. \]
Solution
<1>1. The characteristic polynomial of \(A\) is \[ \chi_A(t)=t(t-1)^2. \]
Direct expansion gives \[ \det(tI-A) = \det\begin{pmatrix} t-1&1&0\\ 1&t-4&1\\ 4&-13&t+3 \end{pmatrix} =(t-1)(t^2-t+1)-(t-1) =t(t-1)^2. \]
<1>2. The eigenspace for the eigenvalue \(0\) is one-dimensional.
The eigenvalue \(0\) has algebraic multiplicity \(1\) by <1>1, so its geometric multiplicity is also \(1\).
<1>3. The eigenspace for the eigenvalue \(1\) is one-dimensional.
One has \[ A-I= \begin{pmatrix} 0&-1&0\\ -1&3&-1\\ -4&13&-4 \end{pmatrix}. \] The first two rows are linearly independent, while the third row equals \(-13\) times the first row plus \(4\) times the second row. Hence \(\operatorname{rank}(A-I)=2\), so by rank-nullity \[ \dim\ker(A-I)=1. \]
<1>4. The Jordan blocks for the eigenvalue \(1\) consist of one block of size \(2\).
By <1>1, the eigenvalue \(1\) has algebraic multiplicity \(2\). By <1>3, its eigenspace has dimension \(1\), which equals the number of Jordan blocks for that eigenvalue. Therefore there is exactly one such block, and its total size must be \(2\).
<1>5. Therefore \[ \JCF(A)=J_2(1)\oplus J_1(0), \] up to permutation of Jordan blocks; explicitly, \[ \JCF(A)\sim \begin{pmatrix} 1&1&0\\ 0&1&0\\ 0&0&0 \end{pmatrix}. \]
Combine <1>2 and <1>4.
Cayley-Hamilton via Jordan canonical form (Tag E-I7ZD4)
Prove Cayley-Hamilton using the JCF.
Solution
Let \(A\in\Endo_k(V)\) with \(V\) finite-dimensional.
<1>1. It is enough to prove the identity after extending scalars to an algebraic closure \(\overline{k}\).
Set \[ \overline{V}=V\otimes_k\overline{k}, \qquad \overline{A}=A\otimes1. \] The characteristic polynomial is unchanged by scalar extension, so \(\chi_{\overline A}(t)=\chi_A(t)\) viewed in \(\overline{k}[t]\). Moreover, the natural map \[ \Endo_k(V)\longrightarrow\Endo_{\overline{k}}(\overline V), \qquad T\longmapsto T\otimes1, \] is injective. Hence if \(\chi_A(A)\otimes1=0\), then \(\chi_A(A)=0\) already over \(k\).
<1>2. Over \(\overline{k}\), choose a basis in which \(\overline A\) has Jordan form \[ J=\bigoplus_r J_{m_r}(\lambda_r). \]
The characteristic polynomial splits over \(\overline{k}\), so the Jordan canonical form theorem applies. Similarity preserves polynomial evaluation: if \(\overline A=PJP^{-1}\), then \[ p(\overline A)=Pp(J)P^{-1} \] for every polynomial \(p\).
<1>3. On a Jordan block \(J_m(\lambda)\) one has \[ (J_m(\lambda)-\lambda I)^m=0. \]
Write \[ J_m(\lambda)=\lambda I+N, \] where \(N\) has \(1\) on the superdiagonal and \(0\) elsewhere. The matrix \(N\) shifts each basis vector at most one step along the Jordan chain, so \(N^m=0\). Thus \[ (J_m(\lambda)-\lambda I)^m=N^m=0. \]
<1>4. The characteristic polynomial \(\chi_J\) annihilates every Jordan block of \(J\).
For each eigenvalue \(\lambda\), let \(a_\lambda\) be its algebraic multiplicity. Then \[ \chi_J(t)=\prod_\lambda(t-\lambda)^{a_\lambda}. \] If \(J_m(\lambda)\) is one of the blocks, then \(m\le a_\lambda\). Hence the factor \[ (J_m(\lambda)-\lambda I)^{a_\lambda} \] is zero by <1>3. Since all factors in the polynomial \(\chi_J(J_m(\lambda))\) commute, this gives \[ \chi_J(J_m(\lambda))=0. \] Therefore \(\chi_J(J)=0\) block by block.
<1>5. Hence \(\chi_A(A)=0\).
By <1>2 and <1>4, \[ \chi_{\overline A}(\overline A)=0. \] Using <1>1, \[ \chi_A(A)\otimes1=0 \] implies \(\chi_A(A)=0\) over the original field \(k\).
Rank-nullity theorem via Jordan canonical form (Tag E-GEVBZ)
Let \(T:V\to W\) be a linear map of vector spaces over a field, with \(V\) finite-dimensional. Prove using Jordan canonical form that \[ \dim V=\dim\ker T+\dim\operatorname{im}T. \]
Solution
Set \(U=\operatorname{im}T\), which is finite-dimensional because \(V\) is finite-dimensional.
<1>1. Define an endomorphism \(S\) of the finite-dimensional space \(V\oplus U\) by \[ S(v,u)=(0,Tv). \] Then \(S^2=0\).
For \((v,u)\in V\oplus U\), \[ S^2(v,u)=S(0,Tv)=(0,T0)=(0,0). \] Thus the minimal polynomial of \(S\) divides \(x^2\). In particular it splits over the base field, so \(S\) has Jordan canonical form consisting only of nilpotent Jordan blocks of sizes \(1\) and \(2\).
<1>2. For every Jordan block of \(S\), its rank plus its nullity equals its size.
A size-\(1\) block is \(J_1(0)=[0]\), which has rank \(0\) and nullity \(1\). A size-\(2\) block is \[ J_2(0)=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \] which has rank \(1\) and nullity \(1\). Thus each block contributes its full dimension to rank plus nullity. Since rank and nullity add across block-diagonal direct sums, \[ \operatorname{rank}S+\operatorname{nullity}S=\dim(V\oplus U). \]
<1>3. The image and kernel of \(S\) are \[ \operatorname{im}S=\{0\}\oplus U, \qquad \ker S=\ker T\oplus U. \]
By definition, \[ \operatorname{im}S=\{(0,Tv):v\in V\}=\{0\}\oplus\operatorname{im}T=\{0\}\oplus U. \] Also \(S(v,u)=0\) exactly when \(Tv=0\), with no restriction on \(u\in U\). Hence \[ \ker S=\ker T\oplus U. \]
<1>4. Therefore \[ \dim V=\dim\ker T+\dim\operatorname{im}T. \]
By <1>3, \[ \operatorname{rank}S=\dim U, \qquad \operatorname{nullity}S=\dim\ker T+\dim U. \] Substituting these into <1>2 gives \[ \dim V+\dim U =\dim U+\dim\ker T+\dim U. \] Cancelling \(\dim U\) from both sides gives \[ \dim V=\dim\ker T+\dim U =\dim\ker T+\dim\operatorname{im}T. \]
Jordan form of $\begin{pmatrix}5&-1\\9&-1\end{pmatrix}\oplus\begin{pmatrix}7&-2\\12&-3\end{pmatrix}$ (Tag E-2KJEL)
Determine \(\operatorname{JCF}(B)\) for \[ B := \begin{pmatrix} 5 & -1 & 0 & 0 \\ 9 & -1 & 0 & 0 \\ 0 & 0 & 7 & -2 \\ 0 & 0 & 12 & -3 \end{pmatrix} .\]
Solution
Write \[ B=B_1\oplus B_2, \qquad B_1=\begin{pmatrix}5&-1\\9&-1\end{pmatrix}, \quad B_2=\begin{pmatrix}7&-2\\12&-3\end{pmatrix}. \]
For \(B_1\), \[ \chi_{B_1}(t)=(t-5)(t+1)+9=(t-2)^2. \] Moreover \[ B_1-2I=\begin{pmatrix}3&-1\\9&-3\end{pmatrix} \] has rank \(1\), so the eigenspace for \(2\) is one-dimensional. Hence \(B_1\) contributes one Jordan block \(J_2(2)\).
For \(B_2\), \[ \chi_{B_2}(t)=(t-7)(t+3)+24=t^2-4t+3=(t-1)(t-3), \] so \(B_2\) has two distinct eigenvalues and is diagonalizable.
Therefore, up to ordering of Jordan blocks, \[ \boxed{ \operatorname{JCF}(B)= \begin{pmatrix} 2&1&0&0\\ 0&2&0&0\\ 0&0&1&0\\ 0&0&0&3 \end{pmatrix}.} \]