11.2: The Jordan Canonical Form

Two \(m \times n\) matrices \(A\) and \(B\) are called row-equivalent if \(A\) can be carried to \(B\) using row operations and, equivalently, if \(B = UA\) for some invertible matrix \(U\). We know (Theorem [thm:006021]) that each \(m \times n\) matrix is row-equivalent to a unique matrix in reduced row-echelon form, and we say that these reduced row-echelon matrices are canonical forms for \(m \times n\) matrices using row operations. If we allow column operations as well, then \(A \to UAV = \left[ \begin{array}{cc} I_r & 0 \\ 0 & 0 \end{array} \right]\) for invertible \(U\) and \(V\), and the canonical forms are the matrices \(\left[ \begin{array}{rr} I_r & 0 \\ 0 & 0 \end{array} \right]\) where \(r\) is the \(rank \;\) (this is the Smith normal form and is discussed in Theorem [thm:005918]). In this section, we discover the canonical forms for square matrices under similarity: \(A \to P^{-1}AP\).

If \(A\) is an \(n \times n\) matrix with distinct real eigenvalues \(\lambda_{1}, \lambda_{2}, \dots, \lambda_{k}\), we saw in Theorem [thm:032952] that \(A\) is similar to a block triangular matrix; more precisely, an invertible matrix \(P\) exists such that

\[\label{eq:11_2Jordan} P^{-1}AP = \left[ \begin{array}{cccc} U_1 & 0 & \cdots & 0 \\ 0 & U_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & U_k \end{array} \right] = \diag(U_1, U_2, \dots, U_k) \]

where, for each \(i\), \(U_{i}\) is upper triangular with \(\lambda_{i}\) repeated on the main diagonal. The Jordan canonical form is a refinement of this theorem. The proof we gave of ([eq:11_2Jordan]) is matrix theoretic because we wanted to give an algorithm for actually finding the matrix \(P\). However, we are going to employ abstract methods here. Consequently, we reformulate Theorem [thm:032952] as follows:

033492 Let \(T : V \to V\) be a linear operator where \(dim \;V = n\). Assume that \(\lambda_{1}, \lambda_{2}, \dots, \lambda_{k}\) are the distinct eigenvalues of \(T\), and that the \(\lambda_{i}\) are all real. Then there exists a basis \(F\) of \(V\) such that \(M_{F}(T) = \diag(U_{1}, U_{2}, \dots, U_{k})\) where, for each \(i\), \(U_{i}\) is square, upper triangular, with \(\lambda_{i}\) repeated on the main diagonal.

Choose any basis \(B = \{\mathbf{b}_{1}, \mathbf{b}_{2}, \dots, \mathbf{b}_{n}\}\) of \(V\) and write \(A = M_{B}(T)\). Since \(A\) has the same eigenvalues as \(T\), Theorem [thm:032952] shows that an invertible matrix \(P\) exists such that \(P^{-1}AP = \diag(U_{1}, U_{2}, \dots, U_{k})\) where the \(U_{i}\) are as in the statement of the Theorem. If \(\mathbf{p}_{j}\) denotes column \(j\) of \(P\) and \(C_{B} : V \to \mathbb{R}^n\) is the coordinate isomorphism, let \(\mathbf{f}_j = C^{-1}_B (\mathbf{p}_j)\) for each \(j\). Then \(F = \{\mathbf{f}_{1}, \mathbf{f}_{2}, \dots, \mathbf{f}_{n}\}\) is a basis of \(V\) and \(C_{B}(\mathbf{f}_{j}) = \mathbf{p}_{j}\) for each \(j\). This means that \(P_{B \gets F} = \left[ C_{B}(\mathbf{f}_{j}) \right] = \left[ \mathbf{p}_{j} \right] = P\), and hence (by Theorem [thm:028683]) that \(P_{F \gets B} = P^{-1}\). With this, column \(j\) of \(M_{F}(T)\) is

\[C_F(T(\mathbf{f}_j)) = P_{F \leftarrow B}C_B(T(\mathbf{f}_j)) = P^{-1}M_B(T)C_B(\mathbf{f}_j) = P^{-1}A\mathbf{p}_j \nonumber \]

for all \(j\). Hence

\[M_F(T) = \left[ C_F(T(\mathbf{f}_j)) \right] = \left[ P^{-1}A\mathbf{p}_j \right] = P^{-1}A \left[ \mathbf{p}_j \right] = P^{-1}AP = \diag(U_1, U_2, \dots, U_k) \nonumber \]

as required.

Jordan Blocks033541 If \(n \geq 1\), define the Jordan block \(J_{n}(\lambda)\) to be the \(n \times n\) matrix with \(\lambda\)s on the main diagonal, \(1\)s on the diagonal above, and \(0\)s elsewhere. We take \(J_{1}(\lambda) = \left[ \lambda \right]\).

Hence

\[J_1(\lambda) = \left[ \lambda \right], \quad J_2(\lambda) = \left[ \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right], \quad J_3(\lambda) = \left[ \begin{array}{ccc} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{array} \right], \quad J_4(\lambda) = \left[ \begin{array}{cccc} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} \right], \quad \dots \nonumber \]

We are going to show that Theorem [thm:033492] holds with each block \(U_{i}\) replaced by Jordan blocks corresponding to eigenvalues. It turns out that the whole thing hinges on the case \(\lambda = 0\). An operator \(T\) is called nilpotent if \(T^{m} = 0\) for some \(m \geq 1\), and in this case \(\lambda = 0\) for every eigenvalue \(\lambda\) of \(T\). Moreover, the converse holds by Theorem [thm:032952]. Hence the following lemma is crucial.

033551 Let \(T : V \to V\) be a linear operator where \(dim \;V = n\), and assume that \(T\) is nilpotent; that is, \(T^{m} = 0\) for some \(m \geq 1\). Then \(V\) has a basis \(B\) such that

\[M_B(T) = \diag(J_1, J_2, \dots, J_k) \nonumber \]

where each \(J_{i}\) is a Jordan block corresponding to \(\lambda = 0\).

A proof is given at the end of this section.

Real Jordan Canonical Form033558 Let \(T : V \to V\) be a linear operator where \(dim \;V = n\), and assume that \(\lambda_{1}, \lambda_{2}, \dots, \lambda_{m}\) are the distinct eigenvalues of \(T\) and that the \(\lambda_{i}\) are all real. Then there exists a basis \(E\) of \(V\) such that

\[M_E(T) = \diag(U_1, U_2, \dots, U_k) \nonumber \]

in block form. Moreover, each \(U_{j}\) is itself block diagonal:

\[U_j = \diag(J_1, J_2, \dots, J_k) \nonumber \]

where each \(J_{i}\) is a Jordan block corresponding to some \(\lambda_{i}\).

Let \(E = \{\mathbf{e}_{1}, \mathbf{e}_{2}, \dots, \mathbf{e}_{n}\}\) be a basis of \(V\) as in Theorem [thm:033492], and assume that \(U_{i}\) is an \(n_{i} \times n_{i}\) matrix for each \(i\). Let

\[E_1 = \{\mathbf{e}_1, \dots, \mathbf{e}_{n_1}\}, \quad E_2 = \{\mathbf{e}_{n_1 + 1}, \dots, \mathbf{e}_{n_2}\}, \quad \dots, \quad E_k = \{\mathbf{e}_{n_{k-1} + 1}, \dots, \mathbf{e}_{n_k}\} \quad \nonumber \]

where \(n_{k} = n\), and define \(V_{i} = span \;\{E_{i}\}\) for each \(i\). Because the matrix \(M_{E}(T) = \diag(U_{1}, U_{2}, \dots, U_{m})\) is block diagonal, it follows that each \(V_{i}\) is \(T\)-invariant and \(M_{E_{i}}(T) = U_{i}\) for each \(i\). Let \(U_{i}\) have \(\lambda_{i}\) repeated along the main diagonal, and consider the restriction \(T : V_{i} \to V_{i}\). Then \(M_{E_{i}}(T - \lambda_{i}I_{n_{i}})\) is a nilpotent matrix, and hence \((T-\lambda_{i}I_{n_{i}})\) is a nilpotent operator on \(V_{i}\). But then Lemma [lem:033551] shows that \(V_{i}\) has a basis \(B_{i}\) such that \(M_{B_i}(T - \lambda_{i}I_{n_{i}}) = \diag(K_{1}, K_{2}, \dots, K_{t_i})\) where each \(K_{i}\) is a Jordan block corresponding to \(\lambda = 0\). Hence

\[\begin{aligned} M_{B_i}(T) &= M_{B_i}(\lambda_iI_{n_i}) + M_{B_i}(T - \lambda_iI_{n_i}) \\ &= \lambda_iI_{n_i} + \diag(K_1, K_2, \dots, K_{t_i}) = \diag(J_1, J_2, \dots, J_k)\end{aligned} \nonumber \]

where \(J_{i} = \lambda_{i}I_{f_i} + K_{i}\) is a Jordan block corresponding to \(\lambda_{i}\) (where \(K_{i}\) is \(f_{i} \times f_{i}\)). Finally,

\[B = B_{1} \cup B_{2} \cup \cdots \cup B_{k} \nonumber \]

is a basis of \(V\) with respect to which \(T\) has the desired matrix.

033630 If \(A\) is an \(n \times n\) matrix with real eigenvalues, an invertible matrix \(P\) exists such that \(P^{-1}AP = \diag(J_{1}, J_{2}, \dots, J_{k})\) where each \(J_{i}\) is a Jordan block corresponding to an eigenvalue \(\lambda_{i}\).

Apply Theorem [thm:033558] to the matrix transformation \(T_{A} : \mathbb{R}^n \to \mathbb{R}^n\) to find a basis \(B\) of \(\mathbb{R}^n\) such that \(M_{B}(T_{A})\) has the desired form. If \(P\) is the (invertible) \(n \times n\) matrix with the vectors of \(B\) as its columns, then \(P^{-1}AP = M_{B}(T_{A})\) by Theorem [thm:028841].

Of course if we work over the field \(\mathbb{C}\) of complex numbers rather than \(\mathbb{R}\), the characteristic polynomial of a (complex) matrix \(A\) splits completely as a product of linear factors. The proof of Theorem [thm:033558] goes through to give

Jordan Canonical Form033655 Let \(T : V \to V\) be a linear operator where \(dim \;V = n\), and assume that \(\lambda_{1}, \lambda_{2}, \dots, \lambda_{m}\) are the distinct eigenvalues of \(T\). Then there exists a basis \(F\) of \(V\) such that

\[M_F(T) = \diag(U_1, U_2, \dots, U_k) \nonumber \]

in block form. Moreover, each \(U_{j}\) is itself block diagonal:

\[U_j = \diag(J_1, J_2, \dots, J_{t_j}) \nonumber \]

where each \(J_{i}\) is a Jordan block corresponding to some \(\lambda_{i}\).

Except for the order of the Jordan blocks \(J_{i}\), the Jordan canonical form is uniquely determined by the operator \(T\). That is, for each eigenvalue \(\lambda\) the number and size of the Jordan blocks corresponding to \(\lambda\) is uniquely determined. Thus, for example, two matrices (or two operators) are similar if and only if they have the same Jordan canonical form. We omit the proof of uniqueness; it is best presented using modules in a course on abstract algebra.

Proof of Lemma 1

033671 Let \(T : V \to V\) be a linear operator where \(dim \;V = n\), and assume that \(T\) is nilpotent; that is, \(T^{m} = 0\) for some \(m \geq 1\). Then \(V\) has a basis \(B\) such that

\[M_B(T) = \diag(J_1, J_2, \dots, J_k) \nonumber \]

where each \(J_{i} = J_{n_{i}}(0)\) is a Jordan block corresponding to \(\lambda = 0\).

The proof proceeds by induction on \(n\). If \(n = 1\), then \(T\) is a scalar operator, and so \(T = 0\) and the lemma holds. If \(n \geq 1\), we may assume that \(T \neq 0\), so \(m \geq 1\) and we may assume that \(m\) is chosen such that \(T^{m} = 0\), but \(T^{m-1} \neq 0\). Suppose \(T^{m-1}\mathbf{u} \neq \mathbf{0}\) for some \(\mathbf{u}\) in \(V\).¹

Claim. \(\{\mathbf{u}, T\mathbf{u}, T^{2}\mathbf{u}, \dots, T^{m-1}\mathbf{u}\}\) is independent.

Proof. Suppose \(a_{0}\mathbf{u} + a_{1}T\mathbf{u} + a_{2}T^{2}\mathbf{u} + \cdots + a_{m-1}T^{m-1}\mathbf{u} = \mathbf{0}\) where each \(a_{i}\) is in \(\mathbb{R}\). Since \(T^{m} = 0\), applying \(T^{m-1}\) gives \(\mathbf{0} = T^{m-1}\mathbf{0} = a_{0} T^{m-1}\mathbf{u}\), whence \(a_{0} = 0\). Hence \(a_{1}T\mathbf{u} + a_{2}T^{2}\mathbf{u} + \cdots + a_{m-1}T^{m-1}\mathbf{u} = \mathbf{0}\) and applying \(T^{m-2}\) gives \(a_{1} = 0\) in the same way. Continue in this fashion to obtain \(a_{i} = 0\) for each \(i\). This proves the Claim.

Now define \(P = span \;\{\mathbf{u}, T\mathbf{u}, T^{2}\mathbf{u}, \dots, T^{m-1}\mathbf{u}\}\). Then \(P\) is a \(T\)-invariant subspace (because \(T^{m} = 0\)), and \(T : P \to P\) is nilpotent with matrix \(M_{B}(T) = J_{m}(0)\) where \(B = \{\mathbf{u}, T\mathbf{u}, T^{2}\mathbf{u}, \dots, T^{m-1}\mathbf{u}\}\). Hence we are done, by induction, if \(V = P \oplus Q\) where \(Q\) is \(T\)-invariant (then \(dim \;Q = n- dim \;P < n\) because \(P \neq 0\), and \(T : Q \to Q\) is nilpotent). With this in mind, choose a \(T\)-invariant subspace \(Q\) of maximal dimension such that \(P \cap Q = \{\mathbf{0}\}\).² We assume that \(V \neq P \oplus Q\) and look for a contradiction.

Choose \(\mathbf{x} \in V\) such that \(\mathbf{x} \notin P \oplus Q\). Then \(T^{m}\mathbf{x} = \mathbf{0} \in P \oplus Q\) while \(T^{0}\mathbf{x} = \mathbf{x} \notin P \oplus Q\). Hence there exists \(k\), \(1 \leq k \leq m\), such that \(T^{k}\mathbf{x} \in P \oplus Q\) but \(T^{k-1}\mathbf{x} \notin P \oplus Q\). Write \(\mathbf{v} = T^{k-1}\mathbf{x}\), so that

\[\mathbf{v} \notin P \oplus Q \quad \mbox{and} \quad T\mathbf{v} \in P \oplus Q \nonumber \]

Let \(T\mathbf{v} = \mathbf{p} + \mathbf{q}\) with \(\mathbf{p}\) in \(P\) and \(\mathbf{q}\) in \(Q\). Then \(\mathbf{0} = T^{m-1}(T\mathbf{v}) = T^{m-1}\mathbf{p} + T^{m-1}\mathbf{q}\) so, since \(P\) and \(Q\) are \(T\)-invariant, \(T^{m-1}\mathbf{p} = -T^{m-1}\mathbf{q} \in P \cap Q = \{\mathbf{0}\}\). Hence

\[T^{m-1}\mathbf{p} = \mathbf{0} \nonumber \]

Since \(\mathbf{p} \in P\) we have \(\mathbf{p} = a_{0}\mathbf{u} + a_{1}T\mathbf{u} + a_{2}T^{2}\mathbf{u} + \cdots + a_{m-1}T^{m-1}\mathbf{u}\) for \(a_{i} \in \mathbb{R}\). Since \(T^{m} = 0\), applying \(T^{m-1}\) gives \(\mathbf{0} = T^{m-1}\mathbf{p} = a_{0}T^{m-1}\mathbf{u}\), whence \(a_{0} = 0\). Thus \(\mathbf{p} = T(\mathbf{p}_{1})\) where

\[\mathbf{p}_{1} = a_{1}\mathbf{u} + a_{2}T\mathbf{u} + \cdots + a_{m-1}T^{m-2}\mathbf{u} \in P \nonumber \]

If we write \(\mathbf{v}_{1} = \mathbf{v} - \mathbf{p}_{1}\) we have

\[T(\mathbf{v}_1) = T(\mathbf{v} - \mathbf{p}_1) = T\mathbf{v} - \mathbf{p} = \mathbf{q} \in Q \nonumber \]

Since \(T(Q) \subseteq Q\), it follows that \(T(Q + \mathbb{R}\mathbf{v}_{1}) \subseteq Q \subseteq Q + \mathbb{R}\mathbf{v}_{1}\). Moreover \(\mathbf{v}_{1} \notin Q\) (otherwise \(\mathbf{v} = \mathbf{v}_{1} + \mathbf{p}_{1} \in P \oplus Q\), a contradiction). Hence \(Q \subset Q + \mathbb{R}\mathbf{v}_{1}\) so, by the maximality of \(Q\), we have \((Q + \mathbb{R}\mathbf{v}_{1}) \cap P \neq \{\mathbf{0}\}\), say

\[\mathbf{0} \neq \mathbf{p}_2 = \mathbf{q}_1 + a\mathbf{v}_1 \quad \mbox{where} \quad \mathbf{p}_2 \in P, \quad \mathbf{q}_1 \in Q, \quad \mbox{and} \quad a \in \mathbb{R} \nonumber \]

Thus \(a\mathbf{v}_{1} = \mathbf{p}_{2} - \mathbf{q}_{1} \in P \oplus Q\). But since \(\mathbf{v}_{1} = \mathbf{v} - \mathbf{p}_{1}\) we have

\[a\mathbf{v} = a\mathbf{v}_1 + a\mathbf{p}_1 \in (P \oplus Q) + P = P \oplus Q \nonumber \]

Since \(\mathbf{v} \notin P \oplus Q\), this implies that \(a = 0\). But then \(\mathbf{p}_{2} = \mathbf{q}_{1} \in P \cap Q = \{\mathbf{0}\}\), a contradiction. This completes the proof.

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1. If \(S : V \to V\) is an operator, we abbreviate \(S(\mathbf{u})\) by \(S\mathbf{u}\) for simplicity.↩
2. Observe that there is at least one such subspace: \(Q = \{\mathbf{0}\}\).↩