Why Jordan Form Is Needed
For a finite-state continuous-time Markov chain, the transition matrix at time \(t\) is \(P(t)=e^{tQ}\), where \(Q\) is the generator. When \(Q\) is diagonalizable, its eigenvalues and eigenvectors reduce the matrix exponential to scalar exponentials. But a repeated eigenvalue need not have enough independent eigenvectors for diagonalization. In that case, Jordan form provides the missing structure: its nilpotent part produces powers of \(t\), so entries of \(P(t)\) can include terms such as \(t e^{-\alpha t}\).
Throughout, let \((X_t)_{t\ge0}\) be a finite-state CTMC on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(S\). Its generator \(Q=(q_{ij})\) has nonnegative off-diagonal entries and rows summing to zero. We use row transition probabilities: \(P(t)=(p_{ij}(t))\), where \(p_{ij}(t)=\mathbb{P}(X_t=j\mid X_0=i)\). The matrix exponential solution of the Kolmogorov equations is \(P(t)=e^{tQ}\).
The Jordan canonical form theorem from linear algebra states that every finite square matrix over the complex numbers is similar to a Jordan-form matrix. We use that standard result here without reproving it. A real matrix may have complex Jordan blocks; conjugate blocks occur together, and the resulting exponential is real. The examples below need only real eigenvalues.
Proof. Define \(F(t)=e^{\lambda t}\sum_{r=0}^{m-1}t^rN^r/r!\). Differentiating the finite sum gives $$ F'(t)=\lambda F(t)+e^{\lambda t}\sum_{r=1}^{m-1}\frac{t^{r-1}}{(r-1)!}N^r =(\lambda I+N)F(t). $$ The last equality follows because \(N\sum_{r=0}^{m-1}t^rN^r/r!=\sum_{r=1}^{m-1}t^{r-1}N^r/(r-1)!\); the omitted final term is zero since \(N^m=0\). Also \(F(0)=I\). The matrix exponential is the unique solution to \(Y'(t)=J_m(\lambda)Y(t)\), \(Y(0)=I\), so \(F(t)=e^{tJ_m(\lambda)}\). For the similarity statement, the power-series definition gives \((TJT^{-1})^k=TJ^kT^{-1}\) for each integer \(k\ge0\), by cancellation of each adjacent \(T^{-1}T\). Substitution term by term into the exponential series yields \(e^{tQ}=T e^{tJ}T^{-1}\). \(\square\)
For a block of size \(m\), the highest possible polynomial power is \(t^{m-1}\): \(N^{m-1}\) has a nonzero entry, while \(N^m=0\). Thus the polynomial factors are not an approximation or an extra modeling effect. They follow exactly from the block’s nilpotent part. A repeated eigenvalue alone does not guarantee such factors: if all its Jordan blocks have size one, its contribution consists only of scalar exponentials.
Computing a Defective Generator
Worked Example: Three Sequential States with Equal Rates
Consider a CTMC on \(S=\{1,2,3\}\): state \(1\) moves to \(2\) at rate \(1\), state \(2\) moves to \(3\) at rate \(1\), and state \(3\) is absorbing. On a probability space carrying this chain, its generator is $$ Q=\begin{bmatrix}-1&1&0\\0&-1&1\\0&0&0\end{bmatrix}. $$ The row sums are \(-1+1+0=0\), \(0-1+1=0\), and \(0+0+0=0\); all off-diagonal entries are nonnegative.
The eigenvalues, read from this triangular matrix, are \(-1,-1,0\). To find the Jordan basis, take \(v_1=(1,0,0)^{\mathsf T}\), \(v_2=(0,1,0)^{\mathsf T}\), and \(v_0=(1,1,1)^{\mathsf T}\). Direct multiplication gives \(Qv_1=-v_1\), \(Qv_2=v_1-v_2\), and \(Qv_0=0\). With these vectors as the columns of \(T\), $$ T=\begin{bmatrix}1&0&1\\0&1&1\\0&0&1\end{bmatrix}, \qquad T^{-1}=\begin{bmatrix}1&0&-1\\0&1&-1\\0&0&1\end{bmatrix}. $$ The displayed inverse is verified by multiplying it by \(T\), which gives the identity. The relations for the three columns show that \(Q=TJT^{-1}\), where $$ J=\begin{bmatrix}-1&1&0\\0&-1&0\\0&0&0\end{bmatrix}. $$
For the size-two block, \(N=\begin{bmatrix}0&1\\0&0\end{bmatrix}\) and \(N^2=0\), so its exponential is \(e^{-t}\begin{bmatrix}1&t\\0&1\end{bmatrix}\). The zero block exponentiates to \(1\). Transforming back gives $$ P(t)=e^{tQ} =\begin{bmatrix}e^{-t}&t e^{-t}&1-(1+t)e^{-t}\\0&e^{-t}&1-e^{-t}\\0&0&1\end{bmatrix}. $$ For example, the first row follows from multiplying \(T\begin{bmatrix}e^{-t}&t e^{-t}&0\\0&e^{-t}&0\\0&0&1\end{bmatrix}\) by \(T^{-1}\).
Check. The row sums are \(e^{-t}+t e^{-t}+1-(1+t)e^{-t}=1\), \(e^{-t}+1-e^{-t}=1\), and \(1\). For \(t\ge0\), \(e^t\ge1+t\), so \(1-(1+t)e^{-t}\ge0\); the other entries are also nonnegative. At \(t=0\), the matrix is the identity. The term \(t e^{-t}\) is the probability of completing the first transition but not the second by time \(t\).
The Jordan calculation also explains why a direct diagonalization attempt fails here. The eigenvalue \(-1\) has algebraic multiplicity two, but only one independent eigenvector, \(v_1\). The vector \(v_2\) is a generalized eigenvector: \((Q+I)v_2=v_1\). Replacing that two-dimensional block by two ordinary eigenvectors would discard the \(t e^{-t}\) term.
Worked Example: Equal Rates Other Than One
Now let state \(1\) move to \(2\) at rate \(2\), state \(2\) move to absorbing state \(3\) at rate \(2\), and state \(3\) remain fixed. The chain is on \(S=\{1,2,3\}\), and $$ Q=\begin{bmatrix}-2&2&0\\0&-2&2\\0&0&0\end{bmatrix}. $$ Each row sums to zero: \(-2+2+0=0\), \(0-2+2=0\), and \(0+0+0=0\); its off-diagonal entries are nonnegative.
The repeated eigenvalue is \(-2\). The vectors \(v_1=(1,0,0)^{\mathsf T}\), \(v_2=(0,1,0)^{\mathsf T}\), \(v_0=(1,1,1)^{\mathsf T}\) satisfy \(Qv_1=-2v_1\), \(Qv_2=2v_1-2v_2\), and \(Qv_0=0\). Thus the Jordan block for \(-2\) has superdiagonal entry \(2\), and its exponential is \(e^{-2t}\begin{bmatrix}1&2t\\0&1\end{bmatrix}\). Transforming back, or checking the resulting matrix against the exponential differential equation, gives $$ P(t)=\begin{bmatrix}e^{-2t}&2t e^{-2t}&1-(1+2t)e^{-2t}\\0&e^{-2t}&1-e^{-2t}\\0&0&1\end{bmatrix}. $$
Check. The first row sums to \(e^{-2t}+2t e^{-2t}+1-(1+2t)e^{-2t}=1\); the second sums to \(e^{-2t}+1-e^{-2t}=1\); the third sums to \(1\). Since \(e^{2t}\ge1+2t\), the first-row third entry is nonnegative. At \(t=0\) the matrix is the identity. The factor \(2t\), rather than \(t\), comes from the rate \(2\) in the nilpotent part of this generator.
Repeated Eigenvalues Without a Jordan Defect
Worked Example: Two Independent Routes to an Absorbing State
Let \(S=\{1,2,3\}\), with transitions from state \(1\) to \(3\) at rate \(1\), from state \(2\) to \(3\) at rate \(1\), and with state \(3\) absorbing. Its generator is $$ Q=\begin{bmatrix}-1&0&1\\0&-1&1\\0&0&0\end{bmatrix}. $$ Its rows sum to zero: \(-1+0+1=0\), \(0-1+1=0\), and \(0+0+0=0\); all off-diagonal entries are nonnegative.
The repeated eigenvalue \(-1\) has two independent eigenvectors, \((1,0,0)^{\mathsf T}\) and \((0,1,0)^{\mathsf T}\). The vector \((1,1,1)^{\mathsf T}\) is an eigenvector for \(0\). Thus \(Q\) is already diagonal in this basis, with diagonal entries \(-1,-1,0\). Applying the exponential and transforming back gives $$ P(t)=\begin{bmatrix}e^{-t}&0&1-e^{-t}\\0&e^{-t}&1-e^{-t}\\0&0&1\end{bmatrix}. $$
Check. Each row sums to one: \(e^{-t}+1-e^{-t}=1\), \(e^{-t}+1-e^{-t}=1\), and \(1=1\). Since \(t\ge0\), \(0<e^{-t}\le1\), so all entries are nonnegative. At \(t=0\), the matrix is the identity. There is no \(t e^{-t}\) term because the repeated eigenvalue has only size-one Jordan blocks.
Reading and Checking a Jordan Calculation
For a finite-state generator that is not diagonalizable, a reliable calculation has three stages: find the Jordan blocks, exponentiate each block, and transform back to the original state coordinates. The transition probabilities belong to the original state ordering, not to the Jordan basis, so the final similarity transformation is essential.
Determine algebraic multiplicities and enough eigenvectors or generalized eigenvectors to form a basis. A repeated eigenvalue is defective only if its eigenspace is too small.
For \(J_m(\lambda)=\lambda I+N\), use \(e^{\lambda t}\sum_{r=0}^{m-1}t^rN^r/r!\). The sum terminates because \(N^m=0\).
If \(Q=TJT^{-1}\), calculate \(T e^{tJ}T^{-1}\), preserving the original row and column state order.
At \(t=0\) it must equal the identity; for \(t\ge0\), its entries must be nonnegative and each row must sum to one.
Common mistake. In the first example, one might see the repeated eigenvalue \(-1\) and incorrectly write a diagonal exponential with two entries \(e^{-t}\), thereby treating the defective block as diagonal. That calculation would miss the probability \(p_{12}(t)=t e^{-t}\). The correction is to include the generalized eigenvector and use \(e^{-t}(I+tN)\) for the size-two block. Substitution into the verified transition matrix confirms that \(p_{12}(0)=0\), while for \(t>0\) it is positive; it also contributes to the first-row sum, which remains exactly one.
These computations matter whenever a generator has fewer eigenvectors than its repeated eigenvalues require. Jordan form still gives an exact expression for \(e^{tQ}\), but its polynomial factors affect transient probabilities and cannot be recovered by merely listing the eigenvalues. Conversely, the third example shows why multiplicity alone is not enough to infer such factors: a repeated eigenvalue with a full set of independent eigenvectors contributes ordinary exponentials.
Check Your Understanding
Use the Jordan-block formula and verify your results against the defining properties of a transition matrix.
- For a Jordan block of size three with eigenvalue \(\lambda\), write its exponential in terms of \(N\), and state why the sum stops where it does.
- In the first worked example, what equation shows that \(v_2\) is a generalized eigenvector rather than an eigenvector for \(-1\)?
- For the generator in the second example, identify the coefficient of \(t e^{-2t}\) in \(p_{12}(t)\) and explain its source.
- Why does a repeated eigenvalue not by itself imply a polynomial-times-exponential term?
- List two checks that the computed \(P(t)\) should satisfy, and carry them out for the third example.