Why Polynomial Evaluation Simplifies on a Jordan Block
A Jordan block has a particularly useful form: it is a scalar multiple of the identity matrix plus a nilpotent matrix. Nilpotency means that some power of the second matrix is zero. Consequently, when a polynomial is evaluated at the block, all sufficiently high powers disappear. The result is an exact finite expansion, not an approximation, and its coefficients are determined by derivatives of the polynomial at the block's eigenvalue.
Throughout, scalars and polynomial coefficients may be in \(\mathbb{R}\) or \(\mathbb{C}\). For a polynomial \(p(t)=a_0+a_1t+\cdots+a_dt^d\) and a square matrix \(\mathbf{A}\), define \(p(\mathbf{A})=a_0\mathbf{I}+a_1\mathbf{A}+\cdots+a_d\mathbf{A}^d\), where \(\mathbf{A}^0=\mathbf{I}\). A matrix \(\mathbf{N}\) is nilpotent if \(\mathbf{N}^q=\mathbf{O}\) for some positive integer \(q\). For a Jordan block of size \(m\), the nilpotent part has its \(m\)th power equal to \(\mathbf{O}\).
For example, for a block of size \(3\), the nilpotent part and the block are
The crucial point is that \(\lambda\mathbf{I}\) and \(\mathbf{N}\) commute: \((\lambda\mathbf{I})\mathbf{N}=\lambda\mathbf{N}=\mathbf{N}(\lambda\mathbf{I})\). This allows the binomial expansion to apply. Since powers \(\mathbf{N}^k\) vanish once \(k\ge m\), only the first \(m\) terms can contribute.
The Finite Taylor Formula
Proof. Write \(p(t)=\sum_{r=0}^{d}a_rt^r\). By the binomial formula for commuting quantities, the coefficient of \(x^k\) in \(p(\lambda+x)\) is \(\sum_{r=k}^{d}a_r\binom{r}{k}\lambda^{r-k}\). Differentiating \(p\) \(k\) times shows that this coefficient equals \(p^{(k)}(\lambda)/k!\). Thus the following is a polynomial identity in \(x\): $$ p(\lambda+x)=\sum_{k=0}^{d}\frac{p^{(k)}(\lambda)}{k!}x^k. $$ Substitute the matrix \(\mathbf{N}\) for \(x\). This substitution is valid because both sides are polynomial expressions in the same matrix. Every term with \(k\ge m\) is zero because \(\mathbf{N}^m=\mathbf{O}\), and hence \(\mathbf{N}^k=\mathbf{O}\) for all \(k\ge m\). The remaining terms give the stated formula. The factorials are nonzero scalars in \(\mathbb{R}\) and \(\mathbb{C}\), so each division is defined. \(\square\)
The sum ends at \(m-1\), even if \(p\) has higher degree. If the degree is less than \(m-1\), the higher derivatives are zero instead. These are two distinct reasons that a term may contribute nothing: its derivative may vanish, or its matrix power may vanish.
Worked Example: Cubing a Three-by-Three Jordan Block
Let \(\mathbf{J}=\mathbf{J}_3(2)\), and take \(p(t)=t^3\). Compute \(p(\mathbf{J})\) using the finite Taylor formula, then verify it by multiplication.
Here \(p(2)=8\), \(p'(t)=3t^2\), so \(p'(2)=12\); \(p''(t)=6t\), so \(p''(2)/2!=12/2=6\); and \(p'''(t)=6\), so \(p'''(2)/3!=6/6=1\). The nilpotent part satisfies \(\mathbf{N}^3=\mathbf{O}\). Therefore,
To check by direct multiplication, first calculate \(\mathbf{J}^2\), then multiply by \(\mathbf{J}\) once more:
The direct computation agrees with the Taylor formula, as required since \(p(\mathbf{J})=\mathbf{J}^3\).
Reading the Entries from Derivatives
The powers of the nilpotent part have a simple pattern: \(\mathbf{N}^k\) has \(1\)s exactly \(k\) places above the main diagonal and zeros everywhere else. The finite Taylor formula therefore gives the entries of \(p(\mathbf{J}_m(\lambda))\) directly.
Proof. First, \(\mathbf{N}^0=\mathbf{I}_m\) has \(1\)s precisely where \(j-i=0\). Suppose \(0\le k<m-1\) and \(\mathbf{N}^k\) has a \(1\) in position \((i,j)\) exactly when \(j-i=k\), and zero otherwise. By matrix multiplication, \((\mathbf{N}^{k+1})_{ij}=\sum_{\ell=1}^m(\mathbf{N}^k)_{i\ell}N_{\ell j}\). The factor \(N_{\ell j}\) is nonzero only when \(j=\ell+1\), while \((\mathbf{N}^k)_{i\ell}\) is nonzero only when \(\ell-i=k\). Both conditions hold exactly when \(j-i=k+1\); in that case the sum equals \(1\). This proves the pattern for every \(k\) by induction. Applying the finite Taylor formula, the term involving \(\mathbf{N}^{j-i}\) contributes \(p^{(j-i)}(\lambda)/(j-i)!\) in position \((i,j)\) when \(j\ge i\). No power of \(\mathbf{N}\) contributes below the main diagonal, where \(j<i\). This gives the formula. \(\square\)
In particular, the diagonal entries are all \(p(\lambda)\), and the first superdiagonal entries are all \(p'(\lambda)\). The next superdiagonal is determined by \(p''(\lambda)/2!\). Derivatives beyond the block's available superdiagonals do not appear.
Worked Example: A Polynomial with Vanishing First Derivative
Let \(\mathbf{J}=\mathbf{J}_3(1)\) and \(p(t)=(t-1)^2(t+2)\). Find \(p(\mathbf{J})\) using the finite Taylor formula and check the result directly.
Set \(x=t-1\), so \(t+2=x+3\) and \(p(1+x)=x^2(x+3)=3x^2+x^3\). Thus \(p(1)=0\), \(p'(1)=0\), and \(p''(1)/2!=3\); terms of order \(3\) and higher vanish on this block because \(\mathbf{N}^3=\mathbf{O}\). It follows that
For a direct check, \(\mathbf{J}-\mathbf{I}_3=\mathbf{N}\) and \(\mathbf{J}+2\mathbf{I}_3=3\mathbf{I}_3+\mathbf{N}\). Since \(p(\mathbf{J})=(\mathbf{J}-\mathbf{I}_3)^2(\mathbf{J}+2\mathbf{I}_3)\),
The direct factorization verifies the computed matrix. In particular, the zero first derivative removes the first superdiagonal, while the nonzero second derivative leaves a nonzero second superdiagonal.
When a Polynomial Becomes a Scalar Matrix
A polynomial can act on a Jordan block in a much simpler way than its degree might suggest. If its derivatives vanish at the eigenvalue, the corresponding superdiagonals disappear. A block of size \(2\), for instance, has only one superdiagonal, so only \(p(\lambda)\) and \(p'(\lambda)\) can affect the result. The next example illustrates this in a case where the derivative term is zero.
Worked Example: A Quadratic on a Two-by-Two Block
Let \(\mathbf{J}=\mathbf{J}_2(-1)\) and \(p(t)=t^2+2t+5\). Compute \(p(\mathbf{J})\) using derivatives, then check by multiplying matrices.
We have \(p(-1)=1-2+5=4\), \(p'(t)=2t+2\), and \(p'(-1)=0\). Since the block has size \(2\), its nilpotent part satisfies \(\mathbf{N}^2=\mathbf{O}\). The formula gives \(p(\mathbf{J})=4\mathbf{I}_2\). Directly,
This agrees with the derivative calculation. The off-diagonal entry cancels because the first derivative at \(-1\) is zero.
Common Pitfall: Forgetting the Factorial
The coefficient of \(\mathbf{N}^k\) is \(p^{(k)}(\lambda)/k!\), not just \(p^{(k)}(\lambda)\). For a concrete check, take \(p(t)=t^2\) and \(\mathbf{J}_3(2)=2\mathbf{I}_3+\mathbf{N}\). The derivatives are \(p(2)=4\), \(p'(2)=4\), and \(p''(2)=2\). Omitting the factorial would give the incorrect expression \(4\mathbf{I}_3+4\mathbf{N}+2\mathbf{N}^2\). Since \(2!=2\), the correct expression is
Indeed, direct multiplication gives \(\mathbf{J}_3(2)^2=\begin{bmatrix}4&4&1\\0&4&4\\0&0&4\end{bmatrix}\). The incorrect coefficient would put \(2\), rather than \(1\), in position \((1,3)\). The factorial is essential even when the calculation is otherwise correct.
When using the formula, first identify the block size and its nilpotent part, then compute the required derivatives at \(\lambda\), and finally discard powers that are zero. This keeps the polynomial's scalar information separate from the block's nilpotent structure. It also provides a direct check on a proposed answer: the result must be upper triangular, have \(p(\lambda)\) on the diagonal, and have the indicated derivative coefficients on its superdiagonals.
Check Your Understanding
Use the finite Taylor formula, and check any computed matrix by direct substitution into the polynomial.
- For \(\mathbf{J}_2(3)\) and \(p(t)=t^2\), compute \(p(\mathbf{J}_2(3))\).
- For \(\mathbf{J}_3(0)\) and \(p(t)=t^3+2t^2+t\), determine which terms remain after evaluation.
- Write the entries of \(p(\mathbf{J}_3(\lambda))\) in terms of \(p(\lambda)\), \(p'(\lambda)\), and \(p''(\lambda)\).
- For a block of size \(2\), explain why derivatives of order \(2\) and higher do not affect the result.
- Identify the error in replacing \(p^{(2)}(\lambda)/2!\) by \(p^{(2)}(\lambda)\), and illustrate it with a quadratic polynomial on a block of size \(3\).
