Why Taylor Data Give a Finite Answer
For a scalar input, the Taylor series of an analytic function may have infinitely many terms. A Jordan block behaves differently: its part above the diagonal is nilpotent, so sufficiently high powers of that part are zero. Consequently, evaluating an analytic function on a Jordan block requires only finitely many derivatives at its eigenvalue. The block’s size tells us exactly how many Taylor terms can contribute.
We work over \(\mathbb{C}\). Let \(f\) be analytic in a neighbourhood of \(\lambda\in\mathbb{C}\), meaning that in some open disk around \(\lambda\) it has a convergent Taylor expansion. A Jordan block of size \(m\) with eigenvalue \(\lambda\) is the matrix \(J_m(\lambda)=\lambda\mathbf{I}_m+\mathbf{N}\), where \(\mathbf{N}\) has ones immediately above the diagonal and zeros elsewhere. The matrix \(\mathbf{N}\) is nilpotent: \(\mathbf{N}^m=\mathbf{O}\). For this block, define \(f(J_m(\lambda))\) by substituting \(\mathbf{N}\) into the Taylor expansion of \(f\) about \(\lambda\). Since the powers \(\mathbf{N}^k\) vanish for \(k\ge m\), this substitution has only finitely many nonzero terms.
Proof. Analyticity at \(\lambda\) gives the scalar expansion \(f(\lambda+w)=\sum_{k=0}^{\infty}\frac{f^{(k)}(\lambda)}{k!}w^k\) for \(w\) sufficiently close to zero. The coefficients in this expansion are the Taylor data used to define \(f(J_m(\lambda))\): replace each scalar power \(w^k\) by the matrix power \(\mathbf{N}^k\). For \(k\ge m\), \(\mathbf{N}^k=\mathbf{O}\), since \(\mathbf{N}^m=\mathbf{O}\). Thus every term from \(k=m\) onward contributes the zero matrix, leaving exactly the displayed finite sum. \(\square\)
The constant term is \(f(\lambda)\mathbf{I}_m\); it is not, in general, the entire answer. Each further term records how \(f\) changes near \(\lambda\), with the \(k\)th derivative multiplying the \(k\)th power of the nilpotent part. The factorial is required by the scalar Taylor coefficients. In particular, the term multiplying \(\mathbf{N}^k\) is \(f^{(k)}(\lambda)/k!\), not simply \(f^{(k)}(\lambda)\).
Reading the Matrix from Its Derivatives
For the Jordan block above, \(\mathbf{N}^k\) has ones on the \(k\)th superdiagonal and zeros elsewhere. For \(k=0\), this means the main diagonal of the identity matrix. Therefore the derivative of order \(k\) contributes only to the \(k\)th superdiagonal of \(f(J_m(\lambda))\).
Proof. The \((i,j)\) entry of \(\mathbf{N}^k\) is \(1\) exactly when \(j-i=k\), and is \(0\) otherwise: multiplying by \(\mathbf{N}\) shifts each nonzero superdiagonal one place upward, starting with \(\mathbf{N}^0=\mathbf{I}_m\). If \(i>j\), there is no nonnegative integer \(k\) with \(j-i=k\), so every term in the Taylor formula has zero \((i,j)\) entry. If \(i\le j\), only \(k=j-i\) contributes to that entry, and its contribution is \(f^{(j-i)}(\lambda)/(j-i)!\). This proves both cases. \(\square\)
Thus the main diagonal is constant, equal to \(f(\lambda)\). The first superdiagonal is constant, equal to \(f'(\lambda)\); the second is \(f''(\lambda)/2!\), and so on. No derivative of order \(m\) or higher can appear in a block of size \(m\). This entrywise description is another way to check a computation: the result must be upper triangular, and each superdiagonal must have the value prescribed by the corresponding derivative.
Computing the Finite Taylor Sum
Worked Example: The Exponential of a Size-Three Block
Compute \(f(J_3(2))\) for \(f(z)=e^z\). In this case, $$ \mathbf{N}=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}, \qquad \mathbf{N}^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}, \qquad \mathbf{N}^3=\mathbf{O}. $$
Every derivative of \(e^z\) is \(e^z\), so \(f^{(k)}(2)=e^2\) for \(k=0,1,2\). The Taylor formula for this size-three block gives $$ f(J_3(2))=e^2\mathbf{I}_3+e^2\mathbf{N}+\frac{e^2}{2}\mathbf{N}^2. $$
Substituting the displayed matrices gives $$ f(J_3(2))= \begin{bmatrix} e^2&e^2&\frac{e^2}{2}\\ 0&e^2&e^2\\ 0&0&e^2 \end{bmatrix}. $$
Check: the diagonal entries are \(f(2)=e^2\), the first superdiagonal entries are \(f'(2)=e^2\), and the second superdiagonal entry is \(f''(2)/2!=e^2/2\). The next possible contribution contains \(\mathbf{N}^3\), which is zero, so no higher derivative changes this matrix.
Worked Example: The Reciprocal of a Size-Three Block
Let \(f(z)=1/z\), which is analytic near \(\lambda=2\), and compute \(f(J_3(2))\). For \(k\ge0\), $$ f^{(k)}(z)=(-1)^k k! z^{-(k+1)}, \qquad \frac{f^{(k)}(2)}{k!}=\frac{(-1)^k}{2^{k+1}}. $$
The coefficients for \(k=0,1,2\) are \(1/2\), \(-1/4\), and \(1/8\), respectively. Hence $$ f(J_3(2))=\frac12\mathbf{I}_3-\frac14\mathbf{N}+\frac18\mathbf{N}^2 =\begin{bmatrix} \frac12&-\frac14&\frac18\\ 0&\frac12&-\frac14\\ 0&0&\frac12 \end{bmatrix}. $$
Check: since \(f(z)=1/z\), the computed matrix should be an inverse of \(J_3(2)\). Multiplication gives $$ \begin{bmatrix}2&1&0\\0&2&1\\0&0&2\end{bmatrix} \begin{bmatrix}\frac12&-\frac14&\frac18\\0&\frac12&-\frac14\\0&0&\frac12\end{bmatrix} = \begin{bmatrix}1&-\frac12+\frac12&\frac14-\frac14\\0&1&-\frac12+\frac12\\0&0&1\end{bmatrix} =\mathbf{I}_3. $$
The multiplication confirms the result, including the positive \(1/8\) in the second superdiagonal. Omitting the second-order Taylor term would fail this check in the \((1,3)\) entry.
Worked Example: The Sine of a Size-Four Nilpotent Block
Take \(f(z)=\sin z\), \(\lambda=0\), and \(J_4(0)=\mathbf{N}\). The first four Taylor coefficients of \(\sin z\) at zero are \(f(0)=0\), \(f'(0)=1\), \(f''(0)/2!=0\), and \(f'''(0)/3!=-1/6\). Since \(\mathbf{N}^4=\mathbf{O}\), only these terms can contribute: $$ f(J_4(0))=\mathbf{N}-\frac16\mathbf{N}^3. $$
For this block, $$ \mathbf{N}=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}, \qquad \mathbf{N}^3=\begin{bmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}. $$ Therefore $$ f(J_4(0))=\begin{bmatrix}0&1&0&-\frac16\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}. $$
Check: the first superdiagonal equals \(f'(0)=1\), the second equals \(f''(0)/2!=0\), and the third equals \(f'''(0)/3!=-1/6\). The next power is \(\mathbf{N}^4=\mathbf{O}\), so Taylor terms of order four and higher contribute zero. This agrees with the displayed matrix.
Changing Coordinates Does Not Change the Formula
A matrix may be presented in a basis in which it is similar to a Jordan block. The same Taylor computation can be transferred back to the original coordinates. Here, similarity means replacing a matrix \(\mathbf{J}\) by \(\mathbf{P}\mathbf{J}\mathbf{P}^{-1}\) for an invertible matrix \(\mathbf{P}\). For an analytic function evaluated by its Taylor series at the block’s eigenvalue, the resulting matrix function transforms by the same change of coordinates.
Proof. Write \(\mathbf{A}=\mathbf{P}\mathbf{J}\mathbf{P}^{-1}\). For each integer \(k\ge0\), \(\mathbf{A}-\lambda\mathbf{I}_m=\mathbf{P}(\mathbf{J}-\lambda\mathbf{I}_m)\mathbf{P}^{-1}\), and therefore \((\mathbf{A}-\lambda\mathbf{I}_m)^k=\mathbf{P}(\mathbf{J}-\lambda\mathbf{I}_m)^k\mathbf{P}^{-1}\); the case \(k=0\) is \(\mathbf{I}_m=\mathbf{P}\mathbf{I}_m\mathbf{P}^{-1}\), and higher powers follow by multiplication and cancellation of \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}_m\). Substitute these powers into the finite Taylor sum. Factoring \(\mathbf{P}\) on the left and \(\mathbf{P}^{-1}\) on the right gives \(f(\mathbf{A})=\mathbf{P}f(\mathbf{J})\mathbf{P}^{-1}\), as claimed. \(\square\)
This result explains how the block formula is used when a matrix has been put into Jordan form: compute the function on each Jordan block, assemble those block results, and transfer the result through the similarity matrix. For a block diagonal Jordan matrix, powers of its nilpotent parts act within their own blocks, so the Taylor computation can be carried out block by block.
Common Pitfall: Keeping Only the Diagonal Term
A frequent error is to replace \(f(J_m(\lambda))\) by \(f(\lambda)\mathbf{I}_m\). That keeps the constant Taylor term but discards the contributions from the nilpotent part. For instance, for \(f(z)=1/z\) and \(J_3(2)\), this mistaken calculation gives $$ f(2)\mathbf{I}_3=\begin{bmatrix}\frac12&0&0\\0&\frac12&0\\0&0&\frac12\end{bmatrix}. $$
But multiplying by \(J_3(2)\) gives $$ \begin{bmatrix}2&1&0\\0&2&1\\0&0&2\end{bmatrix} \begin{bmatrix}\frac12&0&0\\0&\frac12&0\\0&0&\frac12\end{bmatrix} = \begin{bmatrix}1&\frac12&0\\0&1&\frac12\\0&0&1\end{bmatrix} \ne\mathbf{I}_3. $$
The correction is to include every nonzero power of \(\mathbf{N}\). The first derivative supplies the first superdiagonal, and the second derivative supplies the second superdiagonal. For this reciprocal example the full result is \(\frac12\mathbf{I}_3-\frac14\mathbf{N}+\frac18\mathbf{N}^2\), which the earlier multiplication check verifies is the inverse.
To apply the formula reliably, identify \(\lambda\) and \(\mathbf{N}\), find the first power of \(\mathbf{N}\) that is zero, calculate the required derivatives at \(\lambda\), and include the factorial in each coefficient. Then check the superdiagonals against the derivative values. The calculation is finite because of nilpotency, not because the scalar Taylor series has only finitely many terms.
Check Your Understanding
Use the Taylor formula and the nilpotent powers of each block to answer the questions.
- For a size-two Jordan block \(J_2(\lambda)\), which derivatives of \(f\) can affect \(f(J_2(\lambda))\)?
- Write the entries of \(f(J_3(\lambda))\) in terms of \(f(\lambda)\), \(f'(\lambda)\), and \(f''(\lambda)\).
- Compute \(f(J_2(1))\) for \(f(z)=z^2\), and verify the result by direct multiplication.
- Why does the Taylor formula for a size-four Jordan block have no contribution from the fourth derivative?
- For \(f(z)=1/z\), explain why \(f(2)\mathbf{I}_3\) alone cannot be the inverse of \(J_3(2)\).
