Evaluating Polynomials Using Jordan Form

From a Matrix to Its Jordan Form

Evaluating a polynomial at a matrix means substituting the matrix for the polynomial’s variable, with the constant term interpreted as a multiple of the identity matrix. For a matrix in Jordan form, this substitution has a useful structure: each Jordan block can be evaluated separately. The result on a block depends not only on its eigenvalue, but also on its size.

We work over \(\mathbb{C}\), so every square matrix has a Jordan form. When a real matrix has a real Jordan form, the same computations apply over \(\mathbb{R}\). Let \(p(x)=a_0+a_1x+\cdots+a_dx^d\) be a polynomial. For a square matrix \(\mathbf{A}\) of the same size, define \[ p(\mathbf{A})=a_0\mathbf{I}+a_1\mathbf{A}+\cdots+a_d\mathbf{A}^d. \] Here \(\mathbf{A}^0=\mathbf{I}\). A Jordan block of size \(r\) with eigenvalue \(\lambda\) is \(\mathbf{J}_r(\lambda)=\lambda\mathbf{I}_r+\mathbf{N}_r\), where \(\mathbf{N}_r\) has ones on the superdiagonal and zeros elsewhere. In particular, \(\mathbf{N}_r^r=\mathbf{O}\).

The purpose of the calculation is to find \(p(\mathbf{A})\), not merely \(p(\mathbf{J})\). The Jordan form makes the calculation manageable, and similarity then transfers the result back to the original coordinates.

The Polynomial Formula on One Block

For a polynomial \(p\), let \(p^{(j)}(\lambda)\) denote its \(j\)th derivative evaluated at \(\lambda\), with \(p^{(0)}(\lambda)=p(\lambda)\). The Taylor expansion of a polynomial about \(\lambda\) is an exact finite identity. Since \(\mathbf{N}_r^r=\mathbf{O}\), only its first \(r\) terms can contribute when the variable offset is replaced by \(\mathbf{N}_r\).

Proof. Expand \(p(\lambda+t)\) as a polynomial in \(t\): \[ p(\lambda+t)=\sum_{j=0}^{d}\frac{p^{(j)}(\lambda)}{j!}t^j. \] This is an identity of polynomials, so substituting \(\mathbf{N}_r\) for \(t\) gives \(p(\lambda\mathbf{I}_r+\mathbf{N}_r)=\sum_{j=0}^{d}\frac{p^{(j)}(\lambda)}{j!}\mathbf{N}_r^j\). For every \(j\ge r\), \(\mathbf{N}_r^j=\mathbf{O}\), since \(\mathbf{N}_r^r=\mathbf{O}\). Removing these zero terms leaves precisely the sum from \(j=0\) to \(r-1\). \(\square\)

This formula explains what block size changes: a larger block may retain more derivative terms. The eigenvalue supplies the evaluation point, while the nilpotent part \(\mathbf{N}_r\) supplies the terms that sit above the diagonal. The formula does not require \(p(\lambda)\ne0\); if some derivative values vanish, their terms are simply zero.

Evaluating Every Block and Returning to the Original Matrix

Suppose a Jordan form is \(\mathbf{J}=\operatorname{diag}(\mathbf{J}_{r_1}(\lambda_1),\dots,\mathbf{J}_{r_k}(\lambda_k))\). A polynomial in a block-diagonal matrix remains block diagonal: each power of \(\mathbf{J}\) has the corresponding powers of its blocks on the diagonal. The next result states this fact together with the change back to the original matrix.

Proof. From \(\mathbf{A}=\mathbf{S}\mathbf{J}\mathbf{S}^{-1}\), induction on \(m\ge1\) gives \(\mathbf{A}^m=\mathbf{S}\mathbf{J}^m\mathbf{S}^{-1}\): multiplying the expression for the \(m\)th power by \(\mathbf{A}\) gives \(\mathbf{S}\mathbf{J}^m(\mathbf{S}^{-1}\mathbf{S})\mathbf{J}\mathbf{S}^{-1}=\mathbf{S}\mathbf{J}^{m+1}\mathbf{S}^{-1}\). For \(m=0\), the same identity holds because \(\mathbf{I}=\mathbf{S}\mathbf{I}\mathbf{S}^{-1}\). Thus, writing \(p(x)=\sum_{m=0}^{d}a_mx^m\), we obtain \[ p(\mathbf{A})=\sum_{m=0}^{d}a_m\mathbf{S}\mathbf{J}^m\mathbf{S}^{-1} =\mathbf{S}\left(\sum_{m=0}^{d}a_m\mathbf{J}^m\right)\mathbf{S}^{-1} =\mathbf{S}p(\mathbf{J})\mathbf{S}^{-1}. \] For the block statement, each \(\mathbf{J}^m\) has diagonal blocks equal to the \(m\)th powers of the corresponding Jordan blocks and zero off-diagonal blocks. Taking the indicated linear combination of these powers gives the asserted diagonal blocks of \(p(\mathbf{J})\). \(\square\)

Common Mistake: Keeping Only the Eigenvalue

A Jordan block is not just its diagonal eigenvalue repeated. Replacing it by \(\lambda\mathbf{I}\) discards the nilpotent part, which can contribute nonzero terms to the polynomial. For example, for \(p(x)=x^2\) and \(\mathbf{J}_2(3)=\begin{bmatrix}3&1\\0&3\end{bmatrix}\), an incorrect calculation gives \(p(3)\mathbf{I}_2=9\mathbf{I}_2\). That would leave out the off-diagonal contribution.

The correction is to retain \(\mathbf{N}_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\). Since \(p(3)=9\), \(p'(3)=6\), and \(\mathbf{N}_2^2=\mathbf{O}\), the block formula gives \[ p(\mathbf{J}_2(3))=9\mathbf{I}_2+6\mathbf{N}_2 =\begin{bmatrix}9&6\\0&9\end{bmatrix}. \] Directly squaring the Jordan block confirms it: \[ \begin{bmatrix}3&1\\0&3\end{bmatrix}^2 =\begin{bmatrix}9&6\\0&9\end{bmatrix}. \] The eigenvalue determines the derivative values, but the block size determines which of those derivative terms may remain.

Why Block-by-Block Evaluation Is Useful

The procedure separates the work into two tasks. First, evaluate the polynomial on each Jordan block using only its eigenvalue, size, and the relevant derivative values. Second, if the question asks for the matrix in the original coordinates, apply the similarity transformation. When several blocks have the same eigenvalue, their polynomial values need not be identical: blocks of different sizes can retain different numbers of terms. Even when a term’s coefficient is zero, the block’s size still specifies the powers that could appear.

This method is especially useful when direct matrix powers would require many multiplications. It also makes the effect of each Jordan block visible rather than hiding it in a long calculation. The next question is how the largest block sizes relate to the minimal polynomial; that connection is developed separately.