The Binomial Formula for Jordan Block Powers

Why the Binomial Expansion Stops

A Jordan block is a scalar matrix plus a nilpotent matrix. That decomposition makes its powers accessible through a binomial expansion, but the expansion is finite in a particularly useful way: sufficiently high powers of the nilpotent part are zero. The cutoff is determined by the size of the block, not by the eigenvalue.

We work over \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\). For a positive integer \(k\), the nilpotent Jordan block \(N_k\) has ones immediately above the main diagonal and zeros elsewhere. A Jordan block with eigenvalue \(\lambda\) is \(J_k(\lambda)=\lambda\mathbf{I}_k+N_k\). The exponent \(m\) below is a nonnegative integer; the case \(m=0\) means the zeroth power is the identity matrix.

The last assertion follows from the positions of the nonzero entries. Multiplication by \(N_k\) shifts the nonzero entries one diagonal above the main diagonal. After \(j\) multiplications, \(N_k^j\) has ones on the \(j\)-th diagonal above the main diagonal, when that diagonal exists. In particular, \(N_k^k=\mathbf{O}\), but \(N_k^{k-1}\) has a one in position \((1,k)\).

The Binomial Expansion and Its Truncation

The ordinary binomial coefficients appear because \(\lambda\mathbf{I}_k\) and \(N_k\) commute: multiplying either order gives \(\lambda N_k\). More generally, the same expansion holds for any two commuting square matrices. The proof is included to justify the coefficients used in the formula.

Proof. We use induction on \(m\). For \(m=0\), both sides are the identity matrix: the sum has just the term \(\binom00\mathbf{X}^0\mathbf{Y}^0=\mathbf{I}\). Suppose the formula holds for \(m\). Multiplying by \(\mathbf{X}+\mathbf{Y}\) gives \((\mathbf{X}+\mathbf{Y})^{m+1}=\sum_{j=0}^{m}\binom{m}{j}\bigl(\mathbf{X}^{m+1-j}\mathbf{Y}^j+\mathbf{X}^{m-j}\mathbf{Y}^{j+1}\bigr)\). The commutativity assumption permits the factors to be grouped in this order. For each interior power \(\mathbf{X}^{m+1-r}\mathbf{Y}^r\), the two contributions have coefficients \(\binom{m}{r}\) and \(\binom{m}{r-1}\), whose sum is \(\binom{m+1}{r}\). The two end terms have coefficient \(1\). Thus the result is the required sum for \(m+1\), completing the induction. \(\square\)

Proof. Apply the lemma with \(\mathbf{X}=\lambda\mathbf{I}\) and \(\mathbf{Y}=\mathbf{N}\); these commute because \((\lambda\mathbf{I})\mathbf{N}=\lambda\mathbf{N}=\mathbf{N}(\lambda\mathbf{I})\). Also, \((\lambda\mathbf{I})^{m-j}=\lambda^{m-j}\mathbf{I}\), so the full expansion has terms \(\binom{m}{j}\lambda^{m-j}\mathbf{N}^j\) for \(0\le j\le m\). For \(j\ge k\), write \(j=k+r\) with \(r\ge0\). Then \(\mathbf{N}^j=\mathbf{N}^k\mathbf{N}^r=\mathbf{O}\mathbf{N}^r=\mathbf{O}\), so all such terms vanish. The terms not removed this way have \(0\le j\le m\) and \(j\le k-1\); these are exactly the indices \(0\le j\le\min(m,k-1)\). This proves the formula, including \(m=0\), when the sum contains only \(\mathbf{I}\). \(\square\)

For a block of size \(k\), nilpotency removes the terms with \(j\ge k\). It does not remove a term merely because \(j\) is smaller than \(k\): for \(N_k\), every \(N_k^j\) with \(0\le j<k\) is nonzero. However, a coefficient can still make a term zero. For example, if \(\lambda=0\) and \(j<m\), the factor \(\lambda^{m-j}\) is zero. Also, terms with \(j>m\) do not occur in the binomial expansion at all. These are distinct reasons a term may be absent or zero.

Worked Powers of Jordan Blocks

Reading the Formula Entry by Entry

For a Jordan block, each \(N_k^j\) places ones on just one diagonal above the main diagonal. Thus the formula gives an entrywise description without needing to multiply the block repeatedly. In position \((r,s)\), a contribution is possible only if \(s-r=j\). This yields the following result.

Proof. The \(j\)-th power \(N_k^j\) has a one in position \((r,s)\) exactly when \(s-r=j\), and otherwise has zero in that position. The finite binomial formula includes the term with that \(j\) precisely when \(0\le j\le\min(m,k-1)\). For \(s<r\), no such nonnegative \(j\) exists. For \(s\ge r\), set \(j=s-r\). If \(j\le m\), the corresponding coefficient is \(\binom{m}{j}\lambda^{m-j}\); if \(j>m\), the term is absent. This proves the entry formula. \(\square\)

The diagonal entries correspond to \(j=0\), so each equals \(\lambda^m\). The first diagonal above it corresponds to \(j=1\), so its entries are \(m\lambda^{m-1}\) when \(m\ge1\). More distant diagonals are governed by higher binomial coefficients, until either the exponent \(m\) or the block size cuts the expansion off.

Common Mistake: Truncating Too Early

A frequent error is to treat \(N_k^k=\mathbf{O}\) as though it implied \(N_k^{k-1}=\mathbf{O}\). For a block of size two, this mistake would discard the \(j=1\) term and incorrectly give \(J_2(1)^3=\mathbf{I}_2\). The correct calculation is

Indeed, multiplying the proposed answer by \(J_2(1)\) gives \(\begin{bmatrix}1&3\\0&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1&4\\0&1\end{bmatrix}\), not the proposed answer. The correction is to use the exact cutoff: for a size-two block, \(N_2^2=\mathbf{O}\), but \(N_2\ne\mathbf{O}\). Therefore the \(j=1\) term remains.

The formula is useful because it replaces repeated matrix multiplication with a short sum whose length is controlled by the block size. When a matrix is already expressed in Jordan form, each block can be powered using this formula. The result depends on keeping three limits separate: the exponent allows only \(j\le m\), nilpotency allows only \(j<k\), and a zero eigenvalue can make some of the remaining scalar coefficients zero.