Why the Entries of a Power Have a Pattern
A Jordan block is nearly diagonal: its diagonal entries are the same eigenvalue, and its superdiagonal entries are \(1\). Taking powers preserves this upper-triangular shape, but the entries above the diagonal change in a structured way. Knowing that structure lets us compute a power without carrying out a long sequence of matrix multiplications, and it makes the zero-eigenvalue case precise.
For an integer \(m\ge 1\) and \(\lambda\in\mathbb{C}\), use the convention
Write \(\mathbf{N}=\mathbf{J}_m(0)\), the matrix with \(1\)s on the superdiagonal and \(0\)s elsewhere. Then \(\mathbf{J}_m(\lambda)=\lambda\mathbf{I}_m+\mathbf{N}\), and \(\mathbf{N}^m=\mathbf{O}\). The entry \((i,j)\) of a matrix is in row \(i\), column \(j\); thus the \(d\)th superdiagonal consists of positions with \(j-i=d\).
The Entrywise Formula for Nonnegative Powers
Proof. First suppose \(\lambda\ne0\). Denote the proposed entry in row \(i\), column \(j\) by \(q(k,j-i)\), where \(q(k,d)=\binom{k}{d}\lambda^{k-d}\) for \(0\le d\le k\), and \(q(k,d)=0\) otherwise. At \(k=0\), \(q(0,0)=1\) and every other entry is zero, so the formula gives \(\mathbf{I}_m\). For the induction step, set \(\mathbf{B}=\mathbf{J}_m(\lambda)^k\). Matrix multiplication gives \((\mathbf{B}\mathbf{J}_m(\lambda))_{ij}=\lambda B_{ij}+B_{i,j-1}\), where the second term is taken to be zero when \(j=1\). The induction hypothesis makes this \(\lambda q(k,d)+q(k,d-1)\), with \(d=j-i\). For \(0\le d\le k+1\), the binomial identity \(\binom{k}{d}+\binom{k}{d-1}=\binom{k+1}{d}\) gives \(\lambda q(k,d)+q(k,d-1)=\binom{k+1}{d}\lambda^{k+1-d}\); at the endpoints the same identity holds with the out-of-range coefficients equal to zero. For \(d<0\) or \(d>k+1\), both terms and the proposed entry at power \(k+1\) are zero. Thus the formula holds at \(k+1\). Now suppose \(\lambda=0\). The matrix \(\mathbf{N}\) sends the standard basis vector \(\mathbf{e}_j\) to \(\mathbf{e}_{j-1}\) for \(j>1\), and sends \(\mathbf{e}_1\) to \(\mathbf{0}\). Repeated application therefore gives \(\mathbf{N}^k\mathbf{e}_j=\mathbf{e}_{j-k}\) if \(j>k\), and \(\mathbf{0}\) if \(j\le k\). This is exactly the stated entry rule. If \(k\ge m\), there is no pair \(1\le i,j\le m\) with \(j-i=k\), so every entry is zero. \(\square\)
The nonnegative-power formula shows why the entries stop after the \(k\)th superdiagonal: in \(k\) multiplications, a nonzero contribution can move at most \(k\) columns to the right. The formula also handles \(k=0\) without exceptions: only the diagonal has \(j-i=0\), and its entry is \(\lambda^0=1\) when \(\lambda\ne0\).
Worked Example: A Nonzero Eigenvalue and a Third Power
Compute \(\mathbf{J}_3(2)^3\). Here \(m=3\), \(\lambda=2\), and \(k=3\). The possible superdiagonal distances are \(d=0,1,2\), so the entries are \(2^3=8\), \(\binom{3}{1}2^2=12\), and \(\binom{3}{2}2=6\), respectively. Distances greater than \(3\) do not occur in this block. Hence
Check by multiplication: First, \(\mathbf{J}_3(2)^2=\begin{bmatrix}4&4&1\\0&4&4\\0&0&4\end{bmatrix}\). Multiplying by \(\mathbf{J}_3(2)\) gives
Zero Eigenvalues and Powers Below the Block Size
When \(\lambda=0\), it is safer to use the separate rule in the theorem rather than substitute zero into the nonzero-eigenvalue expression. That expression can contain a power such as \(0^0\), which should not be used to decide the entries. For \(\mathbf{N}=\mathbf{J}_m(0)\), the \(k\)th power has \(1\)s exactly \(k\) places above the diagonal, provided those positions fit within the matrix. All other entries are zero. Thus a power with \(0<k<m\) is generally not the zero matrix, whereas every power with \(k\ge m\) is zero.
Worked Example: A Zero-Eigenvalue Block Before It Vanishes
Compute \(\mathbf{J}_4(0)^2\). The rule places \(1\)s where \(j-i=2\), namely in positions \((1,3)\) and \((2,4)\). It gives
Check by multiplication: With \(\mathbf{J}_4(0)=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}\), multiplying \(\mathbf{J}_4(0)\) by itself gives
Here \(2<4\), so the power has not vanished. Raising the block to the fourth power does give the zero matrix: there is no fourth superdiagonal position in a \(4\times4\) matrix.
Worked Example: A Power Smaller Than the Block Size
Compute \(\mathbf{J}_4(3)^2\). Since \(k=2<m=4\), only the diagonal, first superdiagonal, and second superdiagonal can be nonzero. Their entries are \(3^2=9\), \(\binom{2}{1}3=6\), and \(\binom{2}{2}=1\). Therefore
Check by multiplication: The block is \(\begin{bmatrix}3&1&0&0\\0&3&1&0\\0&0&3&1\\0&0&0&3\end{bmatrix}\). Multiplying it by itself gives diagonal entries \(9\), first-superdiagonal entries \(3+3=6\), second-superdiagonal entries \(1\), and zero entries below or more than two places above the diagonal. This produces exactly the displayed matrix.
Negative Powers When They Exist
An integer power with a negative exponent requires an inverse. The Jordan block \(\mathbf{J}_m(\lambda)\) is invertible exactly when \(\lambda\ne0\): if \(\lambda=0\), its first column is zero; if \(\lambda\ne0\), its diagonal entries are nonzero and it is invertible. For completeness, the entry formula extends to negative integers using generalized binomial coefficients. For any integer \(k\) and \(d\ge0\), define \(\binom{k}{0}=1\) and, for \(d\ge1\), define \(\binom{k}{d}=k(k-1)\cdots(k-d+1)/d!\).
Proof. Define a matrix \(\mathbf{F}_k\) by the proposed entries for each integer \(k\). The generalized coefficients satisfy \(\binom{k}{d}+\binom{k}{d-1}=\binom{k+1}{d}\) for \(d\ge1\), including negative \(k\); this follows by putting the two falling products over the common denominator \(d!\) and collecting terms. With the convention that entries below the diagonal are zero, matrix multiplication gives \((\mathbf{F}_k\mathbf{J}_m(\lambda))_{ij}=\lambda(\mathbf{F}_k)_{ij}+(\mathbf{F}_k)_{i,j-1}\). The coefficient identity shows that this equals \((\mathbf{F}_{k+1})_{ij}\), including the diagonal case \(d=0\). Also \(\mathbf{F}_0=\mathbf{I}_m\). Since \(\lambda\ne0\), the block is invertible; the recurrence \(\mathbf{F}_{k+1}=\mathbf{F}_k\mathbf{J}_m(\lambda)\), together with \(\mathbf{F}_0=\mathbf{I}_m\), determines the matrices in both directions from \(k=0\). Thus \(\mathbf{F}_k=\mathbf{J}_m(\lambda)^k\) for every integer \(k\). \(\square
For example, when \(m=2\), \(\lambda=2\), and \(k=-1\), the formula gives \(\mathbf{J}_2(2)^{-1}=\begin{bmatrix}\frac12&-\frac14\\0&\frac12\end{bmatrix}\). Multiplication verifies the result:
Common Pitfall: Treating a Zero Block Like a Scalar
A tempting but incorrect calculation is to say that \(\mathbf{J}_4(0)^2\) must be zero because the eigenvalue is zero and \(0^2=0\). This confuses the power of the eigenvalue with the power of the entire block. The block also has superdiagonal \(1\)s: as the worked example shows, \(\mathbf{J}_4(0)^2\) has \(1\)s in positions \((1,3)\) and \((2,4)\). The correct rule tracks both the eigenvalue and the superdiagonal distance. For a zero-eigenvalue block, a positive power below \(m\) shifts entries along that distance; only powers \(k\ge m\) vanish.
A related indexing error is to place the \(1\)s below the diagonal. Under the convention used here, the Jordan block has its \(1\)s above the diagonal, so a power of the nilpotent block has nonzero entries where \(j-i=k\), not where \(i-j=k\).
Why the Formula Is Useful
The entrywise formula gives more than a shortcut for arithmetic. It identifies exactly which superdiagonals can contribute to a power and how their coefficients depend on the exponent and eigenvalue. For a zero eigenvalue, it locates the point at which the power becomes zero. For a nonzero eigenvalue, it also shows that negative powers are available and remain upper triangular. These facts are useful when calculating powers of a matrix already expressed in Jordan form, without needing to multiply the full matrix repeatedly.
Check Your Understanding
Use the entrywise rules and verify computational answers by multiplication.
- Write \(\mathbf{J}_3(1)^2\) entry by entry.
- For which nonnegative integers \(k\) is \(\mathbf{J}_5(0)^k\) the zero matrix?
- In \(\mathbf{J}_4(2)^2\), what is the entry in row \(1\), column \(3\)? What is the entry in row \(1\), column \(4\)?
- Explain why \(\mathbf{J}_3(0)^{-1}\) is undefined, but \(\mathbf{J}_3(2)^{-1}\) exists.
- For \(m=2\), calculate \(\mathbf{J}_2(1)^{-2}\) from the negative-power formula and check by multiplication.