From a Jordan Decomposition to Matrix Powers
Once a matrix has been put into Jordan form, its powers can be computed block by block. The Jordan form records the eigenvalues on the diagonal and the links in each Jordan chain on the superdiagonal. Those links affect the powers, so simply raising each diagonal entry to a power is not enough when a block has size greater than one. The calculation has two stages: compute powers of the Jordan blocks, then use the similarity decomposition to return to the original coordinates.
Throughout, matrices and scalars are over \(\mathbb{R}\), and \(m\) is a nonnegative integer. A Jordan block of size \(k\) with eigenvalue \(\lambda\) is the \(k\times k\) matrix \(\mathbf{J}_k(\lambda)=\lambda\mathbf{I}_k+\mathbf{N}_k\), where \(\mathbf{N}_k\) has \(1\)s on the superdiagonal and \(0\)s elsewhere. In particular, \(\mathbf{N}_k^k=\mathbf{O}\). A Jordan decomposition of \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is a factorization \(\mathbf{A}=\mathbf{P}\mathbf{J}\mathbf{P}^{-1}\), where \(\mathbf{P}\) is invertible and \(\mathbf{J}\) is block diagonal with Jordan blocks. The Jordan Form Existence Theorem gives such a decomposition when the characteristic polynomial splits over the field; here we use a decomposition that is already known.
The exponent \(m=0\) is included, with \(\mathbf{A}^0=\mathbf{I}\). This case is useful as a check: the formula gives \(\mathbf{P}\mathbf{I}\mathbf{P}^{-1}=\mathbf{I}\). The formulas below use exact arithmetic; no numerical approximation is needed.
Why Powers Respect Similarity and Blocks
Proof. Let \(m\) be a nonnegative integer. For \(m=0\), \(\mathbf{A}^0=\mathbf{I}=\mathbf{P}\mathbf{I}\mathbf{P}^{-1}\). Suppose the formula holds for an exponent \(m\). Then
because \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}\). Induction proves the similarity formula for every \(m\ge 0\). For the block statement, write \(\mathbf{J}=\operatorname{diag}(\mathbf{J}_1,\dots,\mathbf{J}_s)\). The product of two block diagonal matrices with matching blocks is block diagonal, with each diagonal block equal to the product of the corresponding blocks. Repeated multiplication therefore gives \(\mathbf{J}^m=\operatorname{diag}(\mathbf{J}_1^m,\dots,\mathbf{J}_s^m)\) for \(m\ge1\); for \(m=0\), both sides are the identity matrix of the relevant size. \(\square\)
Thus the main computation is the power of a single block. The nilpotent matrix \(\mathbf{N}_k\) contributes only finitely many terms, because all powers \(\mathbf{N}_k^r\) with \(r\ge k\) are zero.
Proof. For \(\lambda\ne0\) and \(m\ge1\), \(\lambda\mathbf{I}_k\) commutes with \(\mathbf{N}_k\), so the binomial formula gives \((\lambda\mathbf{I}_k+\mathbf{N}_k)^m=\sum_{r=0}^{m}\binom{m}{r}\lambda^{m-r}\mathbf{N}_k^r\). Every term with \(r\ge k\) vanishes because \(\mathbf{N}_k^k=\mathbf{O}\), leaving exactly the stated finite sum. The case \(m=0\) is the definition of the zeroth power and does not require the sum. If \(\lambda=0\), then \(\mathbf{J}_k(0)=\mathbf{N}_k\), so \(\mathbf{J}_k(0)^m=\mathbf{N}_k^m\) by repeated multiplication, including \(m=0\). This separate case avoids any need to interpret \(0^0\). \(\square\)
For a size-three block with nonzero eigenvalue, the formula becomes \(\lambda^m\mathbf{I}_3+m\lambda^{m-1}\mathbf{N}_3+\binom{m}{2}\lambda^{m-2}\mathbf{N}_3^2\) when \(m\ge2\). The second and third terms record the superdiagonal contributions. When the exponent is smaller than the block size, terms with \(r>m\) are simply absent.
A Full Similarity Calculation
Worked Example: All Powers from a Nontrivial Decomposition
Suppose \(\mathbf{A}=\mathbf{P}\mathbf{J}\mathbf{P}^{-1}\), with \(\mathbf{P}=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}\) and \(\mathbf{J}=\operatorname{diag}(\mathbf{J}_2(2),[-1])\). Find a formula for \(\mathbf{A}^m\), then check it for \(m=2\) and \(m=3\).
The inverse is \(\mathbf{P}^{-1}=\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}\), as multiplying \(\mathbf{P}\mathbf{P}^{-1}\) gives \(\mathbf{I}_3\). The block formula gives, for every \(m\ge0\), \(\mathbf{J}_2(2)^m=\begin{bmatrix}2^m&b_m\\0&2^m\end{bmatrix}\), where \(b_0=0\) and \(b_m=m2^{m-1}\) for \(m\ge1\). The other block has power \(c_m=(-1)^m\). The separate definition of \(b_0\) ensures no negative exponent is needed at \(m=0\).
To identify the original matrix, use \(m=1\), so \(2^m=2\), \(b_m=1\), and \(c_m=-1\). Thus \(\mathbf{A}=\begin{bmatrix}2&1&-1\\0&2&-3\\0&0&-1\end{bmatrix}\).
For \(m=2\), the formula gives \(2^m=4\), \(b_m=4\), and \(c_m=1\), so the proposed power is \(\begin{bmatrix}4&4&-4\\0&4&-3\\0&0&1\end{bmatrix}\). Direct multiplication confirms it: the \((1,2)\) entry is \(2\cdot1+1\cdot2=4\), the \((1,3)\) entry is \(2(-1)+1(-3)+(-1)(-1)=-4\), and the \((2,3)\) entry is \(2(-3)+(-3)(-1)=-3\). The diagonal entries are \(4,4,1\), and the entries below the diagonal are zero. Hence \(\mathbf{A}^2\) is the stated matrix.
For \(m=3\), the formula gives \(2^m=8\), \(b_m=12\), and \(c_m=-1\). Multiplying the verified square by \(\mathbf{A}\) gives \(\mathbf{A}^3=\begin{bmatrix}8&12&-12\\0&8&-9\\0&0&-1\end{bmatrix}\): the \((1,2)\) entry is \(4\cdot1+4\cdot2=12\); the \((1,3)\) entry is \(4(-1)+4(-3)+(-4)(-1)=-12\); and the \((2,3)\) entry is \(4(-3)+(-3)(-1)=-9\). The diagonal entries are \(8,8,-1\), as required. This checks the formula by direct multiplication for two positive exponents.
When the Eigenvalue Is Zero
A zero-eigenvalue block is nilpotent, so its powers eventually become zero. The power formula for nonzero eigenvalues must not be used by substituting \(\lambda=0\): its displayed scalar powers could include a \(0^0\) term. Instead, compute the power of the nilpotent matrix \(\mathbf{N}_k\) directly. In particular, for a size-two zero block, the first power is the block itself and every power with exponent at least two is zero.
Worked Example: A Similarity Decomposition with a Zero Block
Let \(\mathbf{P}=\begin{bmatrix}1&1\\-1&0\end{bmatrix}\), \(\mathbf{J}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), and \(\mathbf{P}^{-1}=\begin{bmatrix}0&-1\\1&1\end{bmatrix}\). Compute the powers of \(\mathbf{A}=\mathbf{P}\mathbf{J}\mathbf{P}^{-1}\).
First, \(\mathbf{P}\mathbf{J}=\begin{bmatrix}0&1\\0&-1\end{bmatrix}\), so \(\mathbf{A}=\begin{bmatrix}1&1\\-1&-1\end{bmatrix}\). Since \(\mathbf{J}^0=\mathbf{I}_2\), \(\mathbf{J}^1=\mathbf{J}\), and \(\mathbf{J}^m=\mathbf{O}\) for \(m\ge2\), similarity gives \(\mathbf{A}^0=\mathbf{I}_2\), \(\mathbf{A}^1=\mathbf{A}\), and \(\mathbf{A}^m=\mathbf{O}\) for \(m\ge2\).
Check: Direct multiplication gives \(\mathbf{A}^2=\begin{bmatrix}1\cdot1+1\cdot(-1)&1\cdot1+1\cdot(-1)\\(-1)\cdot1+(-1)\cdot(-1)&(-1)\cdot1+(-1)\cdot(-1)\end{bmatrix} =\begin{bmatrix}0&0\\0&0\end{bmatrix}\). Therefore all higher powers are also zero, because \(\mathbf{A}^m=\mathbf{A}^2\mathbf{A}^{m-2}\) for \(m\ge2\).
A Larger Block and a Direct Check
Worked Example: The Third Power of a Size-Three Block
Compute \(\mathbf{J}_3(2)^3\) using the block formula, then verify by multiplication. Here \(\mathbf{J}_3(2)=\begin{bmatrix}2&1&0\\0&2&1\\0&0&2\end{bmatrix}\) and \(\mathbf{N}_3=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\).
The formula with \(m=3\) has terms for \(r=0,1,2\), since \(\mathbf{N}_3^3=\mathbf{O}\). The coefficients are \(\binom30 2^3=8\), \(\binom31 2^2=12\), and \(\binom32 2=6\). Also, \(\mathbf{N}_3^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\). Therefore \(\mathbf{J}_3(2)^3=8\mathbf{I}_3+12\mathbf{N}_3+6\mathbf{N}_3^2 =\begin{bmatrix}8&12&6\\0&8&12\\0&0&8\end{bmatrix}\).
For the direct check, first square the block: \(\mathbf{J}_3(2)^2=\begin{bmatrix}4&4&1\\0&4&4\\0&0&4\end{bmatrix}\). Then multiply by \(\mathbf{J}_3(2)\). The first row becomes \((4\cdot2,\;4\cdot1+4\cdot2,\;4\cdot0+4\cdot1+1\cdot2)=(8,12,6)\). The second row becomes \((0,\;4\cdot2,\;4\cdot1+4\cdot2)=(0,8,12)\), and the third row is \((0,0,4\cdot2)=(0,0,8)\). This agrees with the block formula.
A Reliable Calculation and a Common Mistake
For an actual matrix, keep the two stages separate. First record the given \(\mathbf{P}\), \(\mathbf{J}\), and \(\mathbf{P}^{-1}\), and check that \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}\). Next compute each block power, using the nilpotent formula for a zero eigenvalue. Assemble those powers in the same block positions to form \(\mathbf{J}^m\). Finally calculate \(\mathbf{P}\mathbf{J}^m\mathbf{P}^{-1}\). This order reduces the chance of confusing the Jordan coordinates with the original coordinates.
Use the finite binomial sum for a nonzero eigenvalue; for a zero eigenvalue use \(\mathbf{N}_k^m\).
Place each computed block on the diagonal of \(\mathbf{J}^m\), leaving all off-block entries zero.
Compute \(\mathbf{A}^m=\mathbf{P}\mathbf{J}^m\mathbf{P}^{-1}\), preserving the order of the factors.
For \(m=1\), recover \(\mathbf{A}\); for \(m=2\) or \(m=3\), compare with direct multiplication.
Common mistake: dropping the superdiagonal contribution. For \(\mathbf{B}=\begin{bmatrix}2&1\\0&2\end{bmatrix}\), a tempting but incorrect calculation is \(\mathbf{B}^2=\begin{bmatrix}4&1\\0&4\end{bmatrix}\), as if the off-diagonal entry stayed unchanged. Matrix multiplication gives instead \(\mathbf{B}^2=\begin{bmatrix}2\cdot2&2\cdot1+1\cdot2\\0&2\cdot2\end{bmatrix} =\begin{bmatrix}4&4\\0&4\end{bmatrix}\). The block formula gives the same result: \(2^2\mathbf{I}_2+2\cdot2\mathbf{N}_2\). The superdiagonal term accumulates with the exponent; it is not fixed at its original value.
A second useful check is to avoid multiplying the similarity factors in the wrong order. The expression is \(\mathbf{P}\mathbf{J}^m\mathbf{P}^{-1}\), not \(\mathbf{P}^{-1}\mathbf{J}^m\mathbf{P}\). Matrix multiplication is generally not commutative, so changing this order need not give the same matrix. Checking \(m=1\) against the original \(\mathbf{A}\) detects this error immediately.
Check Your Understanding
Use the block formula and the similarity rule, and check computations by multiplication where requested.
- Write \(\mathbf{J}_2(3)^m\) for \(m\ge1\), and calculate its fourth power.
- For \(\mathbf{J}_3(0)\), determine the powers with exponents \(0,1,2,3\).
- If \(\mathbf{A}=\mathbf{P}\mathbf{J}\mathbf{P}^{-1}\), explain why \(\mathbf{A}^0=\mathbf{I}\) follows from the similarity formula.
- Compute the square of \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\) directly and identify the contribution that would be lost by leaving its superdiagonal entry unchanged.
- For a block diagonal Jordan matrix with blocks \(\mathbf{J}_2(1)\) and \(\mathbf{J}_1(-2)\), describe how to form its \(m\)th power without multiplying the full matrix repeatedly.
