Why the Exponents Need Care
For real numbers, the product rule \((ab)^n=a^nb^n\) works because multiplication can be reordered: factors \(a\) and \(b\) commute. Matrix multiplication is associative, but it is not commutative in general. As a result, some familiar exponent laws remain valid for powers of one matrix, while rearranging factors from a product of different matrices can change the answer.
In this tutorial, \(n\) and the other exponents are nonnegative integers, and the matrices in each product are square matrices of the same size over \(\mathbb{R}\). Recall that matrix multiplication is associative, so parentheses can be moved without changing the order of the factors. The order itself, however, cannot generally be changed. We use \(\mathbf{I}_n\) for the \(n\times n\) identity matrix.
The rule for \(\mathbf{A}^{k+1}\) specifies multiplication on the right. Associativity means this is consistent with grouping the \(k+1\) copies in other ways, but the definition does not permit changing the order of different matrices. We first establish the laws that do follow for powers of the same matrix.
Two Laws for Powers of One Matrix
Proof. We prove the first equality by induction on \(n\), with \(m\) fixed. When \(n=0\), the definition of the zeroth power gives \(\mathbf{A}^m\mathbf{A}^0=\mathbf{A}^m\mathbf{I}_n=\mathbf{A}^m=\mathbf{A}^{m+0}\). Now suppose the equality holds for an exponent \(n\). Then, by the recursive definition of powers, associativity, and the induction hypothesis, \[ \mathbf{A}^m\mathbf{A}^{n+1} =\mathbf{A}^m(\mathbf{A}^n\mathbf{A}) =(\mathbf{A}^m\mathbf{A}^n)\mathbf{A} =\mathbf{A}^{m+n}\mathbf{A} =\mathbf{A}^{m+n+1}. \] This proves the first equality for all nonnegative \(n\).
For the second equality, again use induction on \(n\), with \(m\) fixed. At \(n=0\), \((\mathbf{A}^m)^0=\mathbf{I}_n=\mathbf{A}^0=\mathbf{A}^{m\cdot 0}\). Suppose \((\mathbf{A}^m)^n=\mathbf{A}^{mn}\). Then \[ (\mathbf{A}^m)^{n+1} =(\mathbf{A}^m)^n\mathbf{A}^m =\mathbf{A}^{mn}\mathbf{A}^m =\mathbf{A}^{mn+m} =\mathbf{A}^{m(n+1)}. \] The third equality uses the first law, already proved, and \(mn+m=m(n+1)\). This argument includes \(m=0\): every power \((\mathbf{A}^0)^n=\mathbf{I}_n\), as does \(\mathbf{A}^{0\cdot n}\). Thus both laws hold for all the stated exponents. \(\square\)
The first law combines two blocks of copies of the same matrix; the second counts the copies inside repeated blocks. Neither law needs \(\mathbf{A}\) to be invertible. The exponents here are nonnegative, so these statements make no claim about negative powers.
Worked Example: Combining Powers by Adding Exponents
Let \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). Compute \(\mathbf{A}^2\mathbf{A}^3\) and check the addition law by computing \(\mathbf{A}^5\).
First calculate the powers by repeated multiplication: \[ \mathbf{A}^2 =\begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}1&2\\0&1\end{bmatrix}, \qquad \mathbf{A}^3 =\mathbf{A}^2\mathbf{A} =\begin{bmatrix}1&2\\0&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}1&3\\0&1\end{bmatrix}. \] Continuing, \[ \mathbf{A}^4=\mathbf{A}^3\mathbf{A} =\begin{bmatrix}1&3\\0&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}1&4\\0&1\end{bmatrix}, \quad \mathbf{A}^5=\mathbf{A}^4\mathbf{A} =\begin{bmatrix}1&4\\0&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}1&5\\0&1\end{bmatrix}. \]
Now multiply the two blocks: \[ \mathbf{A}^2\mathbf{A}^3 =\begin{bmatrix}1&2\\0&1\end{bmatrix} \begin{bmatrix}1&3\\0&1\end{bmatrix} =\begin{bmatrix}1&5\\0&1\end{bmatrix} =\mathbf{A}^5. \] The entries follow from \(1\cdot1+2\cdot0=1\), \(1\cdot3+2\cdot1=5\), \(0\cdot1+1\cdot0=0\), and \(0\cdot3+1\cdot1=1\). Check: the computed product is exactly the fifth power found by repeated multiplication, as \(\mathbf{A}^2\mathbf{A}^3=\mathbf{A}^{2+3}\) predicts.
Repeated Powers and the Zero Exponent
The power-of-a-power rule has a useful interpretation: \((\mathbf{A}^m)^n\) consists of \(n\) blocks, each containing \(m\) copies of \(\mathbf{A}\). When \(m,n\) are positive, the total is \(mn\) copies. The theorem also handles boundary cases without needing a separate convention: if \(m=0\) or \(n=0\), the result is the identity matrix.
Worked Example: Raising a Power to a Power
Take \(\mathbf{R}=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\). Verify the rule \((\mathbf{R}^2)^3=\mathbf{R}^6\) by computing both sides.
First, \[ \mathbf{R}^2 =\begin{bmatrix}0&1\\-1&0\end{bmatrix} \begin{bmatrix}0&1\\-1&0\end{bmatrix} =\begin{bmatrix}-1&0\\0&-1\end{bmatrix}. \] Thus \[ (\mathbf{R}^2)^3 =\begin{bmatrix}-1&0\\0&-1\end{bmatrix} \begin{bmatrix}-1&0\\0&-1\end{bmatrix} \begin{bmatrix}-1&0\\0&-1\end{bmatrix} =\begin{bmatrix}-1&0\\0&-1\end{bmatrix}. \] For comparison, repeated multiplication gives \[ \mathbf{R}^3=\mathbf{R}^2\mathbf{R} =\begin{bmatrix}0&-1\\1&0\end{bmatrix},\qquad \mathbf{R}^4=\mathbf{R}^3\mathbf{R} =\begin{bmatrix}1&0\\0&1\end{bmatrix}, \] \[ \mathbf{R}^5=\mathbf{R}^4\mathbf{R} =\begin{bmatrix}0&1\\-1&0\end{bmatrix},\qquad \mathbf{R}^6=\mathbf{R}^5\mathbf{R} =\begin{bmatrix}-1&0\\0&-1\end{bmatrix}. \] Therefore \((\mathbf{R}^2)^3=\mathbf{R}^6=\mathbf{R}^{2\cdot3}\). Check: both sides are the same explicitly computed matrix, and multiplying \(\mathbf{R}^5\mathbf{R}\) gives the displayed value of \(\mathbf{R}^6\).
Why a Product of Different Matrices Is Different
For two square matrices \(\mathbf{A}\) and \(\mathbf{B}\), the expression \((\mathbf{A}\mathbf{B})^2\) expands, by associativity, to \(\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\). The expression \(\mathbf{A}^2\mathbf{B}^2\) is \(\mathbf{A}\mathbf{A}\mathbf{B}\mathbf{B}\). These products have the same factors but in a different order. They are guaranteed to agree only when there is an additional reason that the factors can be reordered.
Proof. First note that \(\mathbf{B}^j\mathbf{A}=\mathbf{A}\mathbf{B}^j\) for every nonnegative integer \(j\). For \(j=0\), both sides equal \(\mathbf{A}\). If \(\mathbf{B}^j\mathbf{A}=\mathbf{A}\mathbf{B}^j\), then \[ \mathbf{B}^{j+1}\mathbf{A} =\mathbf{B}(\mathbf{B}^j\mathbf{A}) =\mathbf{B}(\mathbf{A}\mathbf{B}^j) =(\mathbf{B}\mathbf{A})\mathbf{B}^j =(\mathbf{A}\mathbf{B})\mathbf{B}^j =\mathbf{A}\mathbf{B}^{j+1}, \] where the assumed equality \(\mathbf{B}\mathbf{A}=\mathbf{A}\mathbf{B}\) is used. Now prove the claimed formula by induction on \(k\). For \(k=0\), both sides are \(\mathbf{I}_n\). If it holds for \(k\), then \[ (\mathbf{A}\mathbf{B})^{k+1} =(\mathbf{A}\mathbf{B})^k\mathbf{A}\mathbf{B} =\mathbf{A}^k\mathbf{B}^k\mathbf{A}\mathbf{B} =\mathbf{A}^k\mathbf{A}\mathbf{B}^k\mathbf{B} =\mathbf{A}^{k+1}\mathbf{B}^{k+1}. \] The third equality follows from the established relation \(\mathbf{B}^k\mathbf{A}=\mathbf{A}\mathbf{B}^k\); the last uses the addition law for powers of a single matrix. This proves the result. \(\square\)
Commutativity is a sufficient condition for the product-power formula; this theorem does not say that it is necessary in every particular case. For example, the formula holds at \(k=0\) and \(k=1\) for any pair of square matrices, whether or not they commute. The issue can first arise at \(k=2\), when the factor order in the two expressions can differ.
Worked Example: A Product-Power Rule That Fails
Use \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\) to compare \((\mathbf{A}\mathbf{B})^2\) with \(\mathbf{A}^2\mathbf{B}^2\).
First compute the products in both orders: \[ \mathbf{A}\mathbf{B} =\begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}1&0\\1&1\end{bmatrix} =\begin{bmatrix}2&1\\1&1\end{bmatrix}, \quad \mathbf{B}\mathbf{A} =\begin{bmatrix}1&0\\1&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}1&1\\1&2\end{bmatrix}. \] Since these are unequal, the matrices do not commute. Squaring their product gives \[ (\mathbf{A}\mathbf{B})^2 =\begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}2&1\\1&1\end{bmatrix} =\begin{bmatrix}5&3\\3&2\end{bmatrix}. \] Meanwhile, \[ \mathbf{A}^2 =\begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}1&2\\0&1\end{bmatrix}, \quad \mathbf{B}^2 =\begin{bmatrix}1&0\\1&1\end{bmatrix} \begin{bmatrix}1&0\\1&1\end{bmatrix} =\begin{bmatrix}1&0\\2&1\end{bmatrix}, \] so \[ \mathbf{A}^2\mathbf{B}^2 =\begin{bmatrix}1&2\\0&1\end{bmatrix} \begin{bmatrix}1&0\\2&1\end{bmatrix} =\begin{bmatrix}5&2\\2&1\end{bmatrix}. \] Thus \((\mathbf{A}\mathbf{B})^2\ne\mathbf{A}^2\mathbf{B}^2\). Check: direct multiplication gives the entries \(4+1=5\), \(2+1=3\), \(2+1=3\), and \(1+1=2\) in the first square, and \(1+4=5\), \(0+2=2\), \(0+2=2\), and \(1\) in the second product. The differing entries confirm the failure.
Common Mistake: Reordering Factors as if They Were Numbers
A tempting but invalid calculation is \[ (\mathbf{A}\mathbf{B})^2 =\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B} \mathrel{\color{red}{=}} \mathbf{A}\mathbf{A}\mathbf{B}\mathbf{B} =\mathbf{A}^2\mathbf{B}^2. \] The middle step silently moves the middle \(\mathbf{B}\) past \(\mathbf{A}\). Associativity permits changing parentheses, not changing factor order. In the preceding example, that erroneous step predicts \(\begin{bmatrix}5&2\\2&1\end{bmatrix}\), whereas the correct calculation is \((\mathbf{A}\mathbf{B})^2=\begin{bmatrix}5&3\\3&2\end{bmatrix}\). The correction is to keep the alternating order when expanding: \((\mathbf{A}\mathbf{B})^2=\mathbf{A}\mathbf{B}\mathbf{A}\mathbf{B}\). Only after establishing \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}\) may the factors be rearranged to obtain \(\mathbf{A}^2\mathbf{B}^2\).
Keep the valid and conditional rules distinct:
| Expression | Rule | Condition |
|---|---|---|
| \(\mathbf{A}^m\mathbf{A}^n\) | \(\mathbf{A}^{m+n}\) | Same square matrix; nonnegative integer exponents |
| \((\mathbf{A}^m)^n\) | \(\mathbf{A}^{mn}\) | Same square matrix; nonnegative integer exponents |
| \((\mathbf{A}\mathbf{B})^k\) | \(\mathbf{A}^k\mathbf{B}^k\) | Guaranteed if \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}\) |
Why These Conditions Matter
The reliable way to use matrix exponent laws is to inspect the factors before applying a scalar-looking rule. When every factor is a power of one matrix, associativity and the definition of powers give the addition and power-of-a-power laws. When two different matrices appear, expansion reveals the order of the factors; a rearrangement then needs a separate justification, such as commutativity. This distinction is useful in simplifying expressions, checking symbolic calculations, and avoiding false identities in proofs.
For any proposed rule, a small exact example can expose an unjustified rearrangement, but one example cannot prove a general law. The proofs above establish the laws for all square matrices and all nonnegative integer exponents under their stated conditions. Conversely, the noncommuting example is enough to disprove a claim that \((\mathbf{A}\mathbf{B})^2=\mathbf{A}^2\mathbf{B}^2\) holds for every pair of square matrices.
Check Your Understanding
For each question, keep track of both the exponent and the order of the matrix factors.
- State the definition of \(\mathbf{A}^0\). Why is it not the zero matrix?
- Use the addition law to simplify \(\mathbf{A}^4\mathbf{A}^3\), stating the condition on the exponents.
- For a square matrix \(\mathbf{C}\), simplify \((\mathbf{C}^3)^2\). Which proved law applies?
- Expand \((\mathbf{A}\mathbf{B})^3\) as a product of six factors in order. Which step would require commutativity to turn it into \(\mathbf{A}^3\mathbf{B}^3\)?
- Give a sufficient condition under which \((\mathbf{A}\mathbf{B})^k=\mathbf{A}^k\mathbf{B}^k\) holds for every nonnegative integer \(k\).
