Recipe

Compute a matrix power $A^k$

Use when: you have a square matrix $A$ and a positive integer $k$ , and need the matrix $A^k$

Goal: the matrix $A^k$ , found by whichever method is actually fastest for A's shape and k's size

Taught in Powers of a matrix

Quick recipe

Confirm $A$ is square
Check whether $A$ 's shape unlocks a shortcut before multiplying anything
Build the power up one multiplication at a time
Watch for a repeating pattern, and use it to skip ahead

1
Only a square matrix can multiply itself — check that its row count equals its column count before doing anything else.

Example

$A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ is $2\times2$ , so $A^2, A^3,\dots$ are all defined.
- $A$ is square — continue to Step 2.
- $A$ is not square — $A^k$ for $k\geq2$ doesn't exist; stop here.
Why: $AA$ requires $A$ 's column count to match $A$ 's row count, the same inner-dimension rule any matrix product requires — a genuine power only exists when those two counts happen to be equal.
2
A diagonal matrix's powers come from raising each diagonal entry to the $k$ -th power directly; a triangular matrix (upper or lower) is guaranteed to stay triangular of the same kind at every power, though its entries still take real multiplication to find.

Example

$A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ is neither diagonal nor triangular — both off-diagonal entries are nonzero and the diagonal itself is $0$ — so no shortcut skips the arithmetic here; Steps 3–4 have to do the actual multiplying.

⚡ Shortcut

If $A$ actually is diagonal, skip Steps 3–4 entirely. For $D=\begin{pmatrix}2&0&0\\0&-3&0\\0&0&5\end{pmatrix}$ , $D^4=\begin{pmatrix}2^4&0&0\\0&(-3)^4&0\\0&0&5^4\end{pmatrix}=\begin{pmatrix}16&0&0\\0&81&0\\0&0&625\end{pmatrix}$ — no matrix multiplication at all, just three scalar exponents.

Why: these shapes aren't special-cased by accident — every off-diagonal entry of a diagonal matrix's product is a sum of terms that each include at least one zero factor, so it vanishes automatically, and the same zero pattern that makes a matrix triangular reproduces itself under multiplication.
3
Compute $A^2=AA$ first, then $A^3=A^2A$ , and so on, reusing each previous power rather than starting over from scratch.

Example

$A^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I$ $A^3=A^2A=(-I)A=-A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$

Why: associativity guarantees $A^jA^k=A^{j+k}$ no matter how the factors are grouped, so building up gradually from the last power computed always lands on the correct next one.
4
If an early power comes back to $I$ , to $-I$ , or to $A$ itself, every later power is forced to cycle through the same short list, and a large $k$ can be reached by repeated squaring ( $A^2, A^4, A^8, \dots$ ) instead of one multiplication at a time.

Example

Since $A^2=-I$ , squaring again gives $A^4=(A^2)^2=(-I)^2=I$ . Then $A^5=A^4A=IA=A$ — the original matrix, with no further multiplication needed once the $-I,I$ pattern is spotted: $\boxed{A^5=\begin{pmatrix}0&1\\-1&0\end{pmatrix}}$

Why: once a power repeats an earlier one, every later power is forced to repeat too, by the same associativity that makes $A^jA^k=A^{j+k}$ well-defined regardless of how the factors are grouped.

Common mistakes

Squaring each entry individually ( $A^{\circ2}$ ) instead of multiplying $A$ by itself — the entrywise square is a different, far less useful operation.
Trying to compute $A^k$ for a matrix that isn't square.
Recomputing $A^3$ as three fresh multiplications from scratch instead of reusing the already-computed $A^2$ .
Forgetting that $A^0=I$ , not the zero matrix.
Assuming a triangular matrix's powers can be read off entry by entry the way a diagonal matrix's can — triangular only guarantees the zero pattern survives, not that the surviving entries come for free.

Compute a matrix power A^k

Compute a matrix power $A^k$