Matrix Powers by Diagonalization

Squaring $P D P^{-1}$

The matrix A = (4 −1; 2 1) diagonalizes as $A = P D P^{-1}$, with P = (1 1; 1 2) holding the eigenvectors (1, 1) and (1, 2) as its columns, D = (3 0; 0 2) holding the eigenvalues, and the inverse of P equal to the matrix (2 −1; −1 1).

Square it. $A^{2} = \left(P D P^{-1}\right)\left(P D P^{-1}\right)$. Matrix multiplication is associative, so the six factors may be grouped in any way that keeps their order: $A^{2} = P D \left(P^{-1} P\right) D P^{-1}$. The inner pair $P^{-1} P$ is the identity I, and multiplying by I changes nothing, so $A^{2} = P D D P^{-1} = P D^{2} P^{-1}$.

Every power

Multiply once more by A to get the cube: $A^{3} = \left(P D^{2} P^{-1}\right)\left(P D P^{-1}\right) = P D^{2} \left(P^{-1} P\right) D P^{-1} = P D^{3} P^{-1}$. Each extra factor of A brings one more D into the middle and one more $P^{-1} P$, which collapses to I.

With n factors of A there are n − 1 inner pairs, and every one of them is I. What is left is one P at the front, n copies of D in the middle and one $P^{-1}$ at the back: $A^{n} = P D^{n} P^{-1}$, for every whole number n.

Powers of a diagonal matrix

Square D = (3 0; 0 2) by multiplying rows into columns. The top left entry is 3 × 3 + 0 × 0 = 9, the bottom right is 0 × 0 + 2 × 2 = 4, and each off-diagonal entry is a sum of two products that each contain a 0, such as 3 × 0 + 0 × 2 = 0. So D squared is the matrix (9 0; 0 4): still diagonal, with each entry squared.

Every further factor of D does the same, so each entry on the diagonal is raised to the power on its own. $D^{5}$ has $3^{5} = 243$ and $2^{5} = 32$ on its diagonal and zeros elsewhere.

Raising D to the fifth power raises each diagonal entry to the fifth power, $3^{5} = 243$ and $2^{5} = 32$, and the zeros stay zero.

$A^{5}$ in two products

Now $A^{5} = P D^{5} P^{-1}$. Work from the left. Multiplying P on the right by a diagonal matrix scales its columns: the first column (1, 1) by 243 and the second column (1, 2) by 32. So $P D^{5}$ is the matrix (243 32; 243 64).

Multiply that by the inverse of P, the matrix (2 −1; −1 1). Row 1 gives 243 × 2 + 32 × (−1) = 486 − 32 = 454 and 243 × (−1) + 32 × 1 = −211. Row 2 gives 243 × 2 + 64 × (−1) = 486 − 64 = 422 and 243 × (−1) + 64 × 1 = −179.

So $A^{5}$ is the matrix (454 −211; 422 −179). The work was two powers of numbers and two matrix products, and it would be the same amount of work for $A^{20}$ or $A^{100}$.

The columns of P scaled by 243 and 32, then multiplied by the inverse of P, give A to the fifth power.

Check it the long way

Multiplying A by itself step by step gives the same matrices. $A^{2}$ is (14 −5; 10 −1), and through the diagonal it is $P D^{2} P^{-1}$: $P D^{2}$ is (9 4; 9 8), and multiplying by the inverse of P gives the entries 18 − 4 = 14, −9 + 4 = −5, 18 − 8 = 10 and −9 + 8 = −1, which is (14 −5; 10 −1) again.

Carrying on, $A^{3}$ is (46 −19; 38 −11), $A^{4}$ is (146 −65; 130 −49), and $A^{5}$ is (454 −211; 422 −179), the same matrix as before. The long way took four matrix products, and every further power costs one more.

Every entry of $A^{n}$

Keep n as a letter. $D^{n}$ has $3^{n}$ and $2^{n}$ on its diagonal, so $P D^{n}$ scales the first column of P by $3^{n}$ and the second by $2^{n}$: its columns are $\left(3^{n}, 3^{n}\right)$ and $\left(2^{n}, 2 \times 2^{n}\right)$. Multiplying by the inverse of P gives every entry of $A^{n}$ at once. The top row of $A^{n}$ is $2 \times 3^{n} - 2^{n}$ and $2^{n} - 3^{n}$, and the bottom row is $2 \times 3^{n} - 2^{n+1}$ and $2^{n+1} - 3^{n}$.

Check it at n = 1: 6 − 2 = 4, 2 − 3 = −1, 6 − 4 = 2 and 4 − 3 = 1, which is A. At n = 5: 486 − 32 = 454, 32 − 243 = −211, 486 − 64 = 422 and 64 − 243 = −179, which is $A^{5}$.

The larger eigenvalue takes over

The first column of $A^{n}$ is what A does, n times over, to the column (1, 0). Write (1, 0) in eigenvector steps: (1, 0) = 2(1, 1) − (1, 2). Each application of A multiplies the first part by 3 and the second by 2, so after n steps the column is $2 \times 3^{n} \left(1, 1\right) - 2^{n} \left(1, 2\right)$, which is the first column of the formula.

Divide by $3^{n}$ to compare the two parts: $2\left(1, 1\right) - \left(\frac{2}{3}\right)^{n} \left(1, 2\right)$. The factor $\left(\frac{2}{3}\right)^{n}$ shrinks toward 0, so the column points closer and closer along (1, 1), the eigenvector of the larger eigenvalue. This is how a population model behaves: $M^{n}$ times the starting counts gives the counts n years later, and in the long run the counts settle into the proportions of the eigenvector with the largest eigenvalue.

The first column of $A^{n}$ divided by $3^{n}$, for n = 0, 1, 2, 3 and 4: (1, 0), then $\left(\frac{4}{3} , \frac{2}{3}\right)$, $\left(\frac{14}{9} , \frac{10}{9}\right)$, $\left(\frac{46}{27} , \frac{38}{27}\right)$ and $\left(\frac{146}{81} , \frac{130}{81}\right)$. Each point lies on the line y = 2x − 2, which runs in the direction (1, 2), and the gap to (2, 2) shrinks by a factor of $\frac{2}{3}$ at every step.

The usual mistakes

Writing $A^{n} = P^{n} D^{n} \left(P^{-1}\right)^{n}$. Matrices do not commute, so the factors of $\left(P D P^{-1}\right)^{n}$ cannot be gathered up by kind. Only the inner pairs cancel, which leaves one P and one $P^{-1}$.

Raising the entries of A itself. $A^{5}$ is not the matrix of fifth powers (1024 −1; 32 1); only a diagonal matrix can be raised entry by entry.

Multiplying the diagonal entries by the power instead of raising them. $D^{5}$ has 243 and 32 on its diagonal, not 15 and 10.

Forgetting P and $P^{-1}$ at the end. $D^{5}$ is $A^{5}$ written in eigenvector coordinates; it becomes $A^{5}$ only after the two outer products.

Losing the sign of a negative eigenvalue. $\left(-1\right)^{10} = 1$ and $\left(-1\right)^{11} = -1$, so an even power and an odd power give different matrices.

Rabbits, voles and owls

In the applications below, a breeding matrix is diagonalized and only the diagonal matrix is raised to the tenth power, to count the young and adult rabbit pairs ten months ahead. Then the starting numbers of voles and owls are written as a mix of two eigenvectors, each part is multiplied by its own eigenvalue once for each year, and the larger eigenvalue gives the long-run number of voles for each owl.