Find the $n$th power of a 3-by-3 circulant matrix

Question

Consider the matrix given as $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}$$ Write a formula for $A^n$ for $n\in\mathbb{N}$.

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My attempt: The first that comes to mind is to diagonalize it and hence find the formula for $A^n$, but that is very messy, so I tried to do something else it goes as:

bserve that $$A=\begin{bmatrix}a_0 & a_2 & a_1\\ a_1 & a_0 & a_2\\ a_2 & a_1 & a_0\end{bmatrix}=a_0\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}+a_1\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}+a_2\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix}$$

Let $U=\begin{bmatrix}0 & 0 & 1\\1 & 0 & 0\\0 & 1 & 0\end{bmatrix}$ then we will have that $$\begin{matrix}U^2=\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{bmatrix} & \text{ and } & U^3=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\end{matrix}$$

Hence we have $A=a_0I+a_1U+a_2U^2 = a_0U^3+a_1U+a_2U^2=(a_0U^2+a_1I+a_2U)U$

This got me thinking that there might be an easy way to solve the above problem, but I could not make any further progress.

Please Help and thanks in advance.

Answer 1

$A$ is a so-called circulant matrix, which can be denoted as ${\rm circ}(a_0,a_1,a_2)$. Let $\alpha = \exp{2\pi i/3}$; $\alpha$ is a 3rd root of unity, and satisfies $\alpha^3 = 1$ and $1 + \alpha + \alpha^2 = 0$ (as does its conjugate). By the general theory of circulants, we can diagonalize $A$ as follows: $$ A = U \,{\rm diag}(d_0,d_1,d_2)\, U^*,$$ where $U$ is the unitary matrix: $$ U = \frac{1}{\sqrt{3}}\,\pmatrix{1 &1 & 1 \\ 1 &\alpha &\alpha^2 \\ 1 &\alpha^2 &\alpha^4},$$ and $d = \sqrt{3}\,U^* a$. We recover $a$ from the inverse transformation: $a = \frac{1}{\sqrt{3}} U d.$

We can now calculate \begin{align} A^n &= (U D U^*)^n \\ &= U D^n U^* \\ &= U\, {\rm diag}(d_0^n, d_1^n, d_2^n)\, U^*\\ &= {\rm circ}(a^{(n)}_0, a^{(n)}_1, a^{(n)}_2), \end{align} where $a^{(n)} = \frac{1}{\sqrt{3}} U\cdot (d^n_0,d^n_1,d^n_2)^T.$

Answer 2

Circulant matrices are related to Fourier transform.

Consider the DFT matrix of length $N=3$ $$ \mathbf{W} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & \Omega & \Omega^2 \\ 1 & \Omega^2 & \Omega^4 \end{pmatrix} $$ where $\Omega = e^{-2\pi i/N}$.

Consider now the DFT of the signal $$ \mathbf{y}= \mathbf{W} \begin{pmatrix} a_0 \\ a_1 \\ a_2 \end{pmatrix} $$

From here, you can observe that $$ \mathbf{WA}= \mathbf{W} \begin{pmatrix} a_0 & a_2 & a_1 \\ a_1 & a_0 & a_2 \\ a_2 & a_1 & a_0 \end{pmatrix} = \mathrm{Diag}(\mathbf{y}) \mathbf{W} $$ and thus $$ \mathbf{A} = \mathbf{W}^{-1} \mathrm{Diag}(\mathbf{y}) \mathbf{W} = \frac{1}{N} \mathbf{W}^{H} \mathbf{D} \mathbf{W} $$ with $\mathbf{D}=\mathrm{Diag}(\mathbf{y})$.

Finally the $k$th power of $\mathbf{A}$ is $$ \mathbf{A}^k = \frac{1}{N} \mathbf{W}^{H} \mathbf{D}^k \mathbf{W} $$

Answer 3

I think you've reached a decent result and by just using the multinomial theorem you can obtain a generalized solution.

$A=a_0I+ a_1U + a_2 U^2 \\ A^n = (a_0I+ a_1U + a_2 U^2)^n\\$

Then the coeffecients of $U^3$, $U^2$ and $U^1$ are determined using the multinomial theorem and just plugged back into the result matrix.

$$ \begin{matrix} \sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\ \sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\ \sum_{a+b+c=n,\\ b+2c =3k+2} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k+1} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c & \sum_{a+b+c=n,\\ b+2c =3k} \frac{n!}{a!b!c!}(a_0)^ a(a_1)^b(a_2)^c \\ \end{matrix} $$ The answer may seem obvious but unfortunately the multinomial expansion can't be simplified further (as per my knowledge). Even if it could it would just be a huge mess.