How to Find a Formula of the Power of a Matrix

Linear algebra problems and solutions

Problem 8

Let $A= \begin{bmatrix}
1 & 2\\
2& 1
\end{bmatrix}$.
Compute $A^n$ for any $n \in \N$.

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Plan.

We diagonalize the matrix $A$ and use this Problem.

Steps.

  1. Find eigenvalues and eigenvectors of the matrix $A$.
  2. Diagonalize the matrix $A$.
  3. Use the result of  this Problem.

Proof.

We first diagonalize the matrix $A$. We solve
\begin{align*}
\det(A-\lambda I) & =\begin{vmatrix}
1-\lambda & 2\\
2& 1-\lambda
\end{vmatrix} \\
&=(1-\lambda)^2-4=\lambda^2-2\lambda-3 =(\lambda+1)(\lambda-3)=0
\end{align*}
and obtain the eigenvalues $\lambda=-1, 3$.

To find an eigenvector $\mathbf{x}$ corresponding to $\lambda=-1$, we solve $(A+I)\mathbf{x}=\mathbf{0}$ or
\[ \begin{bmatrix}
2 & 2\\
2& 2
\end{bmatrix}
\begin{bmatrix} x_1 \\
x_2
\end{bmatrix}=
\begin{bmatrix} 0 \\
0
\end{bmatrix}.\] We obtain an eigenvector $\mathbf{x}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}$ corresponding to $\lambda=-1$.

Similarly, solving $(A-3I)\mathbf{y}=\mathbf{0}$, we obtain an eigenvector $\mathbf{y}=\begin{bmatrix}
1 \\
1
\end{bmatrix}$
corresponding to $\lambda=3$.

Thus the invertible matrix $S=[\mathbf{x} \, \mathbf{y}]=\begin{bmatrix}
1 & 1 \\
-1& 1
\end{bmatrix}$ diagonalizes the matrix $A$, that is,
\[ S^{-1}AS =\begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix} \text{ or equivalently }
A=S\begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix} S^{-1}.\] Then for each $n \in \N$, we have
\begin{align*}
A^n &= \left (S\begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix} S^{-1} \right)^n
=S \begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix}^n S^{-1}
=S \begin{bmatrix}
(-1)^n & 0\\
0& 3^n
\end{bmatrix} S^{-1} \\[6pt] &=
\begin{bmatrix}
1 & 1 \\
-1& 1
\end{bmatrix}
\begin{bmatrix}
(-1)^n & 0\\
0& 3^n
\end{bmatrix}
\frac{1}{2}
\begin{bmatrix}
1 & -1 \\
1& 1
\end{bmatrix}
=\frac{1}{2} \begin{bmatrix}
(-1)^n+3^n & (-1)^{n+1}+3^n\\
(-1)^{n+1}+3^n& (-1)^n+3^n
\end{bmatrix}.
\end{align*}

(See this problem for the details of these computations.)

Therefore we obtained the formula
\[A^n=\frac{1}{2} \begin{bmatrix}
(-1)^n+3^n & (-1)^{n+1}+3^n\\
(-1)^{n+1}+3^n& (-1)^n+3^n
\end{bmatrix}.\]

Comment.

Another typical method to compute a power of a square matrix is mathematical induction. To use it, we need to first compute several small powers like $A^2$ and $A^3$ and guess the formula for $A^n$.

If you can guess the formula, then the mathematical induction part is not difficult. But for this specific problem, the formula is a bit complicated to guess as you can see from the solution above. Thus we used diagonalization trick.


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