Find the Formula for the Power of a Matrix Using Linear Recurrence Relation

Linear algebra problems and solutions

Problem 323

Suppose that $A$ is $2\times 2$ matrix that has eigenvalues $-1$ and $3$.
Then for each positive integer $n$ find $a_n$ and $b_n$ such that
\[A^{n+1}=a_nA+b_nI,\] where $I$ is the $2\times 2$ identity matrix.

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Since $-1, 3$ are eigenvalues of the matrix $A$, the characteristic polynomial of $A$ is
\[(t+1)(t-3)=t^2-2t-3.\] By Cayley-Hamilton theorem, we have
\[A^2-2A-3I=O,\] where $O$ is the $2\times 2$ zero matrix.
Thus we have
\[A^2=2A+3I, \tag{*}\] and hence $a_1=2, b_1=3$.

Multiplying (*) by $A$, we have
Thus, $a_2=7, b_2=6$.

In general, we have
&=a_n(2A+3I)+b_n A\\
Thus we obtain

This yields the linear recurrence relation
\[a_{n+2}=2a_{n+1}+3a_{n}\] with initial values $a_1=2, a_2=7$.

The general term $a_n$ of the sequence $(a_n)_{i=n}^{\infty}$ is obtained in the previous post Solve a linear recurrence relation using vector space technique:
\[a_n=\frac{1}{4} \big( (-1)^n+3^{n+1} \big).\] (You may alternatively find this using an elementary technique.)

Then we also have
\[b_n=3a_{n-1}=\frac{3}{4} \big( (-1)^{n-1}+3^{n} \big).\] (This formula is true for $n=1$ as well.)

In conclusion, we have obtained
A^n=\frac{1}{4} \big( (-1)^n+3^{n+1} \big)A+\frac{3}{4} \big( (-1)^{n-1}+3^{n} \big)I
for any positive integer $n$.

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