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.

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
\begin{align*}
A^3&=AA^2=2A^2+3A\\
&=2(2A+3I)+3A\\
&=7A+6I.
\end{align*}
Thus, $a_2=7, b_2=6$.

In general, we have
\begin{align*}
A^{n+2}&=AA^{n+1}\\
&=A(a_nA+b_nI)\\
&=a_nA^2+b_nA\\
&=a_n(2A+3I)+b_n A\\
&=(2a_n+b_n)A+(3a_n)I.
\end{align*}
Thus we obtain
\begin{align*}
a_{n+1}&=2a_n+b_n\\
b_{n+1}&=3a_n.
\end{align*}

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
\begin{align*}
A^n=\frac{1}{4} \big( (-1)^n+3^{n+1} \big)A+\frac{3}{4} \big( (-1)^{n-1}+3^{n} \big)I
\end{align*}
for any positive integer $n$.

Solve a Linear Recurrence Relation Using Vector Space Technique
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Let $U$ be a subspace of $V$ defined by
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Let $V$ be a real vector space of all real sequences
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Let $T$ be […]

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Let
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-1 & 2 & 5
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Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.
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Let
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Find the eigenvalues and the eigenvectors of the matrix
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Hint.
Apply the Cayley-Hamilton theorem.
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\[(a_i)_{i=1}^{\infty}=(a_1, a_2, \cdots).\]
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Prove that $U$ is a subspace of […]

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