A Recursive Relationship for a Power of a Matrix

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 685

Suppose that the $2 \times 2$ matrix $A$ has eigenvalues $4$ and $-2$. For each integer $n \geq 1$, there are real numbers $b_n , c_n$ which satisfy the relation
\[ A^{n} = b_n A + c_n I , \] where $I$ is the identity matrix.

Find $b_n$ and $c_n$ for $2 \leq n \leq 5$, and then find a recursive relationship to find $b_n, c_n$ for every $n \geq 1$.

 
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Solution.

Because the eigenvalues of $A$ are $4$ and $-2$, its characteristic polynomial must be
\[ p(\lambda) = (\lambda – 4) ( \lambda + 2) = \lambda^2 – 2 \lambda – 8 . \]

The Cayley-Hamilton Theorem tells us that $p(A) = 0$, the zero matrix. Rearranging terms, we find that
\[ A^2 = 2 A + 8 I . \] So $b_2=2$ and $c_2=8$.


With this relationship, we can reduce the higher powers of $A$:
\begin{align*}
A^3 &= A^2 A= (2A + 8 I)A = 2A^2 + 8 A\\
& = 2(2A+8I)+8A= 12 A + 16 I
\end{align*}
\begin{align*}
A^4 &= A^3 A=(12 A + 16 I)A = 12A^2+16A\\
&=12(2A+8I)+16A = 40 A + 96 I
\end{align*}
\begin{align*}
A^5 &=A^4A= (40 A + 96 I)A = 40 A^2 + 96 A \\
&=40(2A+8I)+96A = 176 A + 320 I
\end{align*}
Hence, we have $b_3=12, c_3=16, b_4=40, c_4=96, b_5=176$, and $c_5=320$.


To find the recursive relationship, suppose we know that $A^n = b_n A + c_n I$. Then
\begin{align*}
A^{n+1} &= A^{n} A = ( b_n A + c_n I ) A = b_n A^2 + c_n A\\
&=b_n(2A+8I)+c_n A = (2 b_n + c_n) A + 8 b_n I.
\end{align*}
This gives the recursive relationships
\[ b_{n+1} = 2 b_n + c_n , \qquad c_{n+1} = 8 b_n . \]

Using the work above, you can quickly verify this for $1 \leq n \leq 4$.


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