2006 SMT/Advanced Topics Problems/Problem 2

Problem

Define $A=\left(\,\begin{matrix}0 & 1\\ 3 &0\end{matrix}\right)$ . Find a vertical vector $v$ such that $(A^8+A^6+A^4+A^2+I)v=\left(\begin{matrix}0\\11\end{matrix}\right)$ (where $I$ is the $2\times2$ identity matrix).

Solution

If we calculate the first couple powers of $A$ , we quickly see a pattern:

Lemma: $A^{2n}=\left(\,\begin{matrix}3^n & 0\\ 0 & 3^n\end{matrix}\right)$

Proof: Clearly this is true for $n=0$ . Thus, we proceed with an induction arguement.

$A^{2(n+1)}=A^{2n}\cdot A\cdot A=\left(\,\begin{matrix}3^n & 0\\ 0 & 3^n\end{matrix}\right)\cdot\left(\,\begin{matrix}0 & 1\\ 3 & 0\end{matrix}\right)\cdot\left(\,\begin{matrix}0 & 1\\ 3 & 0\end{matrix}\right)$

$=\left(\,\begin{matrix}0 & 3^n\\ 3^{n+1} & 0\end{matrix}\right)\cdot\left(\,\begin{matrix}0 & 1\\ 3 & 0\end{matrix}\right)$

$=\left(\,\begin{matrix}3^{n+1} & 0\\ 0 & 3^{n+1}\end{matrix}\right)$

Thus, our proof is complete. $\square$

Therefore, $A^8+A^6+A^4+A^2+I=\left(\,\begin{matrix}81 & 0\\ 0 & 81\end{matrix}\right)+\left(\,\begin{matrix}27 & 0\\ 0 & 27\end{matrix}\right)+\left(\,\begin{matrix}9 & 0\\ 0 & 9\end{matrix}\right)+\left(\,\begin{matrix}3 & 0\\ 0 & 3\end{matrix}\right)+\left(\,\begin{matrix}1 & 0\\ 0 & 1\end{matrix}\right)$ and this is equal to $\left(\,\begin{matrix}121 & 0\\ 0 & 121\end{matrix}\right)$ .

Finally, we want to find a vector $v$ such that $\left(\,\begin{matrix}121 & 0\\ 0 & 121\end{matrix}\right)v=\left(\begin{matrix}0\\11\end{matrix}\right)$ . Letting $v=\left(\begin{matrix}a\\b\end{matrix}\right)$ , we get the system of equations $121a=0$ and $121b=11$ . From this, we have $a=0$ and $b=\frac{1}{11}$ , and so our desired vector is $\boxed{\left(\begin{matrix}0\\ \frac{1}{11}\end{matrix}\right)}$ .

Page

Toolbox

Search

2006 SMT/Advanced Topics Problems/Problem 2

Problem

Solution

See Also