Science:Math Exam Resources/Courses/MATH221/April 2013/Question 09/Solution 1

Determine eigenvalues of $\ B=10A$

We will first find the eigenvalues of the matrix $B=\left[{\begin{array}{cc}8&5\\2&5\end{array}}\right]$ to avoid dealing with fractions. The eigenvalues of the matrix A, will simply be ${\frac {1}{10}}$ of this integer matrix B.

To find the eigenvalues we calculate as usual:

${\begin{aligned}\det \left(\left[{\begin{array}{cc}8-t&5\\2&5-t\end{array}}\right]\right)&=(8-t)(5-t)-(5)(2)\\&=t^{2}-13t+30\\&=(t-3)(t-10)\\&=0\end{aligned}}$

Determine eigenvectors of $\ B=10A$

For the eigenvectors with eigenvalue 3 we look at $\left[{\begin{array}{cc}8-3&5\\2&5-3\end{array}}\right]=\left[{\begin{array}{cc}5&5\\2&2\end{array}}\right]$ which gives the eigenvector $\left[{\begin{array}{cc}1\\-1\end{array}}\right]$.

For the eigenvalue 10, we look at the $\left[{\begin{array}{cc}-2&5\\2&-5\end{array}}\right]$ which gives the eigenvectors $\left[{\begin{array}{cc}5\\2\end{array}}\right]$.

Eigenvalues and eigenvectors of $\ A$

For the matrix A this means that A has the

• eigenvalue ${\frac {3}{10}}$ with eigenvector $v_{1}=\left[{\begin{array}{cc}1\\-1\end{array}}\right]$
• eigenvalue $\ 1$ with eigenvector $v_{2}=\left[{\begin{array}{cc}5\\2\end{array}}\right]$

Find $\ a,b\in \mathbb {R}$ such that $\ x_{0}=av_{1}+bv_{2}$

To find the closed formulae for the components of x_n we decompose the vector $x_{0}$ into a sum of the eigenvectors of A. To do so, we use the extended matrix:

$[v_{1}\ v_{2}\ |\ x_{0}]=\left[{\begin{array}{cc|c}1&5&4\\-1&2&1\\\end{array}}\right]\sim \left[{\begin{array}{cc|c}1&5&4\\0&7&5\\\end{array}}\right]\sim \left[{\begin{array}{cc|c}1&0&3/7\\0&1&5/7\\\end{array}}\right]$

Thus a = 3/7 and b = 5/7 and therefore

$\left[{\begin{array}{cc}4\\1\end{array}}\right]=3/7\left[{\begin{array}{cc}1\\-1\end{array}}\right]+5/7\left[{\begin{array}{cc}5\\2\end{array}}\right]$.

Computing $\ x_{n}$

We can now compute $x_{n}$ as follows:

${\begin{aligned}x_{n}=A^{n}x_{0}&=A^{n}\left({\frac {3}{7}}\left[{\begin{array}{cc}1\\-1\end{array}}\right]+{\frac {5}{7}}\left[{\begin{array}{cc}5\\2\end{array}}\right]\right)\\&={\frac {3}{7}}A^{n}\left[{\begin{array}{cc}1\\-1\end{array}}\right]+{\frac {5}{7}}A^{n}\left[{\begin{array}{cc}5\\2\end{array}}\right]\\&={\frac {3}{7}}{\frac {3^{n}}{10^{n}}}\left[{\begin{array}{cc}1\\-1\end{array}}\right]+{\frac {5}{7}}1^{n}\left[{\begin{array}{cc}5\\2\end{array}}\right]\end{aligned}}$