Science:Math Exam Resources/Courses/MATH307/April 2010/Question 04 (b)

MATH307 April 2010

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Question 04 (b)
Consider the following recursion relation: $x_{n+1}=ax_{n}+bx_{n-1}$ where $n\geq 1$, and $a,b$ $\epsilon$ $\mathbb {R}$ How must the parameters a and b be related if it is given that $A_{1}$ has $\displaystyle \lambda _{1}=1$ as its eigenvalue? What is the other eigenvalue of $A_{1}$ (in terms of a and b )?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Knowing $A_{1}$ has an eigenvalue of 1 we need to solve the equation: $A_{1}x=\lambda x$ where $\lambda$ = $1$ ${\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}={\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$ making it into a system of linear equations we get: 1) $ax_{1}+bx_{0}$ = $x_{1}$ 2) $x_{1}$ = $x_{0}$ using 2) we can substitute into 1) so that we are only working with one variable, $ax_{0}+bx_{0}$ = $x_{0}$ $x_{0}$($a$ $+$ $b$) $=$ $x_{0}$ $a$ $+$ $b$ $=$ $1$ now that we see how $a$ and $b$ have to relate, let’s find the other eigenvalue of $A_{1}$ solve for either $a$ or $b$ and input back into $A_{1}$ and then do our standard $det(A_{1}$ $-$ $\lambda I)$ $=$ $0$ I’m going to continue after having solved for $b$ so our new $A_{1}$ = ${\begin{bmatrix}a&1-a\\1&0\end{bmatrix}}$ $det(A_{1}-\lambda I)=det({\begin{bmatrix}a-\lambda &1-a\\1&-\lambda \end{bmatrix}})=(a-\lambda )(-\lambda )-(1-a)=0$ we end up obtaining the equation: $\lambda ^{2}$ $-$ $\lambda a$ $+$ ($a$ $-$ $1$) $=$ $0$ using the quadratic formula with a = $1$, b = $-a$ and c = $a$ $-$ $1$, our result is: $\lambda ={\frac {a\pm {\sqrt {a^{2}-4(a-1)}}}{2}}={\frac {a\pm {\sqrt {a^{2}-4a+4}}}{2}}={\frac {a\pm {\sqrt {(a-2)^{2}}}}{2}}={\frac {a\pm (a-2)}{2}}$ $\lambda$ $=$ $1$, $a$ $-$ $1$ we knew about the eigenvalue of $1$ already, and so the other eigenvalue of $A_{1}$ is $a$ $-$ $1$ or, equivalently, $-b$