# Question 1

I assume we’re defining (a_n) + (b_n) = (a_n + b_n) and \lambda(a_n) = (\lambda a_n). Note that the zero sequence u_n = 0 has u_n + u_{n - 1} = 2 \times 0 = 0 = u_{n + 1}.

Suppose that (u_n), (v_n) both have u_{n + 1} = u_n + u_{n - 1} and v_{n + 1} = u_n + u_{n - 1}. We also have \lambda u_n + v_n + \lambda u_{n - 1} + v_{n - 1} = \lambda (u_n + u_{n - 1}) + v_n + v_{n - 1} = \lambda u_{n + 1} + v_{n + 1}. So the sequence (\lambda u_n + v_n) also satisfies the recurrence, so this set forms a vector subspace of the space of real sequences, so is a vector space.

We solve the auxilliary equation:

\begin{align*}\lambda^2 - \lambda - 1 & = \left(\lambda - \frac 1 2\right)^2 - \frac 1 4 - 1 \\ & = \left(\lambda - \frac 1 2\right)^2 - \frac 5 4 \\ &= 0\end{align*}

So:

\displaystyle \lambda = \frac {1 \pm \sqrt 5} 2

giving the general solution for the recurrence as:

\displaystyle u_n = A \left(\frac {1 + \sqrt 5} 2\right)^n + B \left(\frac {1 - \sqrt 5} 2\right)^n

So the space is spanned by the sequences (a_n), (b_n) with \displaystyle a_n = \left(\frac {1 + \sqrt 5} 2\right)^n and \displaystyle b_n = \left(\frac {1 - \sqrt 5} 2\right)^n. Since:

\begin{align*}\frac {a_n} {b_n} &= \left(\frac {1 + \sqrt 5} 2 \times \frac 2 {1 - \sqrt 5}\right)^n \\ & = \left(\frac {1 + \sqrt 5} {1 - \sqrt 5}\right)^n \\ & = \left(-\frac 1 2\right)^n \left(3 + \sqrt 5\right)^n \end{align*}

is not a constant sequence, (a_n) is not a multiple of (b_n), so they are linearly independent. So the vector space has basis \{(a_n), (b_n)\} and so dimension 2.