Question

(1 point) Note: In this problem, use the method from class. See the lecture notes for January 12 on Brightspace. Also see Example 8.54 in the textbook. More examples (using real matrices) are in Section 8.6 of the textbook. Consider the sequence defined recursively by n+2 We can use matrix diagonalization to find an explicit formula for F (a) Find a matrix A that satisfies n+1 n+2 (b) Find the appropriate exponent k such that Fh Fi (c) Find a diagonal matrix D and an invertible matrix P such that A-PDP-

Answer 1

Given, F₀ = 0, F₁ = 3, F_n+2 = 2F_n+1-2F_n

a) Now, = $\begin{bmatrix} 0*F_{n}+1*F_{n+1}\\ -2*F_{n}+2*F_{n+1} \end{bmatrix}$

i.e., $\begin{bmatrix} F_{n+1}\\ F_{n+2} \end{bmatrix}$ = $\begin{bmatrix} 0 & 1\\ -2 & 2 \end{bmatrix}$ $\begin{bmatrix} F_n\\ F_{n+1} \end{bmatrix}$

Therefore, A = $\begin{bmatrix} 0 & 1\\ -2 & 2 \end{bmatrix}$ .

b) Here, $\begin{bmatrix} F_1\\ F_2 \end{bmatrix}$ = A $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$

And, $\begin{bmatrix} F_2\\ F_3 \end{bmatrix}$ = A $\begin{bmatrix} F_1\\ F_2 \end{bmatrix}$ = A*A $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$ ,i.e., $\begin{bmatrix} F_2\\ F_3 \end{bmatrix}$ = A² $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$

And, $\begin{bmatrix} F_3\\ F_4 \end{bmatrix}$ = A $\begin{bmatrix} F_2\\ F_3 \end{bmatrix}$ = A*A² $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$ ,i.e., $\begin{bmatrix} F_3\\ F_4 \end{bmatrix}$ = A³ $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$

Proceeding in this way we get, $\begin{bmatrix} F_n\\ F_{n+1} \end{bmatrix}$ = Aⁿ $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$

Therefore, k = n.

c) Now, the characteristic equation of A is, $\begin{vmatrix} 0-x & 1\\ -2 & 2-x \end{vmatrix}$ = 0

i.e., (-x)*(2-x)-1*(-2) = 0

i.e., x²-2x+2 = 0

i.e., x = 1+i, 1-i

Therefore, the eigenvalues of A are 1+i and 1-i.

Now, the eigenvectors corresponding to the eigenvalue 1+i are $c\begin{bmatrix} 1\\ 1+i \end{bmatrix}$ , where c is nonzero real.

And, the eigenvectors corresponding to the eigenvalue 1-i are $c\begin{bmatrix} 1\\ 1-i \end{bmatrix}$ , where c is nonzero real.

Let, P = $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ . Then P is non-singular matrix.

Now, AP = $\begin{bmatrix} 0 & 1\\ -2 & 2 \end{bmatrix}$ $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ = $\begin{bmatrix} 1+i & 1-i\\ 2i & -2i \end{bmatrix}$ = $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ $\begin{bmatrix} 1+i & 0\\ 0 & 1-i \end{bmatrix}$

i.e., AP = PD

i.e., D = P^-1AP

Therefore, D = $\begin{bmatrix} 1+i & 0\\ 0 & 1-i \end{bmatrix}$ and P = $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ .

d) Here, P = $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ .

Now, adj(P) = $\begin{bmatrix} 1-i & -1\\ -1-i & 1 \end{bmatrix}$ and det(P) = -2i.

Then, P^-1 = adj(P)/det(P) = -(1/2i) $\begin{bmatrix} 1-i & -1\\ -1-i & 1 \end{bmatrix}$

i.e., P^-1 = $\begin{bmatrix} (1+i)/2 & -i/2\\ (1-i)/2 & i/2 \end{bmatrix}$

Therefore, P^-1 = $\begin{bmatrix} (1+i)/2 & -i/2\\ (1-i)/2 & i/2 \end{bmatrix}$ .

e) A⁵ = PD⁵P^-1 = $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ $\begin{bmatrix} 1+i & 0\\ 0 & 1-i \end{bmatrix}^5$ $\begin{bmatrix} (1+i)/2 & -i/2\\ (1-i)/2 & i/2 \end{bmatrix}$

i.e., A⁵ = $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ $\begin{bmatrix} (1+i)^5 & 0\\ 0 & (1-i)^5 \end{bmatrix}$ $\begin{bmatrix} (1+i)/2 & -i/2\\ (1-i)/2 & i/2 \end{bmatrix}$

i.e., A⁵ = $\begin{bmatrix} 1 & 1\\ 1+i & 1-i \end{bmatrix}$ $\begin{bmatrix} (1+i)^6/2 & -i(1+i)^5/2\\ (1-i)^6/2 & i(1-i)^5/2 \end{bmatrix}$

i.e., A⁵ = $\begin{bmatrix} [(1+i)^6+(1-i)^6]/2 & i[(1-i)^5-(1+i)^5]/2\\ [(1+i)^7+(1-i)^7]/2 & i[(1-i)^6-(1+i)^6]/2 \end{bmatrix}$

Therefore, A⁵ = $\begin{bmatrix} [(1+i)^6+(1-i)^6]/2 & i[(1-i)^5-(1+i)^5]/2\\ [(1+i)^7+(1-i)^7]/2 & i[(1-i)^6-(1+i)^6]/2 \end{bmatrix}$ .

f) We know that $\begin{bmatrix} F_n\\ F_{n+1} \end{bmatrix}$ = Aⁿ $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$

Putting n = 5 we get, $\begin{bmatrix} F_5\\ F_{6} \end{bmatrix}$ = A⁵ $\begin{bmatrix} F_0\\ F_1 \end{bmatrix}$

i.e., $\begin{bmatrix} F_5\\ F_{6} \end{bmatrix}$ = $\begin{bmatrix} [(1+i)^6+(1-i)^6]/2 & i[(1-i)^5-(1+i)^5]/2\\ [(1+i)^7+(1-i)^7]/2 & i[(1-i)^6-(1+i)^6]/2 \end{bmatrix}$ $\begin{bmatrix} 0\\ 3 \end{bmatrix}$

i.e., $\begin{bmatrix} F_5\\ F_{6} \end{bmatrix}$ = $\begin{bmatrix} 3i[(1-i)^5-(1+i)^5]/2\\ 3i[(1-i)^6-(1+i)^6]/2 \end{bmatrix}$

Therefore, F₅ = 3i[(1-i)⁵-(1+i)⁵]/2

g) From previous part we get, F₅ = 3i[(1-i)⁵-(1+i)⁵]/2

Replacing 5 by n we get, F_n = 3i[(1-i)ⁿ-(1+i)ⁿ]/2.

Therefore, F_n = 3i[(1-i)ⁿ-(1+i)ⁿ]/2.