Finding a basis $\mathbb{R}_2$ of eigenvectors

Question

Find a basis $\mathbb{R}_2$ of eigenvectors of $A$, $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} $. Next, find the formula of $A^n$, $n \in \mathbb{N}$.

I have no idea how to solve first part of the problem. In the second part should I find $A^2$, $A^3$, $A^4$, try to guess $A^n$ and try to prove my hypothesis?

Answer 1

Let's first find the eigenvalues. To find an eigenvalue, you want to solve the equation $|\lambda I -A| =0$. This is called the characteristic polynomial, and in your case this polynomial is $\lambda^2-\lambda-1$. This has roots $\frac{1}{2}(1 \pm \sqrt{5})$. Then, solve the vector equation $(\lambda I - A)\textbf{v} = 0$ to get the eigenvectors. I'll let you take it from there. For the second part, I would do exactly what you said.

Answer 2

Because $A^2=A+I$, this matrix is the one expressing the Fibonacci recurrence: $$ {F_{n+1} \choose F_{n+1}} =\begin{pmatrix}1&1\\1&0\end{pmatrix}{F_{n} \choose F_{n-1}} $$ This gives $$ {F_{n+1} \choose F_{n}} =\begin{pmatrix}1&1\\1&0\end{pmatrix}^n{F_{1} \choose F_{0}} $$ and $$ \begin{pmatrix}1&1\\1&0\end{pmatrix}^n= \begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix} $$ which can also be written $A^n= F_n A + F_{n-1}I$.

This matrix formulation for the Fibonacci sequence is well worth knowing and easily proved.

By diagonalizing this matrix, we get Binet's formula for the Fibonacci numbers.

Answer 3

Solving $determinant(A-\lambda I)=0$ gives roots will imply that $\lambda=\frac{1 \pm \sqrt{5}}{2}$. You then $rref(A-(\frac{1+\sqrt{5}}{2})I)= \begin{bmatrix} 1 & -\frac{1 + \sqrt{5}}{2} \\ 0 & 0 \end{bmatrix}$. Say $x_2=t$. Thus, $x_1=t(\frac{1 + \sqrt{5}}{2})$.

Hence, $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} t(\frac{1 + \sqrt{5}}{2}) \\ t \end{bmatrix}=\begin{bmatrix} t(\frac{1 + \sqrt{5}}{2})+t\\ t(\frac{1 + \sqrt{5}}{2}) \end{bmatrix}=\frac{1+\sqrt{5}}{2}\begin{bmatrix} t(\frac{1 + \sqrt{5}}{2})\\ t \end{bmatrix}$ (I know the last two matrices was a big jump; I'll leave that to you.)

Analgously to the first root, $rref(A-(\frac{1-\sqrt{5}}{2})I)=\begin{bmatrix} 1 & -\frac{1-\sqrt{5}}{2}\\ 0 & 0 \end{bmatrix}$.

Similarly, we figure (that being the hence part) $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} t(\frac{1 - \sqrt{5}}{2}) \\ t \end{bmatrix}=\frac{1-\sqrt{5}}{2}\begin{bmatrix} \frac{1-\sqrt{5}}{2}t\\ t \end{bmatrix}$.

So find a $t$ that is not linearly dependent (zero would be bad) for this equation like $2$ and this will be your end result: $A^n=A(n,t)=\begin{bmatrix} t\frac{1+\sqrt{5}}{2} & t\frac{1-\sqrt{5}}{2}\\ t & t \end{bmatrix}\begin{bmatrix} (\frac{1+\sqrt{5}}{2})^n & 0\\ 0 & (\frac{1-\sqrt{5}}{2})^n \end{bmatrix} \begin{bmatrix} t(\frac{1+\sqrt{5}}{2}) & t(\frac{1-\sqrt{5}}{2})\\ t & t \end{bmatrix}^{-1}$.

I'll let you simplify that nasty equation. Trust me the end answer is right though I checked. Also, $(\frac{1-\sqrt{5}}{2})^n$ is being raised to the power of $n$. I don't know why but it looks like I am multiplying it here for some weird reason.