Exercise 5.C.16

Proof. (a)

We will prove by induction on $n$. When $n=1$, the result is clearly true.

Assume it is true for $n$. Then

(b)

We need to find $(x,y)$ and $\lambda \in ℝ$ that satisfy $T(x,y)=\lambda (x,y)$. Thus need to solve the following set of equations

Substituting $y$ in the second equation and dividing both sides by $\lambda$, we get $y=\frac{1+\lambda }{\lambda }x$. Therefore, an eigenvector of $T$ has the form $\left(x,\frac{1+\lambda }{\lambda }x\right)$ and we have

We need then, to solve for $\lambda$ in the following equations

You can check that both are equivalent and have the solutions $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$, which we will denote $\phi$ and $\widehat{\phi }$.

(c)

Substituting the eigenvalues found in (b) on the eigenvector form $\left(x,\frac{1+\lambda }{\lambda }x\right)$, we get

Both are linearly independent since they are eigenvectors corresponding to distinct eigenvalues. Thus, they are indeed a basis of ${ℝ}^2$.

(d)

Here we use the fact that if $\lambda$ is an eigenvalue of $T$ and $v$ a corresponding eigenvector then $T^{k}v={\lambda }^{k}v$. Since $(0,1)=\frac{1}{\sqrt{5}}(0,\phi -\widehat{\phi })$, we have

Therefore

(e)

This follows directly from the fact that $\frac{1}{\sqrt{5}}{\left|\frac{1-\sqrt{5}}{2}\right|}^n<\frac{1}{2}$, for all positive integers $n$, which we will prove by induction.

When $n=1$, we have

Assume the result holds true for $n$. Then

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