Lyapunov exponent for 2D map? 
Identify the Lyapunov exponent for the cat map: $C(x,y) = (2x+y , x+y)$.

I am very confused as to finding the Lyapunov exponent for a two-dimensional map. I've come across a resource that states 
$$\lambda(T,(x, y)) = \liminf_{n\to\infty} \frac1n\log\|DT^n(x, y)\|$$ but I have no idea how to interpret this: $\|DT^n(x, y)\|$. I understand how the Lyapunov exponent is calculated in one dimension as it is just the long-term average of the log of the derivative of the trajectory. I am confused as to how to transform this idea into two dimensions, but I know it has to do with the Jacobian matrix. Also, the source I found that equation from also states that the Lyapunov exponent ends up being $\ln\left(\dfrac{3+\sqrt{5}}2\right)$ which is just the natural log of one of the eigenvalues of the Jacobian. Can someone please point me in the right direction?
 A: I would like to provide a more thorough answer.

Let $f:\mathbb{R}^d\to \mathbb{R}^d$ be a continuously differentiable function. For $p\in\mathbb{R}^d$ a point, we may define the (top) Lyapunov exponent of $f$ at $p$ by
$$\chi_p(f)=\lim_{n\to\infty} \dfrac{\log \Vert T_pf^n\Vert}{n}\in[-\infty,\infty].$$
This limit may fail to exist, even as an extended real number; to remedy this similarly we define the lower (top) Lyapunov exponent of $f$ at $p$ by
$$\underline{\chi}_p(f)=\liminf_{n\to\infty}\dfrac{\log \Vert T_pf^n\Vert}{n}\in[-\infty,\infty].$$
(One can also define the upper (top) Lyapunov exponent by using $\limsup$.)
Here we interpret $\Vert T_pf^n\Vert$ as follows: First off, let's fix a $p\in\mathbb{R}^d$ and a time value $n\in\mathbb{Z}_{\geq1}$. By composing $f$ $n$-times with itself we have another continuously differentiable function:
$$f^n=\underbrace{f\circ f\circ\cdots\circ f}_{n\text{ times}}: \mathbb{R}^d\to \mathbb{R}^d.$$
Instead of dealing with $f^n$, we'll use the best linear approximation $T_pf^n$ of $f^n$ in a vicinity of $p$; this is the derivative of $f$ at $p$. Formally $T_pf^n$ is a map from the tangent space $T_p\mathbb{R}^d$ of $\mathbb{R}^d$ at $p$ to the tangent space $T_{f^n(p)}\mathbb{R}^d$ of $\mathbb{R}^d$ at $f^n(p)$. However in this case $\mathbb{R}^d$ is already a linear space, so we can safely identify both $T_p\mathbb{R}^d$ and $T_{f^n(p)}\mathbb{R}^d$ with $\mathbb{R}^d$. Thus our linear approximation can be considered as a linear map
$$T_pf^n:\mathbb{R}^d\to \mathbb{R}^d.$$
(Of course if $f$ is a linear map to begin with, as Matthew remarked it's derivative will be itself. Also note that one can adapt the line of thought above to self-maps of nonlinear spaces; I've arbitrarily decided the level of generalization that might be useful for the OP.)
(Also note that the notation $T_pf^n$ is somewhat ambiguous, it really stands for $T_p(f^n)$. Since we've identified any tangent space with $\mathbb{R}^d$, so even though it is syntactic to consider $(T_pf)^n$ we don't necessarily have $T_p(f^n)=(T_pf)^n$ (though this formula is correct when $p$ is a fixed point, i.e. $f(p)=p$); the relation between these two expressions is governed by the chain rule; see Intuition of cocycles and their use in dynamical systems and Differential of an nth powered function.. )
Now we think of $\mathbb{R}^d$ as a normed vector space; e.g. we can use the standard Euclidean norm
$$|v|= |v|_{\ell^2}=\sqrt{\sum_{i=1}^d v_i^2},$$
where $v_1,v_2,...,v_d$ are the components of the vector $v$ w/r/t the standard basis of $\mathbb{R}^d$ (due to finite dimensionality what norm we use is not going to affect the values of Lyapunov exponents; indeed, any two norms are comparable and the comparison constants will decay in time; see e.g. Any two norms equivalent on a finite dimensional norm linear space., Any two norms on finite dimensional space are equivalent, Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces, Any two norms on a finite dimensional $\Bbb K$-linear space are equivalent.). Once we endow $\mathbb{R}^d$ with a norm, we can make sense of the expression $$\Vert T_pf^n \Vert$$ as an operator norm:
$$\Vert T_pf^n \Vert = \max_{\substack{v\in\mathbb{R}^d\\|v|= 1}} |T_pf^n(v)|.$$
Geometrically, one can think of $\Vert T_pf^n \Vert$ as the maximal stretching amount with which the linear approximation $T_pf^n$ stretches a vector.
All this points toward a heuristic regarding what the (top) Lyapunov exponent measures: we have a point $p$ and another point $q$ that is infinitesimally close to $p$. We compare the orbits $p\mapsto f(p)\mapsto f^2(p)\mapsto\cdots f^n(p)\mapsto\cdots$ and $q\mapsto f(q)\mapsto f^2(q)\mapsto\cdots f^n(q)\mapsto\cdots$ of $p$ and $q$, respectively. Maybe at each time step these two orbits stay at a bounded distance, or maybe the distance grows at most linearly, or it grows exponentially (or it decays). Of course, what type of behavior we'll see is quite possibly dependend on only on $p$ but its perturbation $q$; perhaps if one perturbs $p$ in a special direction a little bit one gets exponential separation or orbits. This is one way to interpret the expression "$\chi_p(f)>0$" . There are similar interpretations of negative or zero Lyapunov exponents. (See e.g. Lyapunov exponent of a stable p-cycle., Largest Lyapunov Exponent, Why does the negative exponent of Lyapunov imply that the orbit is an attractor?, Computing Lyapunov Exponents for more related discussion.)

Finally let's focus on the case at hand. Put $A:\mathbb{R}^2\to \mathbb{R}^2, (x,y)\mapsto (2x+y,x+y)$. Fix a point $p=(x,y)\in\mathbb{R}^2$. Since $A$ is linear, we may identify the derivative $T_p A$ of $f$ at $p$ with itself; whence also $T_pA^n=A^n$.
First, by the submultiplicativity of operator norms (see Show that the operator norm is submultiplicative) we have $\Vert A^n\Vert\leq \Vert A\Vert^n$. Taking logarithms and dividing by $n$, we have:
$$\dfrac{\log\Vert A^n\Vert}{n}\leq \log\Vert A\Vert.$$
Next, note that $A$ has two eigenvalues, $\frac{3-\sqrt{5}}{2}$ and $\frac{3+\sqrt{5}}{2}$, with eigenvectors $s,u\in\mathbb{R}^2$, $|s|=1=|u|$, respectively. We have that $\frac{3-\sqrt{5}}{2}<1<\frac{3+\sqrt{5}}{2}$ and by definition $\frac{3+\sqrt{5}}{2}=|Au|\leq \Vert A \Vert$. Similarly we have $\left(\frac{3+\sqrt{5}}{2}\right)^n=|A^n u |\leq \Vert A^n\Vert$. Again taking logarithms and dividing by $n$ we get:
$$\log\left(\frac{3+\sqrt{5}}{2}\right)\leq \dfrac{\log\Vert A^n\Vert}{n}\leq \log\Vert A\Vert.$$
Thus we have that $\log\left(\frac{3+\sqrt{5}}{2}\right)\leq \underline{\chi}_p(A)\leq \chi_p(A)\leq\log\Vert A\Vert$. By making the above heuristic about $\Vert A\Vert$ more rigorous ($\ast$), one can also see that $\Vert A\Vert=\frac{3+\sqrt{5}}{2}$; thus both the lower (top) Lyapunov exponent and the (top) Lyapunov exponent of $A$ at any point $p$ is $\log\left(\dfrac{3+\sqrt{5}}{2}\right)$.
($\ast$) Any unit vector $v\in\mathbb{R}^2$ decomposes as $v=\sigma s+\upsilon u$; $|Av|$ gets larger as $\upsilon$ gets larger. (...)

To compare my answer with Matthew C's answer, note that the linear map $A$ is symmetric, so that we have multiplicativity of operator norm over powers of $A$ (Norm of a symmetric matrix equals spectral radius, How to calculate explicitly some matrix norm?, Norm of a symmetric matrix?); thus in the case of $A$ the above argument simplifies quite a bit.
A: Here, $\|DT^n\|= \|T^n\| = \|T\|^n$, since the map $T$ is linear: it is its own derivative. Then by calculating the larger of the two eigenvalues, you have $\|T\|=\frac{3+\sqrt{5}}{2}$ and your result follows.
