Math Insight

1. The one-dimensional analogue of this dynamical system is $z_{n+1}=a z_n$. Recall that the behavior of this system is dictated by the parameter $a$: If $|a|<$
   , then $z_n$ approaches zero as $n$ increases, whereas if $|a|>$
   , $|z_n|$ approaches infinity as $n$ increases. We want to find a way to understand the two-dimensional system in a similar way, but we now have four parameters instead of one. What are the four parameters?
   

   The starting point for analyzing the behavior of the two-dimensional system is re-writing it in a format similar to the one-dimensional system. We want to write it as a single equation, where the state variables at time $n+1$ are equal to one "parameter" times the state variables at time $n$. To do this, we use vectors and matrices. We start by writing the state variables as a vector. We will use two formats for writing vectors. Either $$(x_n, y_n)$$ or $$\begin{bmatrix} x_n\\ y_n \end{bmatrix}$$ will denote the vector of state variables. Typically, the first format will be used when we are discussing the vector in isolation. The second format will be used when a matrix is involved.

   Our objective is now to write the two-dimensional dynamical system as $$\begin{bmatrix} x_{n+1}\\ y_{n+1} \end{bmatrix}=A\begin{bmatrix} x_n\\ y_n \end{bmatrix}$$ To determine the matrix $A$, we need to understand matrix-vector multiplication.

2. Recall that a matrix with $m$ rows and $n$ columns can be multiplied by a vector with $n$ components. For each row, we take each entry and multiply by the corresponding component in the vector, then add them up. Doing this for each row yields a vector with $m$ components. For example, consider $$\displaystyle \begin{bmatrix} 4 & 2 & -1 \\ 3 & 2 & 5 \end{bmatrix} \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$$ The resulting product is a vector with
   components. The first component is given by
   
   $\cdot$
   $+$
   $\cdot$
   $+$
   $\cdot$
   $=$
   
   and the second component is given by
   
   $\cdot$
   $+$
   $\cdot$
   $+$
   $\cdot$
   $=$
   
   Note: for these, make sure the first term in each product comes from the matrix, and the order of the products matches the order of the entries in the matrix/vector.
3. Now we can use this to convert the two-dimensional dynamical system into a matrix equation. When we multiply the matrix by the vector $(x_n, y_n)$, we want the first component to be the right-hand side of the first equation, $x_{n+1} = a x_n + b y_n$. Similarly, the second component should be the right-hand side of the second equation, $y_{n+1} = c x_n + d y_n$. Enter the appropriate parameters below to form the matrix equation.
   
   $\displaystyle\begin{bmatrix} x_{n+1}\\ y_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} x_{n}\\ y_{n} \end{bmatrix}$

1. \begin{eqnarray*} u_{n+1} & = & 3u_n+7v_n\\ v_{n+1} & = & 4u_n-3v_n \end{eqnarray*}
   
   $\displaystyle\begin{bmatrix} u_{n+1}\\ v_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} u_{n}\\ v_{n} \end{bmatrix}$
2. \begin{eqnarray*} x_{n+1} & = & 2x_n-3y_n+z_n\\ y_{n+1} & = & -x_n+2y_n+4z_n\\ z_{n+1} & = & -3x_n+y_n-2z_n \end{eqnarray*}
   
   $\displaystyle\begin{bmatrix} x_{n+1}\\ y_{n+1}\\z_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} x_{n}\\ y_{n} \\z_n \end{bmatrix}$
3. \begin{eqnarray*} x_{n+1} & = & x_n+2y_n-3z_n+4w_n\\ y_{n+1} & = & 2x_n-y_n+2z_n-w_n\\ z_{n+1} & = & 3x_n+2y_n-5z_n+w_n\\ w_{n+1} & = & -x_n+3y_n+2z_n+5w_n \end{eqnarray*}
   
   $\displaystyle\begin{bmatrix} x_{n+1}\\ y_{n+1}\\z_{n+1}\\w_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} x_{n}\\ y_{n} \\z_n\\w_n \end{bmatrix}$

Let's revisit the dynamical system from the first problem. above: \begin{align*} x_{n+1} &= 3.5 x_{n} - y_{n}, \qquad \text{for $n=0,1,2, \ldots$}\\ y_{n+1} &= x_{n} + y_{n} \end{align*} Rewrite this dynamical system in terms of a matrix-vector product:

Let's call this matrix $A$, so that we can write the dynamical system as \begin{gather*} \vc{x}_{n+1} = A\vc{x}_n, \qquad \text{for $n=0,1,2,\ldots$} \end{gather*} where \begin{align*} A = \begin{bmatrix}＿&＿\\＿&＿\end{bmatrix} \qquad \text{and} \qquad \vc{x}_n = \begin{bmatrix} x_{n}\\ y_{n} \end{bmatrix}. \end{align*}

The matrix equation $\vc{x}_{n+1} = A \vc{x}_n$ is shorthand for many equations, one for $n=0$, one for $n=1$, etc. The specific versions for $n=0$ and $n=1$ give the equations for $\vc{x}_1$ and $\vc{x}_2$:
$\vc{x}_1=$
, $\vc{x}_2 = $

As in the first problem, above, we'll use the initial condition $\vc{x}_0 = \left[\begin{matrix}1\\1\end{matrix}\right]$ to calculate $\vc{x}_1$ and $\vc{x}_2$. This time, since we've rewritten the dynamical system as a matrix-vector multiplication, we can easily ask the computer to do the work for us with the programing language R.

To enter the matrix $A$ in R, type the following command:

A=matrix(c(＿, ＿, ＿, ＿), 2, 2, byrow=TRUE)

Once you've entered that command in the console, you can just enter A in the console to view the matrix and make sure it turned out as you expected. R also outputs row and column headings like [,1] to show you the index corresponding to each row and column.

Create the initial condition vector by entering x0 = c(1,1) in the console. The next step is to tell R that we want $\vc{x}_1 = A \vc{x}_0$. Caution: we cannot use the command x1 = A*x0 in R. Instead, in R, A*x0 means to multiply the components of $A$ with the components of $\vc{x}_0$ (recycling the components of $\vc{x}_0$ since there are fewer components). The correct command for calculating $\vc{x}_1$ in R is

x1 = A %*% x0

The result is:
$\vc{x}_1 =$
, $\vc{x}_2 =$
, $\vc{x}_{10} =$