first commit

HumphreyYang · HumphreyYang · commit 70b199e9eff7 · 2023-03-30T13:32:05.000+11:00
diff --git a/lectures/_toc.yml b/lectures/_toc.yml
@@ -22,6 +22,7 @@ parts:
   - file: short_path
   - file: scalar_dynam
   - file: linear_equations
+  - file: eigen
   - file: lp_intro
   - file: lln_clt
   - file: monte_carlo
@@ -30,7 +31,7 @@ parts:
   numbered: true
   chapters:
   - file: simple_linear_regression
-  - file: eigen
+  - file: eigen_II
 - caption: Models
   numbered: true
   chapters:
diff --git a/lectures/eigen.md b/lectures/eigen.md
@@ -833,285 +833,4 @@ to one.
 
 The eigenvectors and eigenvalues of a map $A$ determine how a vector $v$ is transformed when we repeatedly multiply by $A$.
 
-This is discussed further below.
-
-
-
-## Nonnegative Matrices 
-
-Often, in economics, the matrix that we are dealing with is nonnegative.
-
-Nonnegative matrices have several special and useful properties.
-
-In this section we discuss some of them --- in particular, the connection
-between nonnegativity and eigenvalues.
-
-
-### Nonnegative Matrices
-
-An $n \times m$ matrix $A$ is called **nonnegative** if every element of $A$
-is nonnegative, i.e., $a_{ij} \geq 0$ for every $i,j$.
-
-We denote this as $A \geq 0$.
-
-
-### Irreducible Matrices
-
-Let $A$ be a square nonnegative matrix and let $A^k$ be the $k^{th}$ power of A.
-
-Let $a^{k}_{ij}$ be element $(i,j)$ of $A^k$.
-
-$A$ is called **irreducible** if for each $(i,j)$ there is an integer $k \geq 0$ such that $a^{k}_{ij} > 0$.
-
-A matrix $A$ that is not irreducible is called reducible.
-
-Here are some examples to illustrate this further.
-
-1. $A = \begin{bmatrix} 0.5 & 0.1 \\ 0.2 & 0.2 \end{bmatrix}$ is irreducible since $a_{ij}>0$ for all $(i,j)$.
-
-2. $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ is irreducible since $a_{12},a_{21} >0$ and $a^{2}_{11},a^{2}_{22} >0$.
-
-3. $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is reducible since $A^k = A$ for all $k \geq 0$ and thus
-   $a^{k}_{12},a^{k}_{21} = 0$ for all $k \geq 0$.
-
-
-### The Perron-Frobenius Theorem
-
-For a nonnegative matrix $A$ the behaviour of $A^k$ as $k \to \infty$ is controlled by the eigenvalue with the largest
-absolute value, often called the **dominant eigenvalue**.
-
-For a matrix $A$, the Perron-Frobenius theorem characterises certain
-properties of the dominant eigenvalue and its corresponding eigenvector when
-$A$ is a nonnegative square matrix.
-
-```{prf:theorem} Perron-Frobenius Theorem
-:label: perron-frobenius
-
-If a matrix $A \geq 0$ then,
-
-1. the dominant eigenvalue of $A$, $r(A)$, is real-valued and nonnegative. 
-2. for any other eigenvalue (possibly complex) $\lambda$ of $A$, $|\lambda| \leq r(A)$.
-3. we can find a nonnegative and nonzero eigenvector $v$ such that $Av = r(A)v$.
-
-Moreover if $A$ is also irreducible then,
-
-4. the eigenvector $v$ associated with the eigenvalue $r(A)$ is strictly positive.
-5. there exists no other positive eigenvector $v$ (except scalar multiples of v) associated with $r(A)$.
-
-```
-
-(This is a relatively simple version of the theorem --- for more details see
-[here](https://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem)).
-
-We will see applications of the theorem below.
-
-
-(la_neumann)=
-## The Neumann Series Lemma 
-
-```{index} single: Neumann's Lemma
-```
-
-In this section we present a famous result about series of matrices that has
-many applications in economics.
-
-### Scalar Series
-
-Here's a fundamental result about series that you surely know:
-
-If $a$ is a number and $|a| < 1$, then 
-
-```{math}
-:label: gp_sum
-    
-    \sum_{k=0}^{\infty} a^k =\frac{1}{1-a} = (1 - a)^{-1} 
-
-```
-
-For a one-dimensional linear equation $x = ax + b$ where x is unknown we can thus conclude that the solution $x^{*}$ is given by:
-
-$$
-    x^{*} = \frac{b}{1-a} = \sum_{k=0}^{\infty} a^k b
-$$
-
-### Matrix Series
-
-A generalization of this idea exists in the matrix setting.
-
-Consider the system of equations $x = Ax + b$ where $A$ is an $n \times n$
-square matrix and $x$ and $b$ are both column vectors in $\mathbb{R}^n$.
-
-Using matrix algebra we can conclude that the solution to this system of equations will be given by:
-
-```{math}
-:label: neumann_eqn
-    
-    x^{*} = (I-A)^{-1}b
-
-```
-
-What guarantees the existence of a unique vector $x^{*}$ that satisfies
-{eq}`neumann_eqn` ?
-
-The following is a fundamental result in functional analysis that generalises
-{eq}`gp_sum` to a multivariate case.
-
-
-```{prf:theorem} Neumann Series Lemma
-:label: neumann_series_lemma
-
-Let $A$ be a square matrix and let $A^k$ be the $k$-th power of $A$.
-
-Let $r(A)$ be the dominant eigenvector or as it is commonly called the *spectral radius*, defined as $\max_i |\lambda_i|$, where 
-
-* $\{\lambda_i\}_i$ is the set of eigenvalues of $A$ and
-* $|\lambda_i|$ is the modulus of the complex number $\lambda_i$
-
-Neumann's theorem states the following: If $r(A) < 1$, then $I - A$ is invertible, and
-
-$$
-(I - A)^{-1} = \sum_{k=0}^{\infty} A^k
-$$
-```
-
-We can see the Neumann series lemma in action in the following example.
-
-```{code-cell} ipython3
-A = np.array([[0.4, 0.1],
-              [0.7, 0.2]])
-
-evals, evecs = eig(A)   # finding eigenvalues and eigenvectors
-
-r = max(abs(λ) for λ in evals)    # compute spectral radius
-print(r)
-```
-
-The spectral radius $r(A)$ obtained is less than 1. 
-
-Thus, we can apply the Neumann Series lemma to find $(I-A)^{-1}$.
-
-```{code-cell} ipython3
-I = np.identity(2)      #2 x 2 identity matrix
-B = I-A
-```
-
-```{code-cell} ipython3
-B_inverse = np.linalg.inv(B)     #direct inverse method
-```
-
-```{code-cell} ipython3
-A_sum = np.zeros((2,2))        #power series sum of A
-A_power = I
-for i in range(50):
-    A_sum += A_power
-    A_power = A_power @ A
-```
-
-Let's check equality between the sum and the inverse methods.
-
-```{code-cell} ipython3
-np.allclose(A_sum, B_inverse)     
-```
-
-Although we truncate the infinite sum at $k = 50$, both methods give us the same
-result which illustrates the result of the Neumann Series lemma.
-
-## Exercises
-
-```{exercise-start} Leontief's Input-Output Model
-:label: eig_ex1
-```
-[Wassily Leontief](https://en.wikipedia.org/wiki/Wassily_Leontief) developed a model of an economy with $n$ sectors producing $n$ different commodities representing the interdependencies of different sectors of an economy.
-
-Under this model some of the output is consumed internally by the industries and the rest is consumed by external consumers.
-
-We define a simple model with 3 sectors - agriculture, industry, and service.
-
-The following table describes how output is distributed within the economy:
-
-|             | Total output | Agriculture | Industry | Service | Consumer |
-|:-----------:|:------------:|:-----------:|:--------:|:-------:|:--------:|
-| Agriculture |     $x_1$    |   0.3$x_1$  | 0.2$x_2$ |0.3$x_3$ |     4    |
-|   Industry  |     $x_2$    |   0.2$x_1$  | 0.4$x_2$ |0.3$x_3$ |     5    |
-|   Service   |     $x_3$    |   0.2$x_1$  | 0.5$x_2$ |0.1$x_3$ |    12    |
-
-The first row depicts how agriculture's total output $x_1$ is distributed 
-
-* $0.3x_1$ is used as inputs within agriculture itself,
-* $0.2x_2$ is used as inputs by the industry sector to produce $x_2$ units
-* $0.3x_3$ is used as inputs by the service sector to produce $x_3$ units and 
-* 4 units is the external demand by consumers.
-
-We can transform this into a system of linear equations for the 3 sectors as
-given below:
-
-$$
-    x_1 = 0.3x_1 + 0.2x_2 + 0.3x_3 + 4 \\
-    x_2 = 0.2x_1 + 0.4x_2 + 0.3x_3 + 5 \\
-    x_3 = 0.2x_1 + 0.5x_2 + 0.1x_3 + 12
-$$
-
-This can be transformed into the matrix equation $x = Ax + d$ where
-
-$$
-x =
-\begin{bmatrix}
-    x_1 \\
-    x_2 \\
-    x_3
-\end{bmatrix}
-, \; A =
-\begin{bmatrix}
-    0.3 & 0.2 & 0.3 \\
-    0.2 & 0.4 & 0.3 \\
-    0.2 & 0.5 & 0.1
-\end{bmatrix}
-\; \text{and} \;
-d =
-\begin{bmatrix}
-    4 \\
-    5 \\
-    12
-\end{bmatrix}
-$$
-
-The solution $x^{*}$ is given by the equation $x^{*} = (I-A)^{-1} d$
-
-1. Since $A$ is a nonnegative irreducible matrix, find the Perron-Frobenius eigenvalue of $A$.
-
-2. Use the Neumann Series lemma to find the solution $x^{*}$ if it exists.
-
-```{exercise-end}
-```
-
-```{solution-start} eig_ex1
-:class: dropdown
-```
-
-```{code-cell} ipython3
-A = np.array([[0.3, 0.2, 0.3],
-              [0.2, 0.4, 0.3],
-              [0.2, 0.5, 0.1]])
-
-evals, evecs = eig(A)
-
-r = max(abs(λ) for λ in evals)   #dominant eigenvalue/spectral radius
-print(r)
-```
-
-Since we have $r(A) < 1$ we can thus find the solution using the Neumann Series lemma.
-
-```{code-cell} ipython3
-I = np.identity(3)
-B = I - A
-
-d = np.array([4, 5, 12])
-d.shape = (3,1)
-
-B_inv = np.linalg.inv(B)
-x_star = B_inv @ d
-print(x_star)
-```
-
-```{solution-end}
-```
+This is discussed further later.
diff --git a/lectures/eigen_II.md b/lectures/eigen_II.md
