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Remove unused graphviz from eigen_2 #196

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68 changes: 30 additions & 38 deletions lectures/eigen_II.md
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Original file line number Diff line number Diff line change
Expand Up @@ -11,7 +11,7 @@ kernelspec:
name: python3
---

+++ {"user_expressions": []}


# Spectral Theory

Expand All @@ -27,19 +27,12 @@ In addition to what's in Anaconda, this lecture will need the following librarie
```{code-cell} ipython3
:tags: [hide-output]

!pip install graphviz quantecon
```

```{admonition} graphviz
:class: warning
If you are running this lecture locally it requires [graphviz](https://www.graphviz.org)
to be installed on your computer. Installation instructions for graphviz can be found
[here](https://www.graphviz.org/download/)
!pip install quantecon
```

In this lecture we will begin with the foundational concepts in spectral theory.

Then we will explore the Perron-Frobenius Theorem and the Neumann Series Lemma, and connect them to applications in Markov chains and networks.
Then we will explore the Perron-Frobenius Theorem and the Neumann Series Lemma, and connect them to applications in Markov chains and networks.

We will use the following imports:

Expand All @@ -48,7 +41,6 @@ import matplotlib.pyplot as plt
import numpy as np
from numpy.linalg import eig
import scipy as sp
import graphviz as gv
import quantecon as qe
```

Expand Down Expand Up @@ -119,7 +111,7 @@ In other words, if $w$ is a left eigenvector of matrix A, then $A^T w = \lambda
This hints at how to compute left eigenvectors

```{code-cell} ipython3
A = np.array([[3, 2],
A = np.array([[3, 2],
[1, 4]])

# Compute right eigenvectors and eigenvalues
Expand Down Expand Up @@ -174,7 +166,7 @@ $A$ is a nonnegative square matrix.

If a matrix $A \geq 0$ then,

1. the dominant eigenvalue of $A$, $r(A)$, is real-valued and nonnegative.
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 $Av = r(A)v$.

Expand Down Expand Up @@ -204,8 +196,8 @@ Now let's consider examples for each case.
Consider the following irreducible matrix A:

```{code-cell} ipython3
A = np.array([[0, 1, 0],
[.5, 0, .5],
A = np.array([[0, 1, 0],
[.5, 0, .5],
[0, 1, 0]])
```

Expand All @@ -228,8 +220,8 @@ Now we can go through our checklist to verify the claims of the Perron-Frobenius
Consider the following primitive matrix B:

```{code-cell} ipython3
B = np.array([[0, 1, 1],
[1, 0, 1],
B = np.array([[0, 1, 1],
[1, 0, 1],
[1, 1, 0]])

np.linalg.matrix_power(B, 2)
Expand All @@ -253,7 +245,7 @@ np.round(dominant_eigenvalue, 2)
eig(B)
```

+++ {"user_expressions": []}


Now let's verify the claims of the Perron-Frobenius Theorem for the primitive matrix B:

Expand Down Expand Up @@ -298,7 +290,7 @@ def check_convergence(M):
n_list = [1, 10, 100, 1000, 10000]

for n in n_list:

# Compute (A/r)^n
M_n = np.linalg.matrix_power(M/r, n)

Expand All @@ -313,8 +305,8 @@ def check_convergence(M):
A1 = np.array([[1, 2],
[1, 4]])

A2 = np.array([[0, 1, 1],
[1, 0, 1],
A2 = np.array([[0, 1, 1],
[1, 0, 1],
[1, 1, 0]])

A3 = np.array([[0.971, 0.029, 0.1, 1],
Expand All @@ -336,8 +328,8 @@ The convergence is not observed in cases of non-primitive matrices.
Let's go through an example

```{code-cell} ipython3
B = np.array([[0, 1, 1],
[1, 0, 0],
B = np.array([[0, 1, 1],
[1, 0, 0],
[1, 0, 0]])

# This shows that the matrix is not primitive
Expand All @@ -358,7 +350,7 @@ In fact we have already seen the theorem in action before in {ref}`the markov ch
(spec_markov)=
#### Example 3: Connection to Markov chains

We are now prepared to bridge the languages spoken in the two lectures.
We are now prepared to bridge the languages spoken in the two lectures.

A primitive matrix is both irreducible (or strongly connected in the language of graph) and aperiodic.

Expand Down Expand Up @@ -410,22 +402,22 @@ $$

This is proven in {cite}`sargent2023economic` and a nice discussion can be found [here](https://math.stackexchange.com/questions/2433997/can-all-matrices-be-decomposed-as-product-of-right-and-left-eigenvector).

In the formula $\lambda_i$ is an eigenvalue of $P$ and $v_i$ and $w_i$ are the right and left eigenvectors corresponding to $\lambda_i$.
In the formula $\lambda_i$ is an eigenvalue of $P$ and $v_i$ and $w_i$ are the right and left eigenvectors corresponding to $\lambda_i$.

Premultiplying $P^t$ by arbitrary $\psi \in \mathscr{D}(S)$ and rearranging now gives

$$
\psi P^t-\psi^*=\sum_{i=1}^{n-1} \lambda_i^t \psi v_i w_i^{\top}
$$

Recall that eigenvalues are ordered from smallest to largest from $i = 1 ... n$.
Recall that eigenvalues are ordered from smallest to largest from $i = 1 ... n$.

As we have seen, the largest eigenvalue for a primitive stochastic matrix is one.

This can be proven using [Gershgorin Circle Theorem](https://en.wikipedia.org/wiki/Gershgorin_circle_theorem),
This can be proven using [Gershgorin Circle Theorem](https://en.wikipedia.org/wiki/Gershgorin_circle_theorem),
but it is out of the scope of this lecture.

So by the statement (6) of Perron-Frobenius Theorem, $\lambda_i<1$ for all $i<n$, and $\lambda_n=1$ when $P$ is primitive (strongly connected and aperiodic).
So by the statement (6) of Perron-Frobenius Theorem, $\lambda_i<1$ for all $i<n$, and $\lambda_n=1$ when $P$ is primitive (strongly connected and aperiodic).


Hence, after taking the Euclidean norm deviation, we obtain
Expand All @@ -438,7 +430,7 @@ Thus, the rate of convergence is governed by the modulus of the second largest e


(la_neumann)=
## The Neumann Series Lemma
## The Neumann Series Lemma

```{index} single: Neumann's Lemma
```
Expand All @@ -450,12 +442,12 @@ many applications in economics.

Here's a fundamental result about series that you surely know:

If $a$ is a number and $|a| < 1$, then
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}

\sum_{k=0}^{\infty} a^k =\frac{1}{1-a} = (1 - a)^{-1}

```

Expand All @@ -476,7 +468,7 @@ Using matrix algebra we can conclude that the solution to this system of equatio

```{math}
:label: neumann_eqn

x^{*} = (I-A)^{-1}b

```
Expand All @@ -493,7 +485,7 @@ The following is a fundamental result in functional analysis that generalizes

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
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 $\lambda_i$
Expand All @@ -517,7 +509,7 @@ r = max(abs(λ) for λ in evals) # compute spectral radius
print(r)
```

The spectral radius $r(A)$ obtained is less than 1.
The spectral radius $r(A)$ obtained is less than 1.

Thus, we can apply the Neumann Series lemma to find $(I-A)^{-1}$.

Expand All @@ -541,7 +533,7 @@ for i in range(50):
Let's check equality between the sum and the inverse methods.

```{code-cell} ipython3
np.allclose(A_sum, B_inverse)
np.allclose(A_sum, B_inverse)
```

Although we truncate the infinite sum at $k = 50$, both methods give us the same
Expand All @@ -566,11 +558,11 @@ The following table describes how output is distributed within the economy:
| 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
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
* $0.3x_3$ 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
Expand Down