The glossary from the end of Strand: glossary.pdf
Solve for the variables \(x, y\), and \(z\) in the system
\[ \begin{aligned} z & =2 \\ x+y+z & =2 \\ 2 x+2 y+4 z & =8 \end{aligned} \]
. . .
Make an augmented matrix: Example 1.20. Solve for the variables \(x, y\), and \(z\) in the system
\[ \left[\begin{array}{llll} 0 & 0 & 1 & 2 \\ 1 & 1 & 1 & 2 \\ 2 & 2 & 4 & 8 \end{array}\right] \]
. . .
\[ \left[\begin{array}{llll} 0 & 0 & 1 & 2 \\ 1 & 1 & 1 & 2 \\ 2 & 2 & 4 & 8 \end{array}\right] \stackrel{E_{12}}{\longrightarrow}\left[\begin{array}{llll} (1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ 2 & 2 & 4 & 8 \end{array}\right] \xrightarrow[E_{31}(-2)]{\longrightarrow}\left[\begin{array}{rlll} 1 & 1 & 1 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right] \]
We keep on going… \[ \begin{aligned} & {\left[\begin{array}{rrrr} (1) & 1 & 1 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right] \xrightarrow[E_{2}(1 / 2)]{\longrightarrow}\left[\begin{array}{rrrr} 1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]} \\ & \overrightarrow{E_{32}(-1)}\left[\begin{array}{rrrr} (1) & 1 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow[E_{12}(-1)]{\longrightarrow}\left[\begin{array}{rrrr} (1) & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned} \]
There’s still no information on \(y\).
\[ \begin{aligned} & x=-y \\ & z=2 \\ & y \text { is free. } \end{aligned} \]
Suppose we have another augmented matrix, which after GJ elimination has become:
\[ \left[\begin{array}{rrrrr} x & y & z & w & \mathrm{rhs} \\ 1 & 2 & 0 & -1 & 2 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] . \]
. . .
Solutions are: \[ \begin{aligned} x+2 y-w & =2 \\ z+3 w & =0 . \end{aligned} \]
How can we tell from looking at the matrix which variables are free vs bound?
Columns which contain pivots correspond to bound variables. The others correspond to free variables.
(Columns with pivots are called “basic” columns.)
Solve this system… \[ \begin{array}{r} x+y=1 \\ 2 x+y=2 \\ 3 x+2 y=5 . \end{array} \]
. . .
\[ \left[\begin{array}{lll} 1 & 1 & 1 \\ 2 & 1 & 2 \\ 3 & 2 & 5 \end{array}\right] \xrightarrow[E_{21}(-2)]{E_{31}(-3)}\left[\begin{array}{lrl} 1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & -1 & 2 \end{array}\right] \xrightarrow[E_{32}(-1)]{\longrightarrow}\left[\begin{array}{rrr} 1 & 1 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{array}\right] \]
. . .
System is inconsistent.
For an augmented matrix in RREF, if the only only nonzero entry in a row appears on the right-hand side,
\[\text { Row } i \longrightarrow\left( \begin{array}{cccccc|c} * & * & * & * & * & * & * \\ 0 & 0 & 0 & * & * & * & * \\ 0 & 0 & 0 & 0 & * & * & * \\ 0 & 0 & 0 & 0 & 0 & 0 & \alpha \\ \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet \end{array} \right) \longleftarrow \alpha \neq 0 \]
the \(i^{\text {th }}\) equation of the associated system is
\[ 0 x_{1}+0 x_{2}+\cdots+0 x_{n}=\alpha . \]
If \(\alpha\ne 0\), this has no solution!
This is an inconsistent system.
Suppose a matrix \(A_{m \times n}\) has a reduced row echelon form \(E_{m \times n}\): \[ E_A=\left(\begin{array}{cccccccc}1 & * & 0 & 0 & * & * & 0 & * \\ 0 & 0 & 1 & 0 & * & * & 0 & * \\ 0 & 0 & 0 & 1 & * & * & 0 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right)\]
Suppose a matrix \(A_{m \times n}\) has a reduced row echelon form \(E_{m \times n}\): \[ E_A=\left(\begin{array}{cccccccc}1 & * & 0 & 0 & * & * & 0 & * \\ 0 & 0 & 1 & 0 & * & * & 0 & * \\ 0 & 0 & 0 & 1 & * & * & 0 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right)\]
“Basic columns” correspond to the columns in \(A\) which contain the pivotal positions.
Each non-basic column can be formed as a sum of the basic columns to the left.
Basic columns are the same in \(E\) and \(A\)!
Suppose a matrix \(A_{m \times n}\) has a reduced row echelon form \(E_{m \times n}\): \[ E_A=\left(\begin{array}{cccccccc}1 & * & 0 & 0 & * & * & 0 & * \\ 0 & 0 & 1 & 0 & * & * & 0 & * \\ 0 & 0 & 0 & 1 & * & * & 0 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right)\]
rank(\(A\)) = # of pivots = # of nonzero rows in \(E\) = # bound variables in the system
Suppose a matrix \(A_{m \times n}\) has a reduced row echelon form \(E_{m \times n}\): \[ E_A=\left(\begin{array}{cccccccc}1 & * & 0 & 0 & * & * & 0 & * \\ 0 & 0 & 1 & 0 & * & * & 0 & * \\ 0 & 0 & 0 & 1 & * & * & 0 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & * \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right)\]
nullity(\(A\)) = # of non-pivot columns = # free variables in the system
Equivalent ways of saying \(A\) is consistent:
. . . \[ \left(\begin{array}{llll|l} 0 & 0 & \cdots & 0 & \alpha \tag{2.3.2} \end{array}\right) \text {, where } \alpha \neq 0 \text {. } \]
\(\operatorname{rank}[\mathbf{A} \mid \mathbf{b}]=\operatorname{rank}(\mathbf{A})\), in which case either:
\(\mathbf{b}\) is a nonbasic column in \([\mathbf{A} \mid \mathbf{b}]\).
\(\mathbf{b}\) is a combination of the basic columns in \(\mathbf{A}\).
Meyers p. 51
The general linear system with \(m \times n\) coefficient matrix \(A\) and right-hand-side vector \(\mathbf{b}\) is homogeneous if the entries of \(\mathbf{b}\) are all zero.
Otherwise, the system is inhomogeneous.
Homogeneous systems are always consistent!
Can solve by setting all variables to zero. This is the trivial solution
If rank of the matrix = n, there is only the trivial solution
Goal: rank pages in terms of “importance”. Suppose we have four pages:
\[ \begin{equation*} x_{i}=\sum_{x_{j} \in L_{i}} \frac{x_{j}}{n_{j}} \end{equation*} \]
https://tinyurl.com/stat24320notebook1
Remember this was our ‘second try’: \[ \begin{equation*} x_{i}=\sum_{x_{j} \in L_{i}} x_{j} \end{equation*} \]
This was the system of equations. \[ \begin{aligned} & x_{1}=x_{2}+x_{3} \\ & x_{2}=x_{1}+x_{3} \\ & x_{3}=x_{1}+x_{4} \\ & x_{4}=x_{3} . \end{aligned} \]
What happens if we try to solve this? Try it in Python.
Rod with fixed temperatures \(y_{left}\) and \(y_{right}\) at each end, internal heat source given as \(f(x), 0 \leq x \leq 1\)
What’s the temperature at each point along the length?
. . .
Equations to balance the flow of heat from each node to its neighbors:
\[ \begin{equation*} -y_{i-1}+2 y_{i}-y_{i+1}=\frac{h^{2}}{K} f\left(x_{i}\right) \end{equation*} \]
Suppose:
. . .
What are the discretized steady-state equations?
. . .
\[ \begin{array}{lllllr} 2 y_{1}&-y_{2} & & & & =f(1 / 6) / 36 \\ -y_{1}&+2 y_{2}&-y_{3} & & & =f(2 / 6) / 36 \\ &-y_{2}&+2 y_{3}&-y_{4} & & =f(3 / 6) / 36 \\ &&-y_{3}&+2 y_{4}&-y_{5} & =f(4 / 6) / 36 \\ &&&-y_{4}&+2 y_{5} & =f(5 / 6) / 36 \end{array} \]
Build the matrix representing the system of equations:
N=6
M = sym.zeros(N)
for i in range(N):
if (i>0):
M[i-1,i]=-1
M[i,i]=2
if (i<N-1):
M[i+1,i]=-1
M\(\displaystyle \left[\begin{matrix}2 & -1 & 0 & 0 & 0 & 0\\-1 & 2 & -1 & 0 & 0 & 0\\0 & -1 & 2 & -1 & 0 & 0\\0 & 0 & -1 & 2 & -1 & 0\\0 & 0 & 0 & -1 & 2 & -1\\0 & 0 & 0 & 0 & -1 & 2\end{matrix}\right]\)
\[ \begin{equation*} -y_{i-1}+2 y_{i}-y_{i+1}=\frac{h^{2}}{K} f\left(x_{i}\right) \end{equation*} \]
Decide on a function for f. Let’s say \(\frac{h^{2}}{K} f = x_i^2\).
Say that x is running from 0 to 1. Then we can have
import numpy as np
x=np.arange(0,1,1/N)
f=x**2
x=sym.Matrix(x)
f=sym.Matrix(f)sol=M.gauss_jordan_solve(f)
sol[0]\(\displaystyle \left[\begin{matrix}0.416666666666667\\0.833333333333333\\1.22222222222222\\1.5\\1.52777777777778\\1.11111111111111\end{matrix}\right]\)
https://colab.research.google.com/github/ijmbarr/turing-patterns/blob/master/turing-patterns.ipynb#scrollTo=GsEPBsz33E14