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import numpy as np
import matplotlib.pyplot as plt
import plotutils as pu
from sympy import *
%matplotlib inline
We add vectors to get $v + w$. We multiply them by numbers $c$ and $d$ to get $cv$ and $dw$. Combining those two operations (adding $cw$ to $dw$) gives the linear combination $cv + dw$.
$$cv + dw = c\begin{bmatrix}1\\1\end{bmatrix} + d\begin{bmatrix}2\\3\end{bmatrix} = \begin{bmatrix}c + 2d\\c + 3d\end{bmatrix}$$We can also write a column vector inline as $w = (2, 3)$ but that is not the same as the row vector $\begin{bmatrix}2 & 3\end{bmatrix}$.
In [15]:
fix, axes = plt.subplots(1, figsize=(3, 3))
pu.setup_axes(axes, xlim=(-1, 4), ylim=(-1, 4))
arrowprops = {'head_length': 0.25, 'head_width': 0.18}
plt.arrow(0, 0, 1, 1, **arrowprops, ec='r', fc='r')
plt.arrow(0, 0, 2, 3, **arrowprops, ec='b', fc='b')
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Sympy is a wonderful toolkit for doing symbolic programming. We can also use it to cheat and help it solve systems of equations for us. Let's put it to work on one of the problems of the linear algebra book.
Write down the three equations for $c$, $d$ and $e$ so that $cu + dv + ew = b$ and try to find $c$, $d$ and $e$.
$$ u = \begin{bmatrix}2\\-1\\0\end{bmatrix} v = \begin{bmatrix}-1\\2\\-1\end{bmatrix} w = \begin{bmatrix}0\\-1\\2\end{bmatrix} b = \begin{bmatrix}1\\0\\0\end{bmatrix} $$The equations are:
$$ \begin{cases} 2c - d &= 1\\ -c + 2d - e &= 0\\ -d + 2e &= 0\\ \end{cases} $$And we could try solve them ourselves which is not that hard to do via elimination but alternatively we can also use sympy.
In [22]:
s = [S('2*c - d - 1'), S('-c + 2*d - e'), S('-d + 2*e')]
solve(s)
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